A point P divides the line segment joining the points A(1, 3) and B(5, 9) in the ratio 1 : 2, the co-ordinates of the point P are :
( 7 3 , 5 ) \Big(\dfrac{7}{3}, 5\Big) ( 3 7 , 5 )
( 7 , 5 3 ) \Big(7, \dfrac{5}{3}\Big) ( 7 , 3 5 )
( 7 3 , 5 3 ) \Big(\dfrac{7}{3}, \dfrac{5}{3}\Big) ( 3 7 , 3 5 )
( 7 3 , 13 3 ) \Big(\dfrac{7}{3}, \dfrac{13}{3}\Big) ( 3 7 , 3 13 )
Answer
Let point P be (x, y).
Given,
m1 : m2 = 1 : 2
By section-formula,
(x, y) = ( m 1 x 2 + m 2 x 1 m 1 + m 2 , m 1 y 2 + m 2 y 1 m 1 + m 2 ) \Big(\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}, \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big) ( m 1 + m 2 m 1 x 2 + m 2 x 1 , m 1 + m 2 m 1 y 2 + m 2 y 1 )
Substituting values we get :
⇒ ( x , y ) = ( 1 × 5 + 2 × 1 1 + 2 , 1 × 9 + 2 × 3 1 + 2 ) = ( 5 + 2 3 , 9 + 6 3 ) = ( 7 3 , 15 3 ) = ( 7 3 , 5 ) . \Rightarrow (x, y) = \Big(\dfrac{1 \times 5 + 2 \times 1}{1 + 2}, \dfrac{1 \times 9 + 2 \times 3}{1 + 2}\Big) \\[1em] = \Big(\dfrac{5 + 2}{3}, \dfrac{9 + 6}{3}\Big) \\[1em] = \Big(\dfrac{7}{3}, \dfrac{15}{3}\Big) \\[1em] = \Big(\dfrac{7}{3}, 5\Big). ⇒ ( x , y ) = ( 1 + 2 1 × 5 + 2 × 1 , 1 + 2 1 × 9 + 2 × 3 ) = ( 3 5 + 2 , 3 9 + 6 ) = ( 3 7 , 3 15 ) = ( 3 7 , 5 ) .
Hence, Option 1 is the correct option.
AB is a line segment with A = (2, 4) and B = (6, 12). Point P lies on the line segment AB so that P = (3, x), then the ratio AP : PB is :
3 : 2
2 : 3
3 : 1
1 : 3
Answer
Let ratio in which P divides AB be k : 1.
By section-formula,
(x, y) = ( m 1 x 2 + m 2 x 1 m 1 + m 2 , m 1 y 2 + m 2 y 1 m 1 + m 2 ) \Big(\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}, \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big) ( m 1 + m 2 m 1 x 2 + m 2 x 1 , m 1 + m 2 m 1 y 2 + m 2 y 1 )
Substituting values we get :
⇒ ( 3 , x ) = ( k × 6 + 1 × 2 k + 1 , k × 12 + 1 × 4 k + 1 ) ⇒ ( 3 , x ) = ( 6 k + 2 k + 1 , 12 k + 4 k + 1 ) ⇒ 6 k + 2 k + 1 = 3 ⇒ 6 k + 2 = 3 ( k + 1 ) ⇒ 6 k + 2 = 3 k + 3 ⇒ 6 k − 3 k = 3 − 2 ⇒ 3 k = 1 ⇒ k = 1 3 . \Rightarrow (3, x) = \Big(\dfrac{k \times 6 + 1 \times 2}{k + 1}, \dfrac{k \times 12 + 1 \times 4}{k + 1}\Big) \\[1em] \Rightarrow (3, x) = \Big(\dfrac{6k + 2}{k + 1}, \dfrac{12k + 4}{k + 1}\Big) \\[1em] \Rightarrow \dfrac{6k + 2}{k + 1} = 3 \\[1em] \Rightarrow 6k + 2 = 3(k + 1) \\[1em] \Rightarrow 6k + 2 = 3k + 3 \\[1em] \Rightarrow 6k - 3k = 3 - 2 \\[1em] \Rightarrow 3k = 1 \\[1em] \Rightarrow k = \dfrac{1}{3}. ⇒ ( 3 , x ) = ( k + 1 k × 6 + 1 × 2 , k + 1 k × 12 + 1 × 4 ) ⇒ ( 3 , x ) = ( k + 1 6 k + 2 , k + 1 12 k + 4 ) ⇒ k + 1 6 k + 2 = 3 ⇒ 6 k + 2 = 3 ( k + 1 ) ⇒ 6 k + 2 = 3 k + 3 ⇒ 6 k − 3 k = 3 − 2 ⇒ 3 k = 1 ⇒ k = 3 1 .
Substituting value of k in k : 1, we get :
⇒ 1 3 : 1 \dfrac{1}{3} : 1 3 1 : 1
⇒ 1 : 3.
Hence, Option 4 is the correct option.
The ratio in which the join of (2, 4) and (10, 12) is divided by the line x = 7 is :
3 : 5
5 : 3
1 : 5
3 : 1
Answer
Any point on the line x = 7, can be defined as (7, y).
Let point (7, y) divide line joining (2, 4) and (10, 12) in ratio k : 1.
By section-formula,
(x, y) = ( m 1 x 2 + m 2 x 1 m 1 + m 2 , m 1 y 2 + m 2 y 1 m 1 + m 2 ) \Big(\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}, \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big) ( m 1 + m 2 m 1 x 2 + m 2 x 1 , m 1 + m 2 m 1 y 2 + m 2 y 1 )
Substituting values we get :
⇒ ( 7 , y ) = ( k × 10 + 1 × 2 k + 1 , k × 12 + 1 × 4 k + 1 ) ⇒ ( 7 , y ) = ( 10 k + 2 k + 1 , 12 k + 4 k + 1 ) ⇒ 7 = 10 k + 2 k + 1 ⇒ 7 ( k + 1 ) = 10 k + 2 ⇒ 7 k + 7 = 10 k + 2 ⇒ 10 k − 7 k = 7 − 2 ⇒ 3 k = 5 ⇒ k = 5 3 . \Rightarrow (7, y) = \Big(\dfrac{k \times 10 + 1 \times 2}{k + 1}, \dfrac{k \times 12 + 1 \times 4}{k + 1}\Big) \\[1em] \Rightarrow (7, y) = \Big(\dfrac{10k + 2}{k + 1}, \dfrac{12k + 4}{k + 1}\Big) \\[1em] \Rightarrow 7 = \dfrac{10k + 2}{k + 1} \\[1em] \Rightarrow 7(k + 1) = 10k + 2 \\[1em] \Rightarrow 7k + 7 = 10k + 2 \\[1em] \Rightarrow 10k - 7k = 7 - 2 \\[1em] \Rightarrow 3k = 5 \\[1em] \Rightarrow k = \dfrac{5}{3}. ⇒ ( 7 , y ) = ( k + 1 k × 10 + 1 × 2 , k + 1 k × 12 + 1 × 4 ) ⇒ ( 7 , y ) = ( k + 1 10 k + 2 , k + 1 12 k + 4 ) ⇒ 7 = k + 1 10 k + 2 ⇒ 7 ( k + 1 ) = 10 k + 2 ⇒ 7 k + 7 = 10 k + 2 ⇒ 10 k − 7 k = 7 − 2 ⇒ 3 k = 5 ⇒ k = 3 5 .
Substituting value of k in k : 1, we get :
⇒ 5 3 : 1 \dfrac{5}{3} : 1 3 5 : 1
⇒ 5 : 3.
Hence, Option 2 is the correct option.
The line y = 4 divides the join of points (6, 7) and (4, -1) in the ratio :
3 : 5
5 : 3
1 : 5
5 : 1
Answer
Any point on the line y = 4, can be defined as (x, 4).
Let point (x, 4) divide line joining (6, 7) and (4, -1) in ratio k : 1.
By section-formula,
(x, y) = ( m 1 x 2 + m 2 x 1 m 1 + m 2 , m 1 y 2 + m 2 y 1 m 1 + m 2 ) \Big(\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}, \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big) ( m 1 + m 2 m 1 x 2 + m 2 x 1 , m 1 + m 2 m 1 y 2 + m 2 y 1 )
Substituting values we get :
⇒ ( x , 4 ) = ( k × 4 + 1 × 6 k + 1 , k × − 1 + 1 × 7 k + 1 ) ⇒ ( x , 4 ) = ( 4 k + 6 k + 1 , − k + 7 k + 1 ) ⇒ 4 = − k + 7 k + 1 ⇒ 4 ( k + 1 ) = − k + 7 ⇒ 4 k + 4 = − k + 7 ⇒ 4 k + k = 7 − 4 ⇒ 5 k = 3 ⇒ k = 3 5 . \Rightarrow (x, 4) = \Big(\dfrac{k \times 4 + 1 \times 6}{k + 1}, \dfrac{k \times -1 + 1 \times 7}{k + 1}\Big) \\[1em] \Rightarrow (x, 4) = \Big(\dfrac{4k + 6}{k + 1}, \dfrac{-k + 7}{k + 1}\Big) \\[1em] \Rightarrow 4 = \dfrac{-k + 7}{k + 1} \\[1em] \Rightarrow 4(k + 1) = -k + 7 \\[1em] \Rightarrow 4k + 4 = -k + 7 \\[1em] \Rightarrow 4k + k = 7 - 4 \\[1em] \Rightarrow 5k = 3 \\[1em] \Rightarrow k = \dfrac{3}{5}. ⇒ ( x , 4 ) = ( k + 1 k × 4 + 1 × 6 , k + 1 k × − 1 + 1 × 7 ) ⇒ ( x , 4 ) = ( k + 1 4 k + 6 , k + 1 − k + 7 ) ⇒ 4 = k + 1 − k + 7 ⇒ 4 ( k + 1 ) = − k + 7 ⇒ 4 k + 4 = − k + 7 ⇒ 4 k + k = 7 − 4 ⇒ 5 k = 3 ⇒ k = 5 3 .
Substituting value of k in k : 1, we get :
⇒ 3 5 : 1 \dfrac{3}{5} : 1 5 3 : 1
⇒ 3 : 5.
Hence, Option 1 is the correct option.
The ratio in which the join of points (-2, 5) and (5, -2) is divided by y-axis is :
3 : 5
2 : 5
5 : 3
5 : 2
Answer
Any point on y-axis can be defined as (0, y).
Let ratio in which the join of points (-2, 5) and (5, -2) is divided by (0, y) be k : 1.
⇒ ( 0 , y ) = ( k × 5 + 1 × − 2 k + 1 , k × − 2 + 1 × 5 k + 1 ) ⇒ ( 0 , y ) = ( 5 k − 2 k + 1 , − 2 k + 5 k + 1 ) ⇒ 0 = 5 k − 2 k + 1 ⇒ 5 k − 2 = 0 ⇒ 5 k = 2 ⇒ k = 2 5 . \Rightarrow (0, y) = \Big(\dfrac{k \times 5 + 1 \times -2}{k + 1}, \dfrac{k \times -2 + 1 \times 5}{k + 1}\Big) \\[1em] \Rightarrow (0, y) = \Big(\dfrac{5k - 2}{k + 1}, \dfrac{-2k + 5}{k + 1}\Big) \\[1em] \Rightarrow 0 = \dfrac{5k - 2}{k + 1} \\[1em] \Rightarrow 5k - 2 = 0 \\[1em] \Rightarrow 5k = 2 \\[1em] \Rightarrow k = \dfrac{2}{5}. ⇒ ( 0 , y ) = ( k + 1 k × 5 + 1 × − 2 , k + 1 k × − 2 + 1 × 5 ) ⇒ ( 0 , y ) = ( k + 1 5 k − 2 , k + 1 − 2 k + 5 ) ⇒ 0 = k + 1 5 k − 2 ⇒ 5 k − 2 = 0 ⇒ 5 k = 2 ⇒ k = 5 2 .
Substituting value of k in k : 1, we get :
⇒ 2 5 : 1 \dfrac{2}{5} : 1 5 2 : 1
⇒ 2 : 5.
Hence, Option 2 is the correct option.
In what ratio does the point (1, a) divide the join of (-1, 4) and (4, -1) ?
Also, find the value of a.
Answer
We know that,
x = m 1 x 2 + m 2 x 1 m 1 + m 2 x = \dfrac{m_1x_2 + m_2x_1}{m_1 + m_2} x = m 1 + m 2 m 1 x 2 + m 2 x 1
Substituting values we get,
⇒ 1 = m 1 × 4 + m 2 × − 1 m 1 + m 2 ⇒ m 1 + m 2 = 4 m 1 − m 2 ⇒ m 2 + m 2 = 4 m 1 − m 1 ⇒ 2 m 2 = 3 m 1 ⇒ m 1 m 2 = 2 3 . \Rightarrow 1 = \dfrac{m_1 \times 4 + m_2 \times -1}{m_1 + m_2} \\[1em] \Rightarrow m_1 + m_2 = 4m_1 - m_2 \\[1em] \Rightarrow m_2 + m_2 = 4m_1 - m_1 \\[1em] \Rightarrow 2m_2 = 3m_1 \\[1em] \Rightarrow \dfrac{m_1}{m_2} = \dfrac{2}{3}. ⇒ 1 = m 1 + m 2 m 1 × 4 + m 2 × − 1 ⇒ m 1 + m 2 = 4 m 1 − m 2 ⇒ m 2 + m 2 = 4 m 1 − m 1 ⇒ 2 m 2 = 3 m 1 ⇒ m 2 m 1 = 3 2 .
∴ m1 : m2 = 2 : 3.
We know that,
y = m 1 y 2 + m 2 y 1 m 1 + m 2 y = \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2} y = m 1 + m 2 m 1 y 2 + m 2 y 1
Substituting values we get,
⇒ a = 2 × − 1 + 3 × 4 2 + 3 ⇒ a = − 2 + 12 5 ⇒ a = 10 5 ⇒ a = 2. \Rightarrow a = \dfrac{2 \times -1 + 3 \times 4}{2 + 3} \\[1em] \Rightarrow a = \dfrac{-2 + 12}{5} \\[1em] \Rightarrow a = \dfrac{10}{5} \\[1em] \Rightarrow a = 2. ⇒ a = 2 + 3 2 × − 1 + 3 × 4 ⇒ a = 5 − 2 + 12 ⇒ a = 5 10 ⇒ a = 2.
Hence, ratio = 2 : 3 and a = 2.
In what ratio does the point (a, 6) divide the join of (-4, 3) and (2, 8) ?
Also, find the value of a.
Answer
Let ratio in which point (a, 6) divide the join of (-4, 3) and (2, 8) be m1 : m2 .
By section formula,
y = m 1 y 2 + m 2 y 1 m 1 + m 2 y = \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2} y = m 1 + m 2 m 1 y 2 + m 2 y 1
Substituting values we get,
⇒ 6 = m 1 × 8 + m 2 × 3 m 1 + m 2 ⇒ 6 m 1 + 6 m 2 = 8 m 1 + 3 m 2 ⇒ 8 m 1 − 6 m 1 = 6 m 2 − 3 m 2 ⇒ 2 m 1 = 3 m 2 ⇒ m 1 m 2 = 3 2 . \Rightarrow 6 = \dfrac{m_1 \times 8 + m_2 \times 3}{m_1 + m_2} \\[1em] \Rightarrow 6m_1 + 6m_2 = 8m_1 + 3m_2 \\[1em] \Rightarrow 8m_1 - 6m_1 = 6m_2 - 3m_2 \\[1em] \Rightarrow 2m_1 = 3m_2 \\[1em] \Rightarrow \dfrac{m_1}{m_2} = \dfrac{3}{2}. ⇒ 6 = m 1 + m 2 m 1 × 8 + m 2 × 3 ⇒ 6 m 1 + 6 m 2 = 8 m 1 + 3 m 2 ⇒ 8 m 1 − 6 m 1 = 6 m 2 − 3 m 2 ⇒ 2 m 1 = 3 m 2 ⇒ m 2 m 1 = 2 3 .
∴ m1 : m2 = 3 : 2.
We know that,
x = m 1 x 2 + m 2 x 1 m 1 + m 2 x = \dfrac{m_1x_2 + m_2x_1}{m_1 + m_2} x = m 1 + m 2 m 1 x 2 + m 2 x 1
Substituting values we get,
⇒ a = 3 × 2 + 2 × − 4 3 + 2 ⇒ a = 6 − 8 5 ⇒ a = − 2 5 . \Rightarrow a = \dfrac{3 \times 2 + 2 \times -4}{3 + 2} \\[1em] \Rightarrow a = \dfrac{6 - 8}{5} \\[1em] \Rightarrow a = -\dfrac{2}{5}. ⇒ a = 3 + 2 3 × 2 + 2 × − 4 ⇒ a = 5 6 − 8 ⇒ a = − 5 2 .
Hence, ratio = 3 : 2 and a = − 2 5 -\dfrac{2}{5} − 5 2
In what ratio is the join of (4, 3) and (2, -6) divided by the x-axis. Also, find the co-ordinates of the point of intersection.
Answer
Let the point on x-axis be (x, 0) and required ratio be k : 1.
By formula,
y = m 1 y 2 + m 2 y 1 m 1 + m 2 y = \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2} y = m 1 + m 2 m 1 y 2 + m 2 y 1
Substituting values we get,
⇒ 0 = k × − 6 + 1 × 3 k + 1 ⇒ 0 = − 6 k + 3 ⇒ 6 k = 3 ⇒ k = 3 6 = 1 2 . \Rightarrow 0 = \dfrac{k \times -6 + 1 \times 3}{k + 1} \\[1em] \Rightarrow 0 = -6k + 3 \\[1em] \Rightarrow 6k = 3 \\[1em] \Rightarrow k = \dfrac{3}{6} = \dfrac{1}{2}. ⇒ 0 = k + 1 k × − 6 + 1 × 3 ⇒ 0 = − 6 k + 3 ⇒ 6 k = 3 ⇒ k = 6 3 = 2 1 .
k : 1 = 1 2 : 1 = 1 : 2. \dfrac{1}{2} : 1 = 1 : 2. 2 1 : 1 = 1 : 2.
x = m 1 x 2 + m 2 x 1 m 1 + m 2 x = \dfrac{m_1x_2 + m_2x_1}{m_1 + m_2} x = m 1 + m 2 m 1 x 2 + m 2 x 1
Substituting values we get,
⇒ x = 1 × 2 + 2 × 4 1 + 2 ⇒ x = 2 + 8 3 ⇒ x = 10 3 . \Rightarrow x = \dfrac{1 \times 2 + 2 \times 4}{1 + 2} \\[1em] \Rightarrow x = \dfrac{2 + 8}{3} \\[1em] \Rightarrow x = \dfrac{10}{3}. ⇒ x = 1 + 2 1 × 2 + 2 × 4 ⇒ x = 3 2 + 8 ⇒ x = 3 10 .
P = (x, 0) = ( 10 3 , 0 ) . \Big(\dfrac{10}{3}, 0\Big). ( 3 10 , 0 ) .
Hence, ratio = 1 : 2 and co-ordinates of point of intersection = ( 10 3 , 0 ) . \Big(\dfrac{10}{3}, 0\Big). ( 3 10 , 0 ) .
Find the ratio in which the join of (-4, 7) and (3, 0) is divided by the y-axis. Also, find the co-ordinates of the point of intersection.
Answer
Let the point on y-axis be (0, y) and required ratio be k : 1.
By formula,
x = m 1 x 2 + m 2 x 1 m 1 + m 2 x = \dfrac{m_1x_2 + m_2x_1}{m_1 + m_2} x = m 1 + m 2 m 1 x 2 + m 2 x 1
Substituting values we get,
⇒ 0 = k × 3 + 1 × − 4 k + 1 ⇒ 0 = 3 k − 4 ⇒ 3 k = 4 ⇒ k = 4 3 . \Rightarrow 0 = \dfrac{k \times 3 + 1 \times -4}{k + 1} \\[1em] \Rightarrow 0 = 3k - 4 \\[1em] \Rightarrow 3k = 4 \\[1em] \Rightarrow k = \dfrac{4}{3}. ⇒ 0 = k + 1 k × 3 + 1 × − 4 ⇒ 0 = 3 k − 4 ⇒ 3 k = 4 ⇒ k = 3 4 .
k : 1 = 4 3 : 1 = 4 : 3 \dfrac{4}{3} : 1 = 4 : 3 3 4 : 1 = 4 : 3
y = m 1 y 2 + m 2 y 1 m 1 + m 2 y = \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2} y = m 1 + m 2 m 1 y 2 + m 2 y 1
Substituting values we get,
⇒ y = 4 × 0 + 3 × 7 4 + 3 ⇒ y = 0 + 21 7 ⇒ y = 3. \Rightarrow y = \dfrac{4 \times 0 + 3 \times 7}{4 + 3} \\[1em] \Rightarrow y = \dfrac{0 + 21}{7} \\[1em] \Rightarrow y = 3. ⇒ y = 4 + 3 4 × 0 + 3 × 7 ⇒ y = 7 0 + 21 ⇒ y = 3.
P = (0, y) = (0, 3).
Hence, ratio = 4 : 3 and co-ordinates of point of intersection = (0, 3).
Points A, B, C and D divide the line segment joining the points (5, -10) and the origin in five equal parts. Find the co-ordinates of B and D.
Answer
Let point P = (5, -10) and origin (O) = (0, 0).
Points A, B, C and D divide the line segment PO in 5 equal parts.
From figure,
B divides the line segment PO in the ratio 2 : 3.
Let B be (a, b).
By formula,
x = m 1 x 2 + m 2 x 1 m 1 + m 2 x = \dfrac{m_1x_2 + m_2x_1}{m_1 + m_2} x = m 1 + m 2 m 1 x 2 + m 2 x 1
Substituting values we get,
⇒ a = 2 × 0 + 3 × 5 2 + 3 ⇒ a = 0 + 15 5 ⇒ a = 15 5 = 3. \Rightarrow a = \dfrac{2 \times 0 + 3 \times 5}{2 + 3} \\[1em] \Rightarrow a = \dfrac{0 + 15}{5} \\[1em] \Rightarrow a = \dfrac{15}{5} = 3. ⇒ a = 2 + 3 2 × 0 + 3 × 5 ⇒ a = 5 0 + 15 ⇒ a = 5 15 = 3.
y = m 1 y 2 + m 2 y 1 m 1 + m 2 y = \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2} y = m 1 + m 2 m 1 y 2 + m 2 y 1
Substituting values we get,
⇒ b = 2 × 0 + 3 × − 10 2 + 3 ⇒ b = 0 − 30 5 ⇒ b = − 30 5 = − 6. \Rightarrow b = \dfrac{2 \times 0 + 3 \times -10}{2 + 3} \\[1em] \Rightarrow b = \dfrac{0 - 30}{5} \\[1em] \Rightarrow b = -\dfrac{30}{5} = -6. ⇒ b = 2 + 3 2 × 0 + 3 × − 10 ⇒ b = 5 0 − 30 ⇒ b = − 5 30 = − 6.
B = (a, b) = (3, -6).
D divides the line segment PO in the ratio 4 : 1.
Let D be (c, d).
By formula,
x = m 1 x 2 + m 2 x 1 m 1 + m 2 x = \dfrac{m_1x_2 + m_2x_1}{m_1 + m_2} x = m 1 + m 2 m 1 x 2 + m 2 x 1
Substituting values we get,
⇒ c = 4 × 0 + 1 × 5 4 + 1 ⇒ c = 0 + 5 5 ⇒ c = 5 5 = 1. \Rightarrow c = \dfrac{4 \times 0 + 1 \times 5}{4 + 1} \\[1em] \Rightarrow c = \dfrac{0 + 5}{5} \\[1em] \Rightarrow c = \dfrac{5}{5} = 1. ⇒ c = 4 + 1 4 × 0 + 1 × 5 ⇒ c = 5 0 + 5 ⇒ c = 5 5 = 1.
y = m 1 y 2 + m 2 y 1 m 1 + m 2 y = \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2} y = m 1 + m 2 m 1 y 2 + m 2 y 1
Substituting values we get,
⇒ d = 4 × 0 + 1 × − 10 4 + 1 ⇒ d = − 10 5 ⇒ d = − 2. \Rightarrow d = \dfrac{4 \times 0 + 1 \times -10}{4 + 1} \\[1em] \Rightarrow d = -\dfrac{10}{5} \\[1em] \Rightarrow d = -2. ⇒ d = 4 + 1 4 × 0 + 1 × − 10 ⇒ d = − 5 10 ⇒ d = − 2.
D = (c, d) = (1, -2).
Hence, B = (3, -6) and D = (1, -2).
The line joining the points A (-3, -10) and B (-2, 6) is divided by the point P such that P B A B = 1 5 \dfrac{PB}{AB} = \dfrac{1}{5} A B PB = 5 1
Answer
Let co-ordinates of P be (a, b).
Given,
P B A B = 1 5 \dfrac{PB}{AB} = \dfrac{1}{5} A B PB = 5 1
Let PB = x and AB = 5x.
From figure,
⇒ AB = PA + PB
⇒ 5x = PA + x
⇒ PA = 4x.
P A P B = 4 x x = 4 1 \dfrac{PA}{PB} = \dfrac{4x}{x} = \dfrac{4}{1} PB P A = x 4 x = 1 4
PA : PB = 4 : 1.
∴ P divides the line segment joining A and B in ratio = 4 : 1.
By formula,
x = m 1 x 2 + m 2 x 1 m 1 + m 2 = 4 × − 2 + 1 × − 3 4 + 1 = − 8 − 3 5 = − 11 5 y = m 1 y 2 + m 2 y 1 m 1 + m 2 = 4 × 6 + 1 × − 10 4 + 1 = 24 − 10 5 = 14 5 . x = \dfrac{m_1x_2 + m_2x_1}{m_1 + m_2} \\[1em] = \dfrac{4 \times -2 + 1 \times -3}{4 + 1} \\[1em] = \dfrac{-8 - 3}{5} \\[1em] = -\dfrac{11}{5} \\[1em] y = \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2} \\[1em] = \dfrac{4 \times 6 + 1 \times -10}{4 + 1} \\[1em] = \dfrac{24 - 10}{5} \\[1em] = \dfrac{14}{5}. x = m 1 + m 2 m 1 x 2 + m 2 x 1 = 4 + 1 4 × − 2 + 1 × − 3 = 5 − 8 − 3 = − 5 11 y = m 1 + m 2 m 1 y 2 + m 2 y 1 = 4 + 1 4 × 6 + 1 × − 10 = 5 24 − 10 = 5 14 .
Hence, P = ( − 11 5 , 14 5 ) \Big(-\dfrac{11}{5}, \dfrac{14}{5}\Big) ( − 5 11 , 5 14 )
P is a point on the line joining A (4, 3) and B (-2, 6) such that 5AP = 2BP. Find the co-ordinates of P.
Answer
Given,
⇒ 5AP = 2BP
⇒ A P B P = 2 5 \dfrac{AP}{BP} = \dfrac{2}{5} BP A P = 5 2
⇒ AP : PB = 2 : 5.
Let co-ordinates of P be (x, y).
By formula,
x = m 1 x 2 + m 2 x 1 m 1 + m 2 = 2 × − 2 + 5 × 4 2 + 5 = − 4 + 20 7 = 16 7 y = m 1 y 2 + m 2 y 1 m 1 + m 2 = 2 × 6 + 5 × 3 2 + 5 = 12 + 15 7 = 27 7 . x = \dfrac{m_1x_2 + m_2x_1}{m_1 + m_2} \\[1em] = \dfrac{2 \times -2 + 5 \times 4}{2 + 5} \\[1em] = \dfrac{-4 + 20}{7} \\[1em] = \dfrac{16}{7} \\[1em] y = \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2} \\[1em] = \dfrac{2 \times 6 + 5 \times 3}{2 + 5} \\[1em] = \dfrac{12 + 15}{7} \\[1em] = \dfrac{27}{7}. x = m 1 + m 2 m 1 x 2 + m 2 x 1 = 2 + 5 2 × − 2 + 5 × 4 = 7 − 4 + 20 = 7 16 y = m 1 + m 2 m 1 y 2 + m 2 y 1 = 2 + 5 2 × 6 + 5 × 3 = 7 12 + 15 = 7 27 .
Hence, co-ordinates of P = ( 16 7 , 27 7 ) . \Big(\dfrac{16}{7}, \dfrac{27}{7}\Big). ( 7 16 , 7 27 ) .
Calculate the ratio in which the line joining the points (-3, -1) and (5, 7) is divided by the line x = 2. Also, find the co-ordinates of the point of intersection.
Answer
Let point of intersection be (2, y). [∵ any point on the line x = 2 has x co-ordinate = 2]
By formula,
x = m 1 x 2 + m 2 x 1 m 1 + m 2 ⇒ 2 = m 1 × 5 + m 2 × − 3 m 1 + m 2 ⇒ 2 m 1 + 2 m 2 = 5 m 1 − 3 m 2 ⇒ 5 m 1 − 2 m 1 = 2 m 2 + 3 m 2 ⇒ 3 m 1 = 5 m 2 ⇒ m 1 m 2 = 5 3 . x = \dfrac{m_1x_2 + m_2x_1}{m_1 + m_2} \\[1em] \Rightarrow 2 = \dfrac{m_1 \times 5 + m_2 \times -3}{m_1 + m_2} \\[1em] \Rightarrow 2m_1 + 2m_2 = 5m_1 - 3m_2 \\[1em] \Rightarrow 5m_1 - 2m_1 = 2m_2 + 3m_2 \\[1em] \Rightarrow 3m_1 = 5m_2 \\[1em] \Rightarrow \dfrac{m_1}{m_2} = \dfrac{5}{3}. x = m 1 + m 2 m 1 x 2 + m 2 x 1 ⇒ 2 = m 1 + m 2 m 1 × 5 + m 2 × − 3 ⇒ 2 m 1 + 2 m 2 = 5 m 1 − 3 m 2 ⇒ 5 m 1 − 2 m 1 = 2 m 2 + 3 m 2 ⇒ 3 m 1 = 5 m 2 ⇒ m 2 m 1 = 3 5 .
m1 : m2 = 5 : 3.
y = m 1 y 2 + m 2 y 1 m 1 + m 2 = 5 × 7 + 3 × − 1 5 + 3 = 35 − 3 8 = 32 8 = 4. y = \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2} \\[1em] = \dfrac{5 \times 7 + 3 \times -1}{5 + 3} \\[1em] = \dfrac{35 - 3}{8} \\[1em] = \dfrac{32}{8} \\[1em] = 4. y = m 1 + m 2 m 1 y 2 + m 2 y 1 = 5 + 3 5 × 7 + 3 × − 1 = 8 35 − 3 = 8 32 = 4.
Hence, co-ordinates of point of intersection = (2, 4) and ratio = 5 : 3.
Calculate the ratio in which the line joining A(6, 5) and B(4, -3) is divided by the line y = 2.
Answer
Let point of intersection be (x, 2) [∵ any point on the line y = 2 has y co-ordinate = 2]
By formula,
y = m 1 y 2 + m 2 y 1 m 1 + m 2 ⇒ 2 = m 1 × − 3 + m 2 × 5 m 1 + m 2 ⇒ 2 m 1 + 2 m 2 = − 3 m 1 + 5 m 2 ⇒ 2 m 1 + 3 m 1 = 5 m 2 − 2 m 2 ⇒ 5 m 1 = 3 m 2 ⇒ m 1 m 2 = 3 5 . y = \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2} \\[1em] \Rightarrow 2 = \dfrac{m_1 \times -3 + m_2 \times 5}{m_1 + m_2} \\[1em] \Rightarrow 2m_1 + 2m_2 = -3m_1 + 5m_2 \\[1em] \Rightarrow 2m_1 + 3m_1 = 5m_2 - 2m_2 \\[1em] \Rightarrow 5m_1 = 3m_2 \\[1em] \Rightarrow \dfrac{m_1}{m_2} = \dfrac{3}{5}. y = m 1 + m 2 m 1 y 2 + m 2 y 1 ⇒ 2 = m 1 + m 2 m 1 × − 3 + m 2 × 5 ⇒ 2 m 1 + 2 m 2 = − 3 m 1 + 5 m 2 ⇒ 2 m 1 + 3 m 1 = 5 m 2 − 2 m 2 ⇒ 5 m 1 = 3 m 2 ⇒ m 2 m 1 = 5 3 .
m1 : m2 = 3 : 5.
Hence, ratio = 3 : 5.
The point P(5, -4) divides the line segment AB, as shown in the figure, in the ratio 2 : 5. Find the co-ordinates of points A and B. Given AP is smaller than BP.
Answer
Since, A is on x-axis, let it co-ordinates be (a, 0) and B is on y-axis so it's co-ordinates (0, b).
By formula,
x = m 1 x 2 + m 2 x 1 m 1 + m 2 ⇒ 5 = 2 × 0 + 5 × a 2 + 5 ⇒ 5 × 7 = 0 + 5 a ⇒ 5 a = 35 ⇒ a = 35 5 = 7 y = m 1 y 2 + m 2 y 1 m 1 + m 2 ⇒ − 4 = 2 × b + 5 × 0 2 + 5 ⇒ − 4 × 7 = 2 b + 0 ⇒ 2 b = − 28 ⇒ b = − 14. x = \dfrac{m_1x_2 + m_2x_1}{m_1 + m_2} \\[1em] \Rightarrow 5 = \dfrac{2 \times 0 + 5 \times a}{2 + 5} \\[1em] \Rightarrow 5 \times 7 = 0 + 5a \\[1em] \Rightarrow 5a = 35 \\[1em] \Rightarrow a = \dfrac{35}{5} = 7 \\[1em] y = \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2} \\[1em] \Rightarrow -4 = \dfrac{2 \times b + 5 \times 0}{2 + 5} \\[1em] \Rightarrow -4 \times 7 = 2b + 0 \\[1em] \Rightarrow 2b = -28 \\[1em] \Rightarrow b = -14. x = m 1 + m 2 m 1 x 2 + m 2 x 1 ⇒ 5 = 2 + 5 2 × 0 + 5 × a ⇒ 5 × 7 = 0 + 5 a ⇒ 5 a = 35 ⇒ a = 5 35 = 7 y = m 1 + m 2 m 1 y 2 + m 2 y 1 ⇒ − 4 = 2 + 5 2 × b + 5 × 0 ⇒ − 4 × 7 = 2 b + 0 ⇒ 2 b = − 28 ⇒ b = − 14.
∴ A = (a, 0) = (7, 0) and
B = (0, b) = (0, -14).
Hence, A = (7, 0) and B = (0, -14).
Find the co-ordinates of the points of tri-section of the line joining the points (-3, 0) and (6, 6).
Answer
From figure,
Let A and B be the points of tri-section of the line joining the points (-3, 0) and (6, 6).
So, A and B divides the segment in three equal parts.
A divides the line segment in ratio 1 : 2. Let co-ordinates of A be (a, b).
By formula,
x = m 1 x 2 + m 2 x 1 m 1 + m 2 ⇒ a = 1 × 6 + 2 × − 3 1 + 2 ⇒ a = 6 − 6 3 ⇒ a = 0 y = m 1 y 2 + m 2 y 1 m 1 + m 2 ⇒ b = 1 × 6 + 2 × 0 1 + 2 ⇒ b = 6 3 = 2. x = \dfrac{m_1x_2 + m_2x_1}{m_1 + m_2} \\[1em] \Rightarrow a = \dfrac{1 \times 6 + 2 \times -3}{1 + 2} \\[1em] \Rightarrow a = \dfrac{6 - 6}{3} \\[1em] \Rightarrow a = 0 \\[1em] y = \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2} \\[1em] \Rightarrow b = \dfrac{1 \times 6 + 2 \times 0}{1 + 2} \\[1em] \Rightarrow b = \dfrac{6}{3} = 2. x = m 1 + m 2 m 1 x 2 + m 2 x 1 ⇒ a = 1 + 2 1 × 6 + 2 × − 3 ⇒ a = 3 6 − 6 ⇒ a = 0 y = m 1 + m 2 m 1 y 2 + m 2 y 1 ⇒ b = 1 + 2 1 × 6 + 2 × 0 ⇒ b = 3 6 = 2.
A = (a, b) = (0, 2).
B divides the line segment in ratio 2 : 1. Let co-ordinates of B be (c, d).
By formula,
x = m 1 x 2 + m 2 x 1 m 1 + m 2 ⇒ c = 2 × 6 + 1 × − 3 1 + 2 ⇒ c = 12 − 3 3 ⇒ c = 9 3 = 3. y = m 1 y 2 + m 2 y 1 m 1 + m 2 ⇒ d = 2 × 6 + 1 × 0 2 + 1 ⇒ d = 12 3 = 4. x = \dfrac{m_1x_2 + m_2x_1}{m_1 + m_2} \\[1em] \Rightarrow c = \dfrac{2 \times 6 + 1 \times -3}{1 + 2} \\[1em] \Rightarrow c = \dfrac{12 - 3}{3} \\[1em] \Rightarrow c = \dfrac{9}{3} = 3. \\[1em] y = \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2} \\[1em] \Rightarrow d = \dfrac{2 \times 6 + 1 \times 0}{2 + 1} \\[1em] \Rightarrow d = \dfrac{12}{3} = 4. x = m 1 + m 2 m 1 x 2 + m 2 x 1 ⇒ c = 1 + 2 2 × 6 + 1 × − 3 ⇒ c = 3 12 − 3 ⇒ c = 3 9 = 3. y = m 1 + m 2 m 1 y 2 + m 2 y 1 ⇒ d = 2 + 1 2 × 6 + 1 × 0 ⇒ d = 3 12 = 4.
B = (c, d) = (3, 4).
Hence, points of tri-section are (0, 2) and (3, 4).
Show that the line segment joining the points (-5, 8) and (10, -4) is trisected by the co-ordinate axes.
Answer
Let A = (-5, 8) and B = (10, -4)
Let P and Q be points which trisects AB.
Let P (a, b) divide AB in 1 : 2 and Q (c, d) in 2 : 1.
By section formula (for P),
x = m 1 x 2 + m 2 x 1 m 1 + m 2 ⇒ a = 1 × 10 + 2 × − 5 1 + 2 ⇒ a = 10 − 10 3 ⇒ a = 0 3 = 0. y = m 1 y 2 + m 2 y 1 m 1 + m 2 ⇒ b = 1 × − 4 + 2 × 8 1 + 2 ⇒ b = − 4 + 16 3 ⇒ b = 12 3 = 4. x = \dfrac{m_1x_2 + m_2x_1}{m_1 + m_2} \\[1em] \Rightarrow a = \dfrac{1 \times 10 + 2 \times -5}{1 + 2} \\[1em] \Rightarrow a = \dfrac{10 - 10}{3} \\[1em] \Rightarrow a = \dfrac{0}{3} = 0. \\[1em] y = \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2} \\[1em] \Rightarrow b = \dfrac{1 \times -4 + 2 \times 8}{1 + 2} \\[1em] \Rightarrow b = \dfrac{-4 + 16}{3} \\[1em] \Rightarrow b = \dfrac{12}{3} = 4. x = m 1 + m 2 m 1 x 2 + m 2 x 1 ⇒ a = 1 + 2 1 × 10 + 2 × − 5 ⇒ a = 3 10 − 10 ⇒ a = 3 0 = 0. y = m 1 + m 2 m 1 y 2 + m 2 y 1 ⇒ b = 1 + 2 1 × − 4 + 2 × 8 ⇒ b = 3 − 4 + 16 ⇒ b = 3 12 = 4.
P = (a, b) = (0, 4).
By section formula (for Q),
x = m 1 x 2 + m 2 x 1 m 1 + m 2 ⇒ c = 2 × 10 + 1 × − 5 2 + 1 ⇒ c = 20 − 5 3 ⇒ c = 15 3 = 5. y = m 1 y 2 + m 2 y 1 m 1 + m 2 ⇒ d = 2 × − 4 + 1 × 8 2 + 1 ⇒ d = − 8 + 8 3 ⇒ d = 0 3 = 0. x = \dfrac{m_1x_2 + m_2x_1}{m_1 + m_2} \\[1em] \Rightarrow c = \dfrac{2 \times 10 + 1 \times -5}{2 + 1} \\[1em] \Rightarrow c = \dfrac{20 - 5}{3} \\[1em] \Rightarrow c = \dfrac{15}{3} = 5. \\[1em] y = \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2} \\[1em] \Rightarrow d = \dfrac{2 \times -4 + 1 \times 8}{2 + 1} \\[1em] \Rightarrow d = \dfrac{-8 + 8}{3} \\[1em] \Rightarrow d = \dfrac{0}{3} = 0. x = m 1 + m 2 m 1 x 2 + m 2 x 1 ⇒ c = 2 + 1 2 × 10 + 1 × − 5 ⇒ c = 3 20 − 5 ⇒ c = 3 15 = 5. y = m 1 + m 2 m 1 y 2 + m 2 y 1 ⇒ d = 2 + 1 2 × − 4 + 1 × 8 ⇒ d = 3 − 8 + 8 ⇒ d = 3 0 = 0.
Q = (c, d) = (5, 0).
Since, x co-ordinate of P = 0, it means P lies on y-axis and y co-ordinate of Q = 0, it means Q lies on x-axis.
Hence, proved that line segment joining the points (-5, 8) and (10, -4) is trisected by the co-ordinate axes.
Show that A(3, -2) is a point of trisection of the line-segment joining the points (2, 1) and (5, -8). Also, find the co-ordinates of the other points of trisection.
Answer
Let A divide line-segment joining the points (2, 1) and (5, -8) in m1 : m2 .
By formula,
x = m 1 x 2 + m 2 x 1 m 1 + m 2 ⇒ 3 = m 1 × 5 + m 2 × 2 m 1 + m 2 ⇒ 3 m 1 + 3 m 2 = 5 m 1 + 2 m 2 ⇒ 5 m 1 − 3 m 1 = 3 m 2 − 2 m 2 ⇒ 2 m 1 = m 2 ⇒ m 1 m 2 = 1 2 ⇒ m 1 : m 2 = 1 : 2. x = \dfrac{m_1x_2 + m_2x_1}{m_1 + m_2} \\[1em] \Rightarrow 3 = \dfrac{m_1 \times 5 + m_2 \times 2}{m_1 + m_2} \\[1em] \Rightarrow 3m_1 + 3m_2 = 5m_1 + 2m_2 \\[1em] \Rightarrow 5m_1 - 3m_1 = 3m_2 - 2m_2 \\[1em] \Rightarrow 2m_1 = m_2 \\[1em] \Rightarrow \dfrac{m_1}{m_2} = \dfrac{1}{2} \\[1em] \Rightarrow m_1 : m_2 = 1 : 2. x = m 1 + m 2 m 1 x 2 + m 2 x 1 ⇒ 3 = m 1 + m 2 m 1 × 5 + m 2 × 2 ⇒ 3 m 1 + 3 m 2 = 5 m 1 + 2 m 2 ⇒ 5 m 1 − 3 m 1 = 3 m 2 − 2 m 2 ⇒ 2 m 1 = m 2 ⇒ m 2 m 1 = 2 1 ⇒ m 1 : m 2 = 1 : 2.
Since, A divides line-segment joining the points (2, 1) and (5, -8) in 1 : 2.
Hence, proved A is a point of tri-section.
Let another point of tri-section be B(a, b). So, it will divide the line segment in 2 : 1.
x = m 1 x 2 + m 2 x 1 m 1 + m 2 x = \dfrac{m_1x_2 + m_2x_1}{m_1 + m_2} x = m 1 + m 2 m 1 x 2 + m 2 x 1
Substituting values we get,
⇒ a = 2 × 5 + 1 × 2 2 + 1 ⇒ a = 10 + 2 3 ⇒ a = 4. \Rightarrow a = \dfrac{2 \times 5 + 1 \times 2}{2 + 1} \\[1em] \Rightarrow a = \dfrac{10 + 2}{3} \\[1em] \Rightarrow a = 4. ⇒ a = 2 + 1 2 × 5 + 1 × 2 ⇒ a = 3 10 + 2 ⇒ a = 4.
y = m 1 y 2 + m 2 y 1 m 1 + m 2 y = \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2} y = m 1 + m 2 m 1 y 2 + m 2 y 1
Substituting values we get,
⇒ b = 2 × − 8 + 1 × 1 2 + 1 ⇒ b = − 16 + 1 3 ⇒ b = − 15 3 = − 5. \Rightarrow b = \dfrac{2 \times -8 + 1 \times 1}{2 + 1} \\[1em] \Rightarrow b = \dfrac{-16 + 1}{3} \\[1em] \Rightarrow b = -\dfrac{15}{3} = -5. ⇒ b = 2 + 1 2 × − 8 + 1 × 1 ⇒ b = 3 − 16 + 1 ⇒ b = − 3 15 = − 5.
B = (a, b) = (4, -5).
Hence, co-ordinate of other point of trisection = (4, -5).
The line segment joining the points M(5, 7) and N(-3, 2) is intersected by the y-axis at point L. Write down the abscissa of L. Hence, find the ratio in which L divides MN. Also, find the co-ordinates of L.
Answer
Since, L lies on y-axis let its co-ordinates be (0, y).
Let L divide MN in ratio m1 : m2 .
By formula,
⇒ x = m 1 x 2 + m 2 x 1 m 1 + m 2 ⇒ 0 = m 1 × − 3 + m 2 × 5 m 1 + m 2 ⇒ 0 = − 3 m 1 + 5 m 2 ⇒ 3 m 1 = 5 m 2 ⇒ m 1 m 2 = 5 3 . \Rightarrow x = \dfrac{m_1x_2 + m_2x_1}{m_1 + m_2} \\[1em] \Rightarrow 0 = \dfrac{m_1 \times -3 + m_2 \times 5}{m_1 + m_2} \\[1em] \Rightarrow 0 = -3m_1 + 5m_2 \\[1em] \Rightarrow 3m_1 = 5m_2 \\[1em] \Rightarrow \dfrac{m_1}{m_2} = \dfrac{5}{3}. ⇒ x = m 1 + m 2 m 1 x 2 + m 2 x 1 ⇒ 0 = m 1 + m 2 m 1 × − 3 + m 2 × 5 ⇒ 0 = − 3 m 1 + 5 m 2 ⇒ 3 m 1 = 5 m 2 ⇒ m 2 m 1 = 3 5 .
m1 : m2 = 5 : 3.
By formula,
⇒ y = m 1 y 2 + m 2 y 1 m 1 + m 2 ⇒ y = 5 × 2 + 3 × 7 5 + 3 ⇒ y = 10 + 21 8 ⇒ y = 31 8 . \Rightarrow y = \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2} \\[1em] \Rightarrow y = \dfrac{5 \times 2 + 3 \times 7}{5 + 3} \\[1em] \Rightarrow y = \dfrac{10 + 21}{8} \\[1em] \Rightarrow y = \dfrac{31}{8}. ⇒ y = m 1 + m 2 m 1 y 2 + m 2 y 1 ⇒ y = 5 + 3 5 × 2 + 3 × 7 ⇒ y = 8 10 + 21 ⇒ y = 8 31 .
Hence, abscissa of L = 0, m1 : m2 = 5 : 3 and co-ordinates of L = ( 0 , 31 8 ) \Big(0, \dfrac{31}{8}\Big) ( 0 , 8 31 )
A(-3, 4), B(3, -1) and C(-2, 4) are the vertices of a triangle ABC. Find the length of line segment AP, where point P lies inside BC, such that BP : PC = 2 : 3.
Answer
Given,
BP : PC = 2 : 3.
So, P divides the line segment BC in ratio 2 : 3.
Let co-ordinates of P be (x, y).
∴ x = m 1 x 2 + m 2 x 1 m 1 + m 2 = 2 × − 2 + 3 × 3 2 + 3 = − 4 + 9 5 = 5 5 = 1. \therefore x = \dfrac{m_1x_2 + m_2x_1}{m_1 + m_2} \\[1em] = \dfrac{2 \times -2 + 3 \times 3}{2 + 3} \\[1em] = \dfrac{-4 + 9}{5} \\[1em] = \dfrac{5}{5} = 1. ∴ x = m 1 + m 2 m 1 x 2 + m 2 x 1 = 2 + 3 2 × − 2 + 3 × 3 = 5 − 4 + 9 = 5 5 = 1.
and
y = m 1 y 2 + m 2 y 1 m 1 + m 2 = 2 × 4 + 3 × − 1 2 + 3 = 8 − 3 5 = 5 5 = 1. y = \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2} \\[1em] = \dfrac{2 \times 4 + 3 \times -1}{2 + 3} \\[1em] = \dfrac{8 - 3}{5} \\[1em] = \dfrac{5}{5} = 1. y = m 1 + m 2 m 1 y 2 + m 2 y 1 = 2 + 3 2 × 4 + 3 × − 1 = 5 8 − 3 = 5 5 = 1.
P = (x, y) = (1, 1).
Distance between two points = ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2
A P = [ 1 − ( − 3 ) ] 2 + ( 1 − 4 ) 2 = ( 4 ) 2 + ( − 3 ) 2 = 16 + 9 = 25 = 5 units . AP = \sqrt{[1 - (-3)]^2 + (1 - 4)^2} \\[1em] = \sqrt{(4)^2 + (-3)^2} \\[1em] = \sqrt{16 + 9} \\[1em] = \sqrt{25} \\[1em] = 5 \text{ units}. A P = [ 1 − ( − 3 ) ] 2 + ( 1 − 4 ) 2 = ( 4 ) 2 + ( − 3 ) 2 = 16 + 9 = 25 = 5 units .
Hence, AP = 5 units.
The line segment joining A(2, 3) and B(6, -5) is intercepted by x-axis at the point K. Write down the ordinate of the point K. Hence, find the ratio in which K divides AB. Also, find the co-ordinates of the point K.
Answer
Since, point K lies on x-axis. Its co-ordinates be (x, 0).
Let ratio in which K divides AB be m1 : m2 .
By section-formula,
y = m 1 y 2 + m 2 y 1 m 1 + m 2 ⇒ 0 = m 1 × − 5 + m 2 × 3 m 1 + m 2 ⇒ 0 = − 5 m 1 + 3 m 2 ⇒ 5 m 1 = 3 m 2 ⇒ m 1 m 2 = 3 5 . y = \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2} \\[1em] \Rightarrow 0 = \dfrac{m_1 \times -5 + m_2 \times 3}{m_1 + m_2} \\[1em] \Rightarrow 0 = -5m_1 + 3m_2 \\[1em] \Rightarrow 5m_1 = 3m_2 \\[1em] \Rightarrow \dfrac{m_1}{m_2} = \dfrac{3}{5}. y = m 1 + m 2 m 1 y 2 + m 2 y 1 ⇒ 0 = m 1 + m 2 m 1 × − 5 + m 2 × 3 ⇒ 0 = − 5 m 1 + 3 m 2 ⇒ 5 m 1 = 3 m 2 ⇒ m 2 m 1 = 5 3 .
m1 : m2 = 3 : 5.
Substituting value for x co-ordinate,
⇒ x = m 1 x 2 + m 2 x 1 m 1 + m 2 = 3 × 6 + 5 × 2 3 + 5 = 18 + 10 8 = 28 8 = 7 2 = 3 1 2 . \Rightarrow x = \dfrac{m_1x_2 + m_2x_1}{m_1 + m_2} \\[1em] = \dfrac{3 \times 6 + 5 \times 2}{3 + 5} \\[1em] = \dfrac{18 + 10}{8} \\[1em] = \dfrac{28}{8} \\[1em] = \dfrac{7}{2} = 3\dfrac{1}{2}. ⇒ x = m 1 + m 2 m 1 x 2 + m 2 x 1 = 3 + 5 3 × 6 + 5 × 2 = 8 18 + 10 = 8 28 = 2 7 = 3 2 1 .
Hence, ordinate of K = 0, ratio in which K divides AB = 3 : 5 and K = ( 3 1 2 , 0 ) . \Big(3\dfrac{1}{2}, 0\Big). ( 3 2 1 , 0 ) .
The line segment joining A(4, 7) and B(-6, -2) is intercepted by the y-axis at the point K. Write down the abscissa of the point K. Hence, find the ratio in which K divides AB. Also, find the co-ordinates of the point K.
Answer
Since, point K lies on y-axis. Its co-ordinates be (0, y).
Let ratio in which K divides AB be m1 : m2 .
By section-formula,
x = m 1 x 2 + m 2 x 1 m 1 + m 2 ⇒ 0 = m 1 × − 6 + m 2 × 4 m 1 + m 2 ⇒ 0 = − 6 m 1 + 4 m 2 ⇒ 6 m 1 = 4 m 2 ⇒ m 1 m 2 = 4 6 = 2 3 . x = \dfrac{m_1x_2 + m_2x_1}{m_1 + m_2} \\[1em] \Rightarrow 0 = \dfrac{m_1 \times -6 + m_2 \times 4}{m_1 + m_2} \\[1em] \Rightarrow 0 = -6m_1 + 4m_2 \\[1em] \Rightarrow 6m_1 = 4m_2 \\[1em] \Rightarrow \dfrac{m_1}{m_2} = \dfrac{4}{6} = \dfrac{2}{3}. x = m 1 + m 2 m 1 x 2 + m 2 x 1 ⇒ 0 = m 1 + m 2 m 1 × − 6 + m 2 × 4 ⇒ 0 = − 6 m 1 + 4 m 2 ⇒ 6 m 1 = 4 m 2 ⇒ m 2 m 1 = 6 4 = 3 2 .
m1 : m2 = 2 : 3.
Substituting value for y co-ordinate,
⇒ y = m 1 y 2 + m 2 y 1 m 1 + m 2 = 2 × − 2 + 3 × 7 2 + 3 = − 4 + 21 5 = 17 5 . \Rightarrow y = \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2} \\[1em] = \dfrac{2 \times -2 + 3 \times 7}{2 + 3} \\[1em] = \dfrac{-4 + 21}{5} \\[1em] = \dfrac{17}{5}. ⇒ y = m 1 + m 2 m 1 y 2 + m 2 y 1 = 2 + 3 2 × − 2 + 3 × 7 = 5 − 4 + 21 = 5 17 .
Hence, abscissa of K = 0, ratio in which K divides AB = 2 : 3 and K = ( 0 , 17 5 ) . \Big(0, \dfrac{17}{5}\Big). ( 0 , 5 17 ) .
The line joining P(-4, 5) and Q(3, 2) intersects the y-axis at point R. PM and QN are perpendiculars from P and Q on the x-axis. Find :
(i) the ratio PR : RQ.
(ii) the co-ordinates of R.
(iii) the area of the quadrilateral PMNQ.
Answer
(i) Since, R lies on y-axis. Let its co-ordinates be (0, y).
Let R divide PQ in ratio m1 : m2 .
By section formula,
x = m 1 x 2 + m 2 x 1 m 1 + m 2 ⇒ 0 = m 1 × 3 + m 2 × − 4 m 1 + m 2 ⇒ 3 m 1 − 4 m 2 = 0 ⇒ 3 m 1 = 4 m 2 ⇒ m 1 m 2 = 4 3 . x = \dfrac{m_1x_2 + m_2x_1}{m_1 + m_2} \\[1em] \Rightarrow 0 = \dfrac{m_1 \times 3 + m_2 \times -4}{m_1 + m_2} \\[1em] \Rightarrow 3m_1 - 4m_2 = 0 \\[1em] \Rightarrow 3m_1 = 4m_2 \\[1em] \Rightarrow \dfrac{m_1}{m_2} = \dfrac{4}{3}. x = m 1 + m 2 m 1 x 2 + m 2 x 1 ⇒ 0 = m 1 + m 2 m 1 × 3 + m 2 × − 4 ⇒ 3 m 1 − 4 m 2 = 0 ⇒ 3 m 1 = 4 m 2 ⇒ m 2 m 1 = 3 4 .
m1 : m2 = 4 : 3.
Hence, PR : RQ = 4 : 3.
(ii) Substituting m1 : m2 = 4 : 3 in section formula we get,
y = m 1 y 2 + m 2 y 1 m 1 + m 2 = 4 × 2 + 3 × 5 4 + 3 = 8 + 15 7 = 23 7 = 3 2 7 y = \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2} \\[1em] = \dfrac{4 \times 2 + 3 \times 5}{4 + 3} \\[1em] = \dfrac{8 + 15}{7} \\[1em] = \dfrac{23}{7} \\[1em] = 3\dfrac{2}{7} y = m 1 + m 2 m 1 y 2 + m 2 y 1 = 4 + 3 4 × 2 + 3 × 5 = 7 8 + 15 = 7 23 = 3 7 2
R = (0, y) = ( 0 , 3 2 7 ) \Big(0, 3\dfrac{2}{7}\Big) ( 0 , 3 7 2 )
Hence, co-ordinates of R = ( 0 , 3 2 7 ) \Big(0, 3\dfrac{2}{7}\Big) ( 0 , 3 7 2 )
(iii) From graph,
PMNQ is a trapezium and 1 block = 1 unit.
Area of trapezium = 1 2 × \dfrac{1}{2} \times 2 1 ×
= 1 2 × ( P M + Q N ) × M N = 1 2 × ( 5 + 2 ) × 7 = 1 2 × 7 × 7 = 1 2 × 49 = 24.5 = \dfrac{1}{2} \times (PM + QN) \times MN \\[1em] = \dfrac{1}{2} \times (5 + 2) \times 7 \\[1em] = \dfrac{1}{2} \times 7 \times 7 \\[1em] = \dfrac{1}{2} \times 49 \\[1em] = 24.5 = 2 1 × ( PM + QN ) × MN = 2 1 × ( 5 + 2 ) × 7 = 2 1 × 7 × 7 = 2 1 × 49 = 24.5
Hence, area of PMNQ = 24.5 sq. units.
In the given figure, line APB meets the x-axis at point A and y-axis at point B. P is the point (-4, 2) and AP : PB = 1 : 2. Find the co-ordinates of A and B.
Answer
Since, A lies on x-axis its co-ordinates be (x, 0) and B lies on y-axis its co-ordinates be (0, y).
Given, AP : PB = 1 : 2.
By section formula,
x = m 1 x 2 + m 2 x 1 m 1 + m 2 ⇒ − 4 = 1 × 0 + 2 × x 1 + 2 ⇒ − 4 × 3 = 2 x ⇒ 2 x = − 12 ⇒ x = − 6. y = m 1 y 2 + m 2 y 1 m 1 + m 2 ⇒ 2 = 1 × y + 2 × 0 1 + 2 ⇒ 2 × 3 = y ⇒ y = 6. x = \dfrac{m_1x_2 + m_2x_1}{m_1 + m_2} \\[1em] \Rightarrow -4 = \dfrac{1 \times 0 + 2 \times x}{1 + 2} \\[1em] \Rightarrow -4 \times 3 = 2x \\[1em] \Rightarrow 2x = -12 \\[1em] \Rightarrow x = -6. \\[1em] y = \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2} \\[1em] \Rightarrow 2 = \dfrac{1 \times y + 2 \times 0}{1 + 2} \\[1em] \Rightarrow 2 \times 3 = y \\[1em] \Rightarrow y = 6. x = m 1 + m 2 m 1 x 2 + m 2 x 1 ⇒ − 4 = 1 + 2 1 × 0 + 2 × x ⇒ − 4 × 3 = 2 x ⇒ 2 x = − 12 ⇒ x = − 6. y = m 1 + m 2 m 1 y 2 + m 2 y 1 ⇒ 2 = 1 + 2 1 × y + 2 × 0 ⇒ 2 × 3 = y ⇒ y = 6.
A = (x, 0) = (-6, 0) and B = (0, y) = (0, 6).
Hence, A = (-6, 0) and B = (0, 6).
Given a line segment AB joining the points A(-4, 6) and B(8, 3). Find :
(i) the ratio in which line segment AB is divided by y-axis.
(ii) the co-ordinates of the point of intersection.
(iii) equation of perpendicular bisector of AB.
Answer
(i) Let the y-axis divide AB in the ratio m1 : m2 .
By section-formula, the x-coordinate = ( m 1 x 2 + m 2 x 1 m 1 + m 2 ) \Big(\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}\Big) ( m 1 + m 2 m 1 x 2 + m 2 x 1 )
Since, the x-coordinate on y-axis is 0. Putting value in above formula we get :
⇒ 0 = m 1 × 8 + m 2 × − 4 m 1 + m 2 ⇒ 8 m 1 − 4 m 2 = 0 ⇒ 8 m 1 = 4 m 2 ⇒ m 1 m 2 = 4 8 = 1 2 ⇒ m 1 : m 2 = 1 : 2. \Rightarrow 0 = \dfrac{m_1 \times 8 + m_2 \times -4}{m_1 + m_2} \\[1em] \Rightarrow 8m_1 - 4m_2 = 0 \\[1em] \Rightarrow 8m_1 = 4m_2 \\[1em] \Rightarrow \dfrac{m_1}{m_2} = \dfrac{4}{8} = \dfrac{1}{2} \\[1em] \Rightarrow m_1 : m_2 = 1 : 2. ⇒ 0 = m 1 + m 2 m 1 × 8 + m 2 × − 4 ⇒ 8 m 1 − 4 m 2 = 0 ⇒ 8 m 1 = 4 m 2 ⇒ m 2 m 1 = 8 4 = 2 1 ⇒ m 1 : m 2 = 1 : 2.
Hence, required ratio = 1 : 2.
(ii) The x-coordinate equals to zero on y-axis.
By section formula, the y-coordinate = ( m 1 y 2 + m 2 y 1 m 1 + m 2 ) \Big(\dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big) ( m 1 + m 2 m 1 y 2 + m 2 y 1 )
Substituting value in above formula, we get :
⇒ y = 1 × 3 + 2 × 6 1 + 2 = 3 + 12 3 = 15 3 = 5. \Rightarrow y = \dfrac{1 \times 3 + 2 \times 6}{1 + 2} \\[1em] = \dfrac{3 + 12}{3} \\[1em] = \dfrac{15}{3} \\[1em] = 5. ⇒ y = 1 + 2 1 × 3 + 2 × 6 = 3 3 + 12 = 3 15 = 5.
Hence, the coordinates of the point of intersection are (0, 5).
(iii) By formula,
Mid-point = ( x 1 + x 2 2 , y 1 + y 2 2 ) \Big(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\Big) ( 2 x 1 + x 2 , 2 y 1 + y 2 )
Let M be the mid-point of AB.
M = ( ( − 4 ) + 8 2 , 6 + 3 2 ) = ( 4 2 , 9 2 ) = ( 2 , 9 2 ) \Big(\dfrac{(-4) + 8}{2}, \dfrac{6 + 3}{2}\Big) = \Big(\dfrac{4}{2}, \dfrac{9}{2}\Big) = \Big(2, \dfrac{9}{2}\Big) ( 2 ( − 4 ) + 8 , 2 6 + 3 ) = ( 2 4 , 2 9 ) = ( 2 , 2 9 )
Slope of AB = 3 − 6 8 − ( − 4 ) = − 3 12 = − 1 4 \dfrac{3 - 6}{8 - (-4)} = \dfrac{-3}{12} = -\dfrac{1}{4} 8 − ( − 4 ) 3 − 6 = 12 − 3 = − 4 1
We know that,
Product of slope of perpendicular lines = -1.
Let slope of perpendicular bisector be h.
∴ h × − 1 4 -\dfrac{1}{4} − 4 1
⇒ h = 4.
By point-slope form,
Equation : y - y1 = m(x - x1 )
⇒ y - 9 2 \dfrac{9}{2} 2 9
⇒ 2 y − 9 2 \dfrac{2y - 9}{2} 2 2 y − 9
⇒ 2y - 9 = 8(x - 2)
⇒ 2y - 9 = 8x - 16
⇒ 2y - 8x - 9 + 16 = 0
⇒ 2y - 8x + 7 = 0
⇒ 8x - 2y = 7.
Hence, equation of perpendicular bisector of AB is 8x - 2y = 7.
If P(-b, 9a - 2) divides the line segment joining the points A(-3, 3a + 1) and B(5, 8a) in the ratio 3 : 1, find the values of a and b.
Answer
By section-formula,
x = m 1 x 2 + m 2 x 1 m 1 + m 2 x = \dfrac{m_1x_2 + m_2x_1}{m_1 + m_2} x = m 1 + m 2 m 1 x 2 + m 2 x 1
Substituting values we get,
⇒ − b = 3 × 5 + 1 × − 3 3 + 1 ⇒ − b = 15 − 3 4 ⇒ − b = 12 4 = 3 ⇒ b = − 3 y = m 1 y 2 + m 2 y 1 m 1 + m 2 \Rightarrow -b = \dfrac{3 \times 5 + 1 \times -3}{3 + 1} \\[1em] \Rightarrow -b = \dfrac{15 - 3}{4} \\[1em] \Rightarrow -b = \dfrac{12}{4} = 3 \\[1em] \Rightarrow b = -3 \\[1em] y = \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2} ⇒ − b = 3 + 1 3 × 5 + 1 × − 3 ⇒ − b = 4 15 − 3 ⇒ − b = 4 12 = 3 ⇒ b = − 3 y = m 1 + m 2 m 1 y 2 + m 2 y 1
Substituting values we get,
⇒ 9 a − 2 = 3 × 8 a + 1 × ( 3 a + 1 ) 3 + 1 ⇒ 9 a − 2 = 24 a + 3 a + 1 4 ⇒ 9 a − 2 = 27 a + 1 4 ⇒ 36 a − 8 = 27 a + 1 ⇒ 9 a = 9 ⇒ a = 1. \Rightarrow 9a - 2 = \dfrac{3 \times 8a + 1 \times (3a + 1)}{3 + 1} \\[1em] \Rightarrow 9a - 2 = \dfrac{24a + 3a + 1}{4} \\[1em] \Rightarrow 9a - 2 = \dfrac{27a + 1}{4} \\[1em] \Rightarrow 36a - 8 = 27a + 1 \\[1em] \Rightarrow 9a = 9 \\[1em] \Rightarrow a = 1. ⇒ 9 a − 2 = 3 + 1 3 × 8 a + 1 × ( 3 a + 1 ) ⇒ 9 a − 2 = 4 24 a + 3 a + 1 ⇒ 9 a − 2 = 4 27 a + 1 ⇒ 36 a − 8 = 27 a + 1 ⇒ 9 a = 9 ⇒ a = 1.
Hence, a = 1 and b = -3.
Point A(3, 4) is the center of a circle. If one of its diameters has one end as (7, 8); the other end of this diameter is :
(1, 0)
(0, 1)
(-1, 0)
(0, -1)
Answer
Given,
A(3, 4) is center and let B(7, 8) be one end of the diameter.
Let other end of diameter be C(x, y).
By mid-point formula,
Mid-point = ( x 1 + x 2 2 , y 1 + y 2 2 ) \Big(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\Big) ( 2 x 1 + x 2 , 2 y 1 + y 2 )
From figure,
A is the mid-point of BC.
∴ ( 3 , 4 ) = ( x + 7 2 , y + 8 2 ) ⇒ x + 7 2 = 3 and y + 8 2 = 4 ⇒ x + 7 = 6 and y + 8 = 8 ⇒ x = 6 − 7 and y = 8 − 8 ⇒ x = − 1 and y = 0. \therefore (3, 4) = \Big(\dfrac{x + 7}{2}, \dfrac{y + 8}{2}\Big) \\[1em] \Rightarrow \dfrac{x + 7}{2} = 3 \text{ and } \dfrac{y + 8}{2} = 4 \\[1em] \Rightarrow x + 7 = 6 \text{ and } y + 8 = 8 \\[1em] \Rightarrow x = 6 - 7 \text{ and } y = 8 - 8 \\[1em] \Rightarrow x = -1 \text{ and } y = 0. ∴ ( 3 , 4 ) = ( 2 x + 7 , 2 y + 8 ) ⇒ 2 x + 7 = 3 and 2 y + 8 = 4 ⇒ x + 7 = 6 and y + 8 = 8 ⇒ x = 6 − 7 and y = 8 − 8 ⇒ x = − 1 and y = 0.
∴ C = (-1, 0).
Hence, Option 3 is the correct option.
Point A lies on x-axis and point B lies on y-axis. If P(2, -2) bisects the line segment AB, the co-ordinates of A are :
(4, 0)
(0, 4)
(-4, 0)
(0, -4)
Answer
Given,
Point A lies on x-axis.
∴ Let point A be (x, 0).
Point B lies on y-axis.
∴ Let point B be (0, y).
P(2, -2) bisects the line segment AB or P is the mid-point of AB.
By mid-point formula,
Mid-point = ( x 1 + x 2 2 , y 1 + y 2 2 ) \Big(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\Big) ( 2 x 1 + x 2 , 2 y 1 + y 2 )
Substituting values we get :
⇒ ( 2 , − 2 ) = ( x + 0 2 , 0 + y 2 ) ⇒ ( 2 , − 2 ) = ( x 2 , y 2 ) ⇒ x 2 = 2 and y 2 = − 2 ⇒ x = 4 and y = − 4. \Rightarrow (2, -2) = \Big(\dfrac{x + 0}{2}, \dfrac{0 + y}{2}\Big) \\[1em] \Rightarrow (2, -2) = \Big(\dfrac{x}{2}, \dfrac{y}{2}\Big) \\[1em] \Rightarrow \dfrac{x}{2} = 2 \text{ and } \dfrac{y}{2} = -2 \\[1em] \Rightarrow x = 4 \text{ and } y = -4. ⇒ ( 2 , − 2 ) = ( 2 x + 0 , 2 0 + y ) ⇒ ( 2 , − 2 ) = ( 2 x , 2 y ) ⇒ 2 x = 2 and 2 y = − 2 ⇒ x = 4 and y = − 4.
Co-ordinates of A = (x, 0) = (4, 0).
Hence, Option 1 is the correct option.
In parallelogram ABCD, A = (6, 0), B = (12, -4) and C = (4, -4); then the co-ordinates of vertex D are :
(2, 0)
(-2, 0)
(0, 2)
(0, -2)
Answer
Let co-ordinates of vertex D be (x, y).
We know that,
Diagonal of parallelogram bisect each other.
From figure,
O (a, b) is the mid-point of AC.
By mid-point formula,
Mid-point = ( x 1 + x 2 2 , y 1 + y 2 2 ) \Big(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\Big) ( 2 x 1 + x 2 , 2 y 1 + y 2 )
Substituting values we get :
⇒ ( a , b ) = ( 6 + 4 2 , 0 + ( − 4 ) 2 ) = ( 10 2 , − 4 2 ) = ( 5 , − 2 ) . \Rightarrow (a, b) = \Big(\dfrac{6 + 4}{2}, \dfrac{0 + (-4)}{2}\Big) \\[1em] = \Big(\dfrac{10}{2}, \dfrac{-4}{2}\Big) \\[1em] = (5, -2). ⇒ ( a , b ) = ( 2 6 + 4 , 2 0 + ( − 4 ) ) = ( 2 10 , 2 − 4 ) = ( 5 , − 2 ) .
From figure,
O is also the mid-point of BD.
∴ ( 5 , − 2 ) = ( 12 + x 2 , − 4 + y 2 ) ⇒ 12 + x 2 = 5 and − 4 + y 2 = − 2 ⇒ 12 + x = 10 and − 4 + y = − 4 ⇒ x = 10 − 12 and y = − 4 + 4 ⇒ x = − 2 and y = 0. \therefore (5, -2) = \Big(\dfrac{12 + x}{2}, \dfrac{-4 + y}{2}\Big) \\[1em] \Rightarrow \dfrac{12 + x}{2} = 5 \text{ and } \dfrac{-4 + y}{2} = -2 \\[1em] \Rightarrow 12 + x = 10 \text{ and } -4 + y = -4 \\[1em] \Rightarrow x = 10 - 12 \text{ and } y = -4 + 4 \\[1em] \Rightarrow x = -2 \text{ and } y = 0. ∴ ( 5 , − 2 ) = ( 2 12 + x , 2 − 4 + y ) ⇒ 2 12 + x = 5 and 2 − 4 + y = − 2 ⇒ 12 + x = 10 and − 4 + y = − 4 ⇒ x = 10 − 12 and y = − 4 + 4 ⇒ x = − 2 and y = 0.
D = (-2, 0).
Hence, Option 2 is the correct option.
The point P(2, -7) is reflected in the point (0, 3); the co-ordinates of the image of point P are :
(2, 13)
(2, -13)
(-2, -13)
(-2, 13)
Answer
Let point P on reflection in the point (0, 3) becomes P'(x, y).
So, (0, 3) will be the mid-point of PP'.
By formula,
Mid-point = ( x 1 + x 2 2 , y 1 + y 2 2 ) \Big(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\Big) ( 2 x 1 + x 2 , 2 y 1 + y 2 )
Substituting values we get :
⇒ ( 0 , 3 ) = ( 2 + x 2 , − 7 + y 2 ) ⇒ 0 = 2 + x 2 and 3 = − 7 + y 2 ⇒ 2 + x = 0 and − 7 + y = 6 ⇒ x = − 2 and y = 6 + 7 ⇒ x = − 2 and y = 13. \Rightarrow (0, 3) = \Big(\dfrac{2 + x}{2}, \dfrac{-7 + y}{2}\Big) \\[1em] \Rightarrow 0 = \dfrac{2 + x}{2} \text{ and } 3 = \dfrac{-7 + y}{2} \\[1em] \Rightarrow 2 + x = 0 \text{ and } -7 + y = 6 \\[1em] \Rightarrow x = -2 \text{ and } y = 6 + 7 \\[1em] \Rightarrow x = -2 \text{ and } y = 13. ⇒ ( 0 , 3 ) = ( 2 2 + x , 2 − 7 + y ) ⇒ 0 = 2 2 + x and 3 = 2 − 7 + y ⇒ 2 + x = 0 and − 7 + y = 6 ⇒ x = − 2 and y = 6 + 7 ⇒ x = − 2 and y = 13.
P' = (-2, 13).
Hence, Option 4 is the correct option.
The co-ordinates of the centroid of a triangle with vertices (-6, -3), (0, 0) and (12, -6) are :
(2, 3)
(-2, 3)
(2, -3)
(-2, -3)
Answer
By formula,
Centroid of triangle = ( x 1 + x 2 + x 3 3 , y 1 + y 2 + y 3 3 ) \Big(\dfrac{x_1 + x_2 + x_3}{3}, \dfrac{y_1 + y_2 + y_3}{3}\Big) ( 3 x 1 + x 2 + x 3 , 3 y 1 + y 2 + y 3 )
Substituting values we get :
Centroid = ( − 6 + 0 + 12 3 , − 3 + 0 + ( − 6 ) 3 ) = ( 6 3 , − 9 3 ) = ( 2 , − 3 ) . \text{Centroid } = \Big(\dfrac{-6 + 0 + 12}{3}, \dfrac{-3 + 0 + (-6)}{3}\Big) \\[1em] = \Big(\dfrac{6}{3}, \dfrac{-9}{3}\Big) \\[1em] = (2, -3). Centroid = ( 3 − 6 + 0 + 12 , 3 − 3 + 0 + ( − 6 ) ) = ( 3 6 , 3 − 9 ) = ( 2 , − 3 ) .
Hence, Option 3 is the correct option.
Points A and B have co-ordinates (3, 5) and (x, y) respectively. The mid-point of AB is (2, 3). Find the values of x and y.
Answer
By formula,
Mid-point (M) = ( x 1 + x 2 2 , y 1 + y 2 2 ) \Big(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\Big) ( 2 x 1 + x 2 , 2 y 1 + y 2 )
Given, M = (2, 3)
Substituting values in above formula,
( 2 , 3 ) = ( 3 + x 2 , 5 + y 2 ) ∴ 2 = 3 + x 2 and 3 = 5 + y 2 ⇒ 4 = 3 + x and 6 = 5 + y ⇒ x = 1 and y = 1. (2, 3) = \Big(\dfrac{3 + x}{2}, \dfrac{5 + y}{2}\Big) \\[1em] \therefore 2 = \dfrac{3 + x}{2} \text{ and } 3 = \dfrac{5 + y}{2} \\[1em] \Rightarrow 4 = 3 + x \text{ and } 6 = 5 + y \\[1em] \Rightarrow x = 1 \text{ and } y = 1. ( 2 , 3 ) = ( 2 3 + x , 2 5 + y ) ∴ 2 = 2 3 + x and 3 = 2 5 + y ⇒ 4 = 3 + x and 6 = 5 + y ⇒ x = 1 and y = 1.
Hence, x = 1 and y = 1.
Given M is the mid-point of AB, find the co-ordinates of:
(i) A; if M = (1, 7) and B = (-5, 10),
(ii) B; if A = (3, -1) and M = (-1, 3).
Answer
By formula,
Mid-point (M) = ( x 1 + x 2 2 , y 1 + y 2 2 ) \Big(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\Big) ( 2 x 1 + x 2 , 2 y 1 + y 2 )
(i) Let co-ordinates of A = (x, y).
Substituting values in above formula we get,
( 1 , 7 ) = [ x + ( − 5 ) 2 , y + 10 2 ] ∴ 1 = x − 5 2 and 7 = y + 10 2 ⇒ 2 = x − 5 and 14 = y + 10 ⇒ x = 7 and y = 4. (1, 7) = \Big[\dfrac{x + (-5)}{2}, \dfrac{y + 10}{2}\Big] \\[1em] \therefore 1 = \dfrac{x - 5}{2} \text{ and } 7 = \dfrac{y + 10}{2} \\[1em] \Rightarrow 2 = x - 5 \text{ and } 14 = y + 10 \\[1em] \Rightarrow x = 7 \text{ and } y = 4. ( 1 , 7 ) = [ 2 x + ( − 5 ) , 2 y + 10 ] ∴ 1 = 2 x − 5 and 7 = 2 y + 10 ⇒ 2 = x − 5 and 14 = y + 10 ⇒ x = 7 and y = 4.
A = (x, y) = (7, 4).
Hence, co-ordinates of A = (7, 4).
(ii) Let co-ordinates of B = (x, y).
Substituting values in above formula we get,
( − 1 , 3 ) = ( 3 + x 2 , − 1 + y 2 ) ∴ − 1 = 3 + x 2 and 3 = y − 1 2 ⇒ − 2 = x + 3 and 6 = y − 1 ⇒ x = − 5 and y = 7. (-1, 3) = \Big(\dfrac{3 + x}{2}, \dfrac{-1 + y}{2}\Big) \\[1em] \therefore -1 = \dfrac{3 + x}{2} \text{ and } 3 = \dfrac{y - 1}{2} \\[1em] \Rightarrow -2 = x + 3 \text{ and } 6 = y - 1 \\[1em] \Rightarrow x = -5 \text{ and } y = 7. ( − 1 , 3 ) = ( 2 3 + x , 2 − 1 + y ) ∴ − 1 = 2 3 + x and 3 = 2 y − 1 ⇒ − 2 = x + 3 and 6 = y − 1 ⇒ x = − 5 and y = 7.
B = (x, y) = (-5, 7).
Hence, co-ordinates of B = (-5, 7).
P(-3, 2) is the mid-point of line segment AB as shown in the given figure. Find the co-ordinates of points A and B.
Answer
Since, A lie on y-axis, let its co-ordinates be (0, y) and B lie on x-axis, let its co-ordinates be (x, 0).
P is mid-point of AB.
By formula,
Mid-point = ( x 1 + x 2 2 , y 1 + y 2 2 ) \Big(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\Big) ( 2 x 1 + x 2 , 2 y 1 + y 2 )
Substituting values we get,
P = ( 0 + x 2 , y + 0 2 ) ⇒ ( − 3 , 2 ) = ( x 2 , y 2 ) ∴ − 3 = x 2 and 2 = y 2 ⇒ x = − 6 and y = 4. P = \Big(\dfrac{0 + x}{2}, \dfrac{y + 0}{2}\Big) \\[1em] \Rightarrow (-3, 2) = \Big(\dfrac{x}{2}, \dfrac{y}{2}\Big) \\[1em] \therefore -3 = \dfrac{x}{2} \text{ and } 2 = \dfrac{y}{2} \\[1em] \Rightarrow x = -6 \text{ and } y = 4. P = ( 2 0 + x , 2 y + 0 ) ⇒ ( − 3 , 2 ) = ( 2 x , 2 y ) ∴ − 3 = 2 x and 2 = 2 y ⇒ x = − 6 and y = 4.
A = (0, y) = (0, 4) and B = (x, 0) = (-6, 0).
Hence, the co-ordinates of points A and B are (0, 4) and (-6, 0).
(-5, 2), (3, -6) and (7, 4) are the vertices of a triangle. Find the length of its median through the vertex (3, -6).
Answer
Let A = (3, -6), B = (-5, 2) and C = (7, 4).
From figure, AD is the median.
Since, AD is median so, BD = DC.
Thus, D is mid-point of BC.
By formula,
Mid-point = ( x 1 + x 2 2 , y 1 + y 2 2 ) \Big(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\Big) ( 2 x 1 + x 2 , 2 y 1 + y 2 )
Substituting value we get,
D = ( − 5 + 7 2 , 2 + 4 2 ) = ( 2 2 , 6 2 ) = ( 1 , 3 ) . D = \Big(\dfrac{-5 + 7}{2}, \dfrac{2 + 4}{2}\Big) \\[1em] = \Big(\dfrac{2}{2}, \dfrac{6}{2}\Big) \\[1em] = (1, 3). D = ( 2 − 5 + 7 , 2 2 + 4 ) = ( 2 2 , 2 6 ) = ( 1 , 3 ) .
Distance between two points = ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2
Substituting values we get,
A D = ( 1 − 3 ) 2 + [ 3 − ( − 6 ) ] 2 = ( − 2 ) 2 + ( 9 ) 2 = 4 + 81 = 85 = 9.22 AD = \sqrt{(1 - 3)^2 + [3 - (-6)]^2} \\[1em] = \sqrt{(-2)^2 + (9)^2} \\[1em] = \sqrt{4 + 81} \\[1em] = \sqrt{85} \\[1em] = 9.22 A D = ( 1 − 3 ) 2 + [ 3 − ( − 6 ) ] 2 = ( − 2 ) 2 + ( 9 ) 2 = 4 + 81 = 85 = 9.22
Hence, the length of its median through the vertex (3, -6) = 9.22 units.
Given a line ABCD in which AB = BC = CD, B = (0, 3) and C = (1, 8). Find the co-ordinates of A and D.
Answer
Let co-ordinates of A be (x, y) and D be (p, q).
Since, AB = BC.
B is the mid-point of AC.
By formula,
Mid-point = ( x 1 + x 2 2 , y 1 + y 2 2 ) \Big(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\Big) ( 2 x 1 + x 2 , 2 y 1 + y 2 )
Substituting values we get,
⇒ B = ( x + 1 2 , y + 8 2 ) ⇒ ( 0 , 3 ) = ( x + 1 2 , y + 8 2 ) ⇒ 0 = x + 1 2 and 3 = y + 8 2 ⇒ x + 1 = 0 and y + 8 = 6 ⇒ x = − 1 and y = − 2. \Rightarrow B = \Big(\dfrac{x + 1}{2}, \dfrac{y + 8}{2}\Big) \\[1em] \Rightarrow (0, 3) = \Big(\dfrac{x + 1}{2}, \dfrac{y + 8}{2}\Big) \\[1em] \Rightarrow 0 = \dfrac{x + 1}{2} \text{ and } 3 = \dfrac{y + 8}{2} \\[1em] \Rightarrow x + 1 = 0 \text{ and } y + 8 = 6 \\[1em] \Rightarrow x = -1 \text{ and } y = -2. ⇒ B = ( 2 x + 1 , 2 y + 8 ) ⇒ ( 0 , 3 ) = ( 2 x + 1 , 2 y + 8 ) ⇒ 0 = 2 x + 1 and 3 = 2 y + 8 ⇒ x + 1 = 0 and y + 8 = 6 ⇒ x = − 1 and y = − 2.
A = (x, y) = (-1, -2).
Since, BC = CD.
C is mid-point of BD.
By formula,
Mid-point = ( x 1 + x 2 2 , y 1 + y 2 2 ) \Big(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\Big) ( 2 x 1 + x 2 , 2 y 1 + y 2 )
Substituting values we get,
⇒ C = ( 0 + p 2 , 3 + q 2 ) ⇒ ( 1 , 8 ) = ( p 2 , 3 + q 2 ) ⇒ 1 = p 2 and 8 = 3 + q 2 ⇒ p = 2 and 3 + q = 16 ⇒ p = 2 and q = 13. \Rightarrow C = \Big(\dfrac{0 + p}{2}, \dfrac{3 + q}{2}\Big) \\[1em] \Rightarrow (1, 8) = \Big(\dfrac{p}{2}, \dfrac{3 + q}{2}\Big) \\[1em] \Rightarrow 1 = \dfrac{p}{2} \text{ and } 8 = \dfrac{3 + q}{2} \\[1em] \Rightarrow p = 2 \text{ and } 3 + q = 16 \\[1em] \Rightarrow p = 2 \text{ and } q = 13. ⇒ C = ( 2 0 + p , 2 3 + q ) ⇒ ( 1 , 8 ) = ( 2 p , 2 3 + q ) ⇒ 1 = 2 p and 8 = 2 3 + q ⇒ p = 2 and 3 + q = 16 ⇒ p = 2 and q = 13.
D = (p, q) = (2, 13).
Hence, the co-ordinates of A = (-1, -2) and D = (2, 13).
A (2, 5), B (1, 0), C(-4, 3) and D (-3, 8) are the vertices of quadrilateral ABCD. Find the co-ordinates of the mid-points of AC and BD. Give a special name to the quadrilateral.
Answer
By formula,
Mid-point = ( x 1 + x 2 2 , y 1 + y 2 2 ) \Big(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\Big) ( 2 x 1 + x 2 , 2 y 1 + y 2 )
Let mid-point of AC be E.
Substituting value we get,
E = [ 2 + ( − 4 ) 2 , 5 + 3 2 ] = ( − 2 2 , 8 2 ) = ( − 1 , 4 ) . E = \Big[\dfrac{2 + (-4)}{2}, \dfrac{5 + 3}{2}\Big] \\[1em] = \Big(-\dfrac{2}{2}, \dfrac{8}{2}\Big) \\[1em] = (-1, 4). E = [ 2 2 + ( − 4 ) , 2 5 + 3 ] = ( − 2 2 , 2 8 ) = ( − 1 , 4 ) .
Let mid-point of BD be F.
Substituting value we get,
F = [ 1 + ( − 3 ) 2 , 0 + 8 2 ] = ( − 2 2 , 8 2 ) = ( − 1 , 4 ) . F = \Big[\dfrac{1 + (-3)}{2}, \dfrac{0 + 8}{2}\Big] \\[1em] = \Big(-\dfrac{2}{2}, \dfrac{8}{2}\Big) \\[1em] = (-1, 4). F = [ 2 1 + ( − 3 ) , 2 0 + 8 ] = ( − 2 2 , 2 8 ) = ( − 1 , 4 ) .
Thus, the co-ordinates of the mid-points of AC and BD are same i.e., AC and BD bisect each other.
∴ ABCD is a parallelogram.
Hence, the co-ordinates of the mid-points of AC = (-1, 4) and BD = (-1, 4) and ABCD is a parallelogram.
P (4, 2) and Q (-1, 5) are the vertices of parallelogram PQRS and (-3, 2) are the co-ordinates of the point of intersection of the diagonals. Find co-ordinates of R and S.
Answer
We know that diagonals of a parallelogram bisect each other.
Let co-ordinates of R= (a, b) and S = (c, d).
From figure,
O is the mid-point of PR.
By formula,
Mid-point = ( x 1 + x 2 2 , y 1 + y 2 2 ) \Big(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\Big) ( 2 x 1 + x 2 , 2 y 1 + y 2 )
Substituting value we get,
⇒ ( − 3 , 2 ) = ( 4 + a 2 , 2 + b 2 ) ⇒ − 3 = 4 + a 2 and 2 = 2 + b 2 ⇒ 4 + a = − 6 and 2 + b = 4 ⇒ a = − 10 and b = 2. \Rightarrow (-3, 2) = \Big(\dfrac{4 + a}{2}, \dfrac{2 + b}{2}\Big) \\[1em] \Rightarrow -3 = \dfrac{4 + a}{2} \text{ and } 2 = \dfrac{2 + b}{2} \\[1em] \Rightarrow 4 + a = -6 \text{ and } 2 + b = 4 \\[1em] \Rightarrow a = -10 \text{ and } b = 2. ⇒ ( − 3 , 2 ) = ( 2 4 + a , 2 2 + b ) ⇒ − 3 = 2 4 + a and 2 = 2 2 + b ⇒ 4 + a = − 6 and 2 + b = 4 ⇒ a = − 10 and b = 2.
R = (a, b) = (-10, 2).
O is also the mid-point of QS,
⇒ ( − 3 , 2 ) = ( − 1 + c 2 , 5 + d 2 ) ⇒ − 3 = − 1 + c 2 and 2 = 5 + d 2 ⇒ − 1 + c = − 6 and 5 + d = 4 ⇒ c = − 5 and d = − 1. \Rightarrow (-3, 2) = \Big(\dfrac{-1 + c}{2}, \dfrac{5 + d}{2}\Big) \\[1em] \Rightarrow -3 = \dfrac{-1 + c}{2} \text{ and } 2 = \dfrac{5 + d}{2} \\[1em] \Rightarrow -1 + c = -6 \text{ and } 5 + d = 4 \\[1em] \Rightarrow c = -5 \text{ and } d = -1. ⇒ ( − 3 , 2 ) = ( 2 − 1 + c , 2 5 + d ) ⇒ − 3 = 2 − 1 + c and 2 = 2 5 + d ⇒ − 1 + c = − 6 and 5 + d = 4 ⇒ c = − 5 and d = − 1.
S = (c, d) = (-5, -1).
Hence, the co-ordinates of R = (-10, 2) and S = (-5, -1).
The points (2, -1), (-1, 4) and (-2, 2) are mid-points of the sides of a triangle. Find its vertices.
Answer
Let D = (2, -1), E = (-1, 4) and F = (-2, 2).
Let A(x1 , y1 ), B(x2 , y2 ) and C(x3 , y3 ) be the vertices of △ABC.
Mid-point of AB, i.e. D(2, -1).
∴ 2 = x 1 + x 2 2 and − 1 = y 1 + y 2 2 \therefore 2 = \dfrac{x_1 + x_2}{2} \text{ and } -1 = \dfrac{y_1 + y_2}{2} ∴ 2 = 2 x 1 + x 2 and − 1 = 2 y 1 + y 2
⇒ x1 + x2 = 4 ........(1)
⇒ y1 + y2 = -2 ........(2)
Mid-point of BC, i.e. E(-1, 4).
∴ − 1 = x 2 + x 3 2 and 4 = y 2 + y 3 2 \therefore -1 = \dfrac{x_2 + x_3}{2} \text{ and } 4 = \dfrac{y_2 + y_3}{2} ∴ − 1 = 2 x 2 + x 3 and 4 = 2 y 2 + y 3
⇒ x2 + x3 = -2 ........(3)
⇒ y2 + y3 = 8 ........(4)
Mid-point of AC, i.e. F(-2, 2).
∴ − 2 = x 1 + x 3 2 and 2 = y 1 + y 3 2 \therefore -2 = \dfrac{x_1 + x_3}{2} \text{ and } 2 = \dfrac{y_1 + y_3}{2} ∴ − 2 = 2 x 1 + x 3 and 2 = 2 y 1 + y 3
⇒ x1 + x3 = -4 ........(5)
⇒ y1 + y3 = 4 ........(6)
Adding 1, 3 and 5 we get,
⇒ x1 + x2 + x2 + x3 + x1 + x3 = 4 + (-2) + (-4)
⇒ 2(x1 + x2 + x3 ) = -2
⇒ x1 + x2 + x3 = -1.
From (1),
⇒ 4 + x3 = -1
⇒ x3 = -5.
Substituting value of x3 in (5) we get,
⇒ x1 + (-5) = -4
⇒ x1 = 1.
Substituting value of x3 in (3) we get,
⇒ x2 + (-5) = -2
⇒ x2 = 3.
Adding (2), (4) and (6) we get,
⇒ y1 + y2 + y2 + y3 + y1 + y3 = -2 + 8 + 4
⇒ 2(y1 + y2 + y3 ) = 10
⇒ y1 + y2 + y3 = 5
From (2)
⇒ -2 + y3 = 5
⇒ y3 = 7.
Substituting value of y3 in (4) we get,
⇒ y2 + 7 = 8
⇒ y2 = 1.
Substituting value of y3 in (6) we get,
⇒ y1 + 7 = 4
⇒ y1 = -3.
A = (x1 , y1 ) = (1, -3), B = (x2 , y2 ) = (3, 1), C = (x3 , y3 ) = (-5, 7).
Hence, A = (1, -3), B = (3, 1) and C = (-5, 7).
Points A(-5, x), B(y, 7) and C(1, -3) are collinear (i.e. lie on same straight line) such that AB = BC. Calculate the values of x and y.
Answer
Since, A, B and C are collinear and AB = BC.
We can say that B is the mid-point of AC.
By formula,
Mid-point = ( x 1 + x 2 2 , y 1 + y 2 2 ) \Big(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\Big) ( 2 x 1 + x 2 , 2 y 1 + y 2 )
Substituting value we get,
⇒ B = ( − 5 + 1 2 , x + ( − 3 ) 2 ) ⇒ ( y , 7 ) = ( − 4 2 , x − 3 2 ) ⇒ y = − 4 2 and 7 = x − 3 2 ⇒ y = − 2 and x − 3 = 14 ⇒ y = − 2 and x = 17. \Rightarrow B = \Big(\dfrac{-5 + 1}{2}, \dfrac{x + (-3)}{2}\Big) \\[1em] \Rightarrow (y, 7) = \Big(\dfrac{-4}{2}, \dfrac{x - 3}{2}\Big) \\[1em] \Rightarrow y = \dfrac{-4}{2} \text{ and } 7 = \dfrac{x - 3}{2} \\[1em] \Rightarrow y = -2 \text{ and } x - 3 = 14 \\[1em] \Rightarrow y = -2 \text{ and } x = 17. ⇒ B = ( 2 − 5 + 1 , 2 x + ( − 3 ) ) ⇒ ( y , 7 ) = ( 2 − 4 , 2 x − 3 ) ⇒ y = 2 − 4 and 7 = 2 x − 3 ⇒ y = − 2 and x − 3 = 14 ⇒ y = − 2 and x = 17.
Hence, x = 17 and y = -2.
Points P(a, -4), Q(-2, b) and R(0, 2) are collinear. If Q lies between P and R, such that PR = 2QR, calculate the values of a and b.
Answer
From figure,
PR = PQ + QR
Given,
⇒ PR = 2QR
⇒ PQ +QR = 2QR
⇒ PQ = QR.
∴ Q is the mid-point of P and R.
By formula,
Mid-point = ( x 1 + x 2 2 , y 1 + y 2 2 ) \Big(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\Big) ( 2 x 1 + x 2 , 2 y 1 + y 2 )
Substituting value we get,
⇒ Q = ( a + 0 2 , − 4 + 2 2 ) ⇒ ( − 2 , b ) = ( a 2 , − 2 2 ) ⇒ − 2 = a 2 and b = − 2 2 ⇒ a = − 4 and b = − 1. \Rightarrow Q = \Big(\dfrac{a + 0}{2}, \dfrac{-4 + 2}{2}\Big) \\[1em] \Rightarrow (-2, b) = \Big(\dfrac{a}{2}, \dfrac{-2}{2}\Big) \\[1em] \Rightarrow -2 = \dfrac{a}{2} \text{ and } b = \dfrac{-2}{2} \\[1em] \Rightarrow a = -4 \text{ and } b = -1. ⇒ Q = ( 2 a + 0 , 2 − 4 + 2 ) ⇒ ( − 2 , b ) = ( 2 a , 2 − 2 ) ⇒ − 2 = 2 a and b = 2 − 2 ⇒ a = − 4 and b = − 1.
Hence, a = -4 and b = -1.
The co-ordinates of the centroid of a triangle PQR are (2, -5). If Q = (-6, 5) and R = (11, 8); calculate the co-ordinates of vertex P.
Answer
Let co-ordinates of P = (x, y).
Centroid of the triangle is given by (G) = ( x 1 + x 2 + x 3 3 , y 1 + y 2 + y 3 3 ) \Big(\dfrac{x_1 + x_2 + x_3}{3}, \dfrac{y_1 + y_2 + y_3}{3}\Big) ( 3 x 1 + x 2 + x 3 , 3 y 1 + y 2 + y 3 )
Substituting values we get,
⇒ ( 2 , − 5 ) = ( x + ( − 6 ) + 11 3 , y + 5 + 8 3 ) ⇒ 2 = x − 6 + 11 3 and − 5 = y + 13 3 ⇒ 6 = x + 5 and − 15 = y + 13 ⇒ x = 1 and y = − 28. \Rightarrow (2, -5) = \Big(\dfrac{x + (-6) + 11}{3}, \dfrac{y + 5 + 8}{3}\Big) \\[1em] \Rightarrow 2 = \dfrac{x - 6 + 11}{3} \text{ and } -5 = \dfrac{y + 13}{3} \\[1em] \Rightarrow 6 = x + 5 \text{ and } -15 = y + 13 \\[1em] \Rightarrow x = 1 \text{ and } y = -28. ⇒ ( 2 , − 5 ) = ( 3 x + ( − 6 ) + 11 , 3 y + 5 + 8 ) ⇒ 2 = 3 x − 6 + 11 and − 5 = 3 y + 13 ⇒ 6 = x + 5 and − 15 = y + 13 ⇒ x = 1 and y = − 28.
P = (x, y) = (1, -28).
Hence, co-ordinates of P = (1, -28).
A (5, x), B(-4, 3) and C (y, -2) are the vertices of the triangle ABC whose centroid is the origin. Calculate the values of x and y.
Answer
Centroid of the triangle is given by (G) = ( x 1 + x 2 + x 3 3 , y 1 + y 2 + y 3 3 ) \Big(\dfrac{x_1 + x_2 + x_3}{3}, \dfrac{y_1 + y_2 + y_3}{3}\Big) ( 3 x 1 + x 2 + x 3 , 3 y 1 + y 2 + y 3 )
Substituting values we get,
⇒ ( 0 , 0 ) = [ 5 + ( − 4 ) + y 3 , x + 3 + ( − 2 ) 3 ] ⇒ 0 = 5 − 4 + y 3 and 0 = x + 3 − 2 3 ⇒ 0 = y + 1 and 0 = x + 1 ⇒ y = − 1 and x = − 1. \Rightarrow (0, 0) = \Big[\dfrac{5 + (-4) + y}{3}, \dfrac{x + 3 + (-2)}{3}\Big] \\[1em] \Rightarrow 0 = \dfrac{5 - 4 + y}{3} \text{ and } 0 = \dfrac{x + 3 - 2}{3} \\[1em] \Rightarrow 0 = y + 1 \text{ and } 0 = x + 1 \\[1em] \Rightarrow y = -1 \text{ and } x = -1. ⇒ ( 0 , 0 ) = [ 3 5 + ( − 4 ) + y , 3 x + 3 + ( − 2 ) ] ⇒ 0 = 3 5 − 4 + y and 0 = 3 x + 3 − 2 ⇒ 0 = y + 1 and 0 = x + 1 ⇒ y = − 1 and x = − 1.
Hence, x = -1 and y = -1.
If P(4, 3) is mid-point of line segment AB, then
A = (6, 0), B = (0, 8)
A = (8, 6), B = (8, 0)
A = (6, 8), B = (0, 6)
A = (8, 0) and B = (0, 6)
Answer
From figure,
A lies on x-axis and B lies on y-axis.
Let co-ordinates of A be (x, 0) and B be (0, y).
Given,
P(4, 3) is mid-point of line segment AB.
By mid-point formula,
Mid-point = ( x 1 + x 2 2 , y 1 + y 2 2 ) \Big(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\Big) ( 2 x 1 + x 2 , 2 y 1 + y 2 )
Substituting values we get :
⇒ ( 4 , 3 ) = ( x + 0 2 , 0 + y 2 ) ⇒ ( 4 , 3 ) = ( x 2 , y 2 ) ⇒ x 2 = 4 and y 2 = 3 ⇒ x = 8 and y = 6. \Rightarrow (4, 3) = \Big(\dfrac{x + 0}{2}, \dfrac{0 + y}{2}\Big) \\[1em] \Rightarrow (4, 3) = \Big(\dfrac{x}{2}, \dfrac{y}{2}\Big) \\[1em] \Rightarrow \dfrac{x}{2} = 4 \text{ and } \dfrac{y}{2} = 3 \\[1em] \Rightarrow x = 8 \text{ and } y = 6. ⇒ ( 4 , 3 ) = ( 2 x + 0 , 2 0 + y ) ⇒ ( 4 , 3 ) = ( 2 x , 2 y ) ⇒ 2 x = 4 and 2 y = 3 ⇒ x = 8 and y = 6.
A = (x, 0) = (8, 0),
B = (0, y) = (0, 6).
Hence, Option 4 is the correct option.
Points A and B have co-ordinates (4, y) and (x, 8). The mid-point of AB = (0, 0); the values of x and y are :
x = -4, y = 8
x = 4, y = -8
x = -4, y = -8
x = 4, y = 8
Answer
By mid-point formula,
Mid-point = ( x 1 + x 2 2 , y 1 + y 2 2 ) \Big(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\Big) ( 2 x 1 + x 2 , 2 y 1 + y 2 )
Substituting values we get :
⇒ ( 0 , 0 ) = ( 4 + x 2 , y + 8 2 ) ⇒ x + 4 2 = 0 and y + 8 2 = 0 ⇒ x + 4 = 0 and y + 8 = 0 ⇒ x = − 4 and y = − 8. \Rightarrow (0, 0) = \Big(\dfrac{4 + x}{2}, \dfrac{y + 8}{2}\Big) \\[1em] \Rightarrow \dfrac{x + 4}{2} = 0 \text{ and } \dfrac{y + 8}{2} = 0 \\[1em] \Rightarrow x + 4 = 0 \text{ and } y + 8 = 0 \\[1em] \Rightarrow x = -4 \text{ and } y = -8. ⇒ ( 0 , 0 ) = ( 2 4 + x , 2 y + 8 ) ⇒ 2 x + 4 = 0 and 2 y + 8 = 0 ⇒ x + 4 = 0 and y + 8 = 0 ⇒ x = − 4 and y = − 8.
Hence, Option 3 is the correct option.
Two vertices of a triangle are (-5, 3) and (4, -7). If centroid of the triangle is (-1, 2) then the third vertex is :
(-2, 7)
(-2, 10)
(-2, -10)
(0, -2)
Answer
Let third vertex be (x, y).
By formula,
Centroid of triangle = ( x 1 + x 2 + x 3 3 , y 1 + y 2 + y 3 3 ) \Big(\dfrac{x_1 + x_2 + x_3}{3}, \dfrac{y_1 + y_2 + y_3}{3}\Big) ( 3 x 1 + x 2 + x 3 , 3 y 1 + y 2 + y 3 )
Substituting values we get :
⇒ ( − 1 , 2 ) = ( ( − 5 ) + 4 + x 3 , 3 + ( − 7 ) + y 3 ) ⇒ ( − 1 , 2 ) = ( x − 1 3 , y − 4 3 ) ⇒ − 1 = x − 1 3 and 2 = y − 4 3 ⇒ x − 1 = − 3 and y − 4 = 6 ⇒ x = − 3 + 1 and y = 6 + 4 ⇒ x = − 2 and y = 10. \Rightarrow (-1, 2) = \Big(\dfrac{(-5) + 4 + x}{3}, \dfrac{3 + (-7) + y}{3}\Big) \\[1em] \Rightarrow (-1, 2) = \Big(\dfrac{x - 1}{3}, \dfrac{y - 4}{3}\Big) \\[1em] \Rightarrow -1 = \dfrac{x - 1}{3} \text{ and } 2 = \dfrac{y - 4}{3} \\[1em] \Rightarrow x - 1 = -3 \text{ and } y - 4 = 6 \\[1em] \Rightarrow x = -3 + 1 \text{ and } y = 6 + 4 \\[1em] \Rightarrow x = -2 \text{ and } y = 10. ⇒ ( − 1 , 2 ) = ( 3 ( − 5 ) + 4 + x , 3 3 + ( − 7 ) + y ) ⇒ ( − 1 , 2 ) = ( 3 x − 1 , 3 y − 4 ) ⇒ − 1 = 3 x − 1 and 2 = 3 y − 4 ⇒ x − 1 = − 3 and y − 4 = 6 ⇒ x = − 3 + 1 and y = 6 + 4 ⇒ x = − 2 and y = 10.
Third vertex = (-2, 10).
Hence, Option 2 is the correct option.
Points A, B, C and D divide the join of O(0, 0) and P(0, 10) into five equal parts. The co-ordinates of point D are :
(10, 8)
(-8, 0)
(-10, 0)
(0, 8)
Answer
From figure,
D divides the line segment OP in the ratio 4 : 1.
Let co-ordinates of D be (x, y).
By section-formula,
(x, y) = ( m 1 x 2 + m 2 x 1 m 1 + m 2 , m 1 y 2 + m 2 y 1 m 1 + m 2 ) \Big(\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}, \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big) ( m 1 + m 2 m 1 x 2 + m 2 x 1 , m 1 + m 2 m 1 y 2 + m 2 y 1 )
Substituting values we get :
⇒ ( x , y ) = ( 4 × 0 + 1 × 0 4 + 1 , 4 × 10 + 1 × 0 4 + 1 ) = ( 0 5 , 40 5 ) = ( 0 , 8 ) . \Rightarrow (x, y) = \Big(\dfrac{4 \times 0 + 1 \times 0}{4 + 1}, \dfrac{4 \times 10 + 1\times 0}{4 + 1}\Big) \\[1em] = \Big(\dfrac{0}{5}, \dfrac{40}{5}\Big) \\[1em] = (0, 8). ⇒ ( x , y ) = ( 4 + 1 4 × 0 + 1 × 0 , 4 + 1 4 × 10 + 1 × 0 ) = ( 5 0 , 5 40 ) = ( 0 , 8 ) .
Hence, Option 4 is the correct option.
Line segment joining points (4, 3) and (1, -2) is divided by the point (y, 0) in the ratio :
2 : 3
3 : 2
3 : 4
4 : 2
Answer
Let (y, 0) divide the line segment joining the points (4, 3) and (1, -2) in the ratio k : 1.
By section-formula,
(x, y) = ( m 1 x 2 + m 2 x 1 m 1 + m 2 , m 1 y 2 + m 2 y 1 m 1 + m 2 ) \Big(\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}, \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big) ( m 1 + m 2 m 1 x 2 + m 2 x 1 , m 1 + m 2 m 1 y 2 + m 2 y 1 )
Substituting values for y-coordinate :
⇒ 0 = k × − 2 + 1 × 3 k + 1 ⇒ − 2 k + 3 = 0 ⇒ 2 k = 3 ⇒ k = 3 2 . \Rightarrow 0 = \dfrac{k \times -2 + 1 \times 3}{k + 1} \\[1em] \Rightarrow -2k + 3 = 0 \\[1em] \Rightarrow 2k = 3 \\[1em] \Rightarrow k = \dfrac{3}{2}. ⇒ 0 = k + 1 k × − 2 + 1 × 3 ⇒ − 2 k + 3 = 0 ⇒ 2 k = 3 ⇒ k = 2 3 .
Substituting value of k in k : 1, we get :
⇒ 3 2 : 1 \dfrac{3}{2} : 1 2 3 : 1
⇒ 3 : 2.
Hence, Option 2 is the correct option.
The point P divides the line segment joining the point (1, 2) and (-1, 2) internally in the ratio 1 : 2.
Assertion (A) : The co-ordinates of point P = (1, 6)
Reason (R) : If point P divides the line segment joining the points (x1 , y1 ) and (x2 , y2 ) in the ratio m1 : m2 then :
P = ( m 1 x 2 + m 2 x 1 m 1 + m 2 , m 1 y 2 + m 2 y 1 m 1 + m 2 ) \Big(\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}, \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big) ( m 1 + m 2 m 1 x 2 + m 2 x 1 , m 1 + m 2 m 1 y 2 + m 2 y 1 )
A is true, R is false.
A is false, R is true.
Both A and R are true and R is correct reason for A.
Both A and R are true and R is incorrect reason for A.
Answer
We know that,
If point P divides the line segment joining the points (x1 , y1 ) and (x2 , y2 ) in the ratio m1 : m2 , then :
Co-ordinates of P = ( m 1 x 2 + m 2 x 1 m 1 + m 2 , m 1 y 2 + m 2 y 1 m 1 + m 2 ) \Big(\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}, \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big) ( m 1 + m 2 m 1 x 2 + m 2 x 1 , m 1 + m 2 m 1 y 2 + m 2 y 1 )
So, reason (R) is true.
Here, (x1 , y1 ) = (1, 2) and (x2 , y2 ) = (-1, 2)
m1 : m2 = 1 : 2
Substituting the values, we get :
⇒ P = [ 1 × ( − 1 ) + 2 × 1 1 + 2 , 1 × 2 + 2 × 2 1 + 2 ] = ( − 1 + 2 3 , 2 + 4 3 ) = ( 1 3 , 6 3 ) = ( 1 3 , 2 ) \Rightarrow \text{P} = \Big[\dfrac{1\times (-1) + 2\times 1}{1 + 2}, \dfrac{1 \times 2 + 2 \times 2}{1 + 2}\Big]\\[1em] = \Big(\dfrac{-1 + 2}{3}, \dfrac{2 + 4}{3}\Big)\\[1em] = \Big(\dfrac{1}{3}, \dfrac{6}{3}\Big)\\[1em] = \Big(\dfrac{1}{3}, 2\Big)\\[1em] ⇒ P = [ 1 + 2 1 × ( − 1 ) + 2 × 1 , 1 + 2 1 × 2 + 2 × 2 ] = ( 3 − 1 + 2 , 3 2 + 4 ) = ( 3 1 , 3 6 ) = ( 3 1 , 2 )
So, assertion (A) is false.
Hence, option 2 is the correct option.
Point P(x, y) is equidistant from points A(7, 1) and B(3, 5).
Statement 1: x = 7 + 3 2 \dfrac{7 + 3}{2} 2 7 + 3 1 + 5 2 \dfrac{1 + 5}{2} 2 1 + 5
Statement 2: (x - 7)2 + (y - 1)2 = (x - 3)2 + (y - 5)2 .
Both the statement are true.
Both the statement are false.
Statement 1 is true, and statement 2 is false.
Statement 1 is false, and statement 2 is true.
Answer
It is given that point P(x, y) is equidistant from points A(7, 1) and B(3, 5).
If a point is equidistant from two points it is not necessary that it will be the mid-point of the two points, sp we cannot use mid-point formula to find the co-ordinates of P.
So, statement 1 is false.
Using distance formula,
Distance between two points = ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2
⇒ PA = PB
⇒ ⇒ ( x − 7 ) 2 + ( y − 1 ) 2 = ( x − 3 ) 2 + ( y − 5 ) 2 \Rightarrow \sqrt{(x - 7)^2 + (y - 1)^2} = \sqrt{(x - 3)^2 + (y - 5)^2} ⇒ ( x − 7 ) 2 + ( y − 1 ) 2 = ( x − 3 ) 2 + ( y − 5 ) 2
⇒ (x - 7)2 + (y - 1)2 = (x - 3)2 + (y - 5)2
So, statement 2 is true.
Hence, option 4 is the correct option.
Given a triangle ABC in which A = (4, -4), B = (0, 5) and C = (5, 10). A point P lies on BC such that BP : PC = 3 : 2. Find the length of line segment AP.
Answer
Let the co-ordinates of P be (x, y)
∴ x = m 1 x 2 + m 2 x 1 m 1 + m 2 = 3 × 5 + 2 × 0 3 + 2 = 15 + 0 3 = 15 3 = 5. \therefore x = \dfrac{m_1x_2 + m_2x_1}{m_1 + m_2} \\[1em] = \dfrac{3 \times 5 + 2 \times 0}{3 + 2} \\[1em] = \dfrac{15 + 0}{3} \\[1em] = \dfrac{15}{3} = 5. ∴ x = m 1 + m 2 m 1 x 2 + m 2 x 1 = 3 + 2 3 × 5 + 2 × 0 = 3 15 + 0 = 3 15 = 5.
and,
y = m 1 y 2 + m 2 y 1 m 1 + m 2 = 3 × 10 + 2 × 5 3 + 2 = 30 + 10 5 = 40 5 = 8. y = \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2} \\[1em] = \dfrac{3 \times 10 + 2 \times 5}{3 + 2} \\[1em] = \dfrac{30 + 10}{5} \\[1em] = \dfrac{40}{5} = 8. y = m 1 + m 2 m 1 y 2 + m 2 y 1 = 3 + 2 3 × 10 + 2 × 5 = 5 30 + 10 = 5 40 = 8.
P = (x, y) = (5, 8).
Distance between two points = ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2
A P = ( 5 − 4 ) 2 + ( 8 − ( − 4 ) ) 2 = 1 2 + 12 2 = 1 + 144 = 145 = 12.04 AP = \sqrt{(5 - 4)^2 + (8 - (-4))^2} \\[1em] = \sqrt{1^2 + 12^2} \\[1em] = \sqrt{1 + 144} \\[1em] = \sqrt{145} \\[1em] = 12.04 A P = ( 5 − 4 ) 2 + ( 8 − ( − 4 ) ) 2 = 1 2 + 1 2 2 = 1 + 144 = 145 = 12.04
Hence, AP = 12.04 units.
A(20, 0) and B (10, -20) are two fixed points. Find the co-ordinates of the point P in AB such that : 3PB = AB. Also, find the co-ordinates of some other point Q in AB such that AB = 6AQ.
Answer
From figure,
⇒ AB = AP + PB
⇒ 3PB = AP + PB
⇒ AP = 2PB
⇒ A P P B = 2 1 \dfrac{AP}{PB} = \dfrac{2}{1} PB A P = 1 2
⇒ AP : PB = 2 : 1.
Let the co-ordinates of P be (x, y)
∴ x = m 1 x 2 + m 2 x 1 m 1 + m 2 = 2 × 10 + 1 × 20 2 + 1 = 20 + 20 3 = 40 3 . \therefore x = \dfrac{m_1x_2 + m_2x_1}{m_1 + m_2} \\[1em] = \dfrac{2 \times 10 + 1 \times 20}{2 + 1} \\[1em] = \dfrac{20 + 20}{3} \\[1em] = \dfrac{40}{3}. ∴ x = m 1 + m 2 m 1 x 2 + m 2 x 1 = 2 + 1 2 × 10 + 1 × 20 = 3 20 + 20 = 3 40 .
and,
y = m 1 y 2 + m 2 y 1 m 1 + m 2 = 2 × − 20 + 1 × 0 2 + 1 = − 40 + 0 3 = − 40 3 . y = \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2} \\[1em] = \dfrac{2 \times -20 + 1 \times 0}{2 + 1} \\[1em] = \dfrac{-40 + 0}{3} \\[1em] = \dfrac{-40}{3}. y = m 1 + m 2 m 1 y 2 + m 2 y 1 = 2 + 1 2 × − 20 + 1 × 0 = 3 − 40 + 0 = 3 − 40 .
From figure,
⇒ AB = AQ + QB
⇒ 6AQ = AQ + QB
⇒ QB = 5AQ
⇒ A Q Q B = 1 5 \dfrac{AQ}{QB} = \dfrac{1}{5} QB A Q = 5 1
⇒ AQ : QB = 1 : 5.
Let the co-ordinates of Q be (p, q)
∴ p = m 1 x 2 + m 2 x 1 m 1 + m 2 = 1 × 10 + 5 × 20 1 + 5 = 10 + 100 6 = 110 6 = 55 3 . \therefore p = \dfrac{m_1x_2 + m_2x_1}{m_1 + m_2} \\[1em] = \dfrac{1 \times 10 + 5 \times 20}{1 + 5} \\[1em] = \dfrac{10 + 100}{6} \\[1em] = \dfrac{110}{6} = \dfrac{55}{3}. ∴ p = m 1 + m 2 m 1 x 2 + m 2 x 1 = 1 + 5 1 × 10 + 5 × 20 = 6 10 + 100 = 6 110 = 3 55 .
and,
q = m 1 y 2 + m 2 y 1 m 1 + m 2 = 1 × − 20 + 5 × 0 1 + 5 = − 20 + 0 6 = − 10 3 . q = \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2} \\[1em] = \dfrac{1 \times -20 + 5 \times 0}{1 + 5} \\[1em] = \dfrac{-20 + 0}{6} \\[1em] = \dfrac{-10}{3}. q = m 1 + m 2 m 1 y 2 + m 2 y 1 = 1 + 5 1 × − 20 + 5 × 0 = 6 − 20 + 0 = 3 − 10 .
Hence, P = ( 40 3 , − 40 3 ) and Q = ( 55 3 , − 10 3 ) \Big(\dfrac{40}{3}, -\dfrac{40}{3}\Big) \text{ and Q} = \Big(\dfrac{55}{3}, -\dfrac{10}{3}\Big) ( 3 40 , − 3 40 ) and Q = ( 3 55 , − 3 10 )
A(-8, 0), B(0, 16) and C(0, 0) are the vertices of a triangle ABC. Point P lies on AB and Q lies on AC such that AP : PB = 3 : 5 and AQ : QC = 3 : 5.
Show that : PQ = 3 8 \dfrac{3}{8} 8 3
Answer
The triangle is shown in the figure below:
Let the co-ordinates of P be (x, y)
∴ x = m 1 x 2 + m 2 x 1 m 1 + m 2 = 3 × 0 + 5 × − 8 3 + 5 = 0 − 40 8 = − 40 8 = − 5. \therefore x = \dfrac{m_1x_2 + m_2x_1}{m_1 + m_2} \\[1em] = \dfrac{3 \times 0 + 5 \times -8}{3 + 5} \\[1em] = \dfrac{0 - 40}{8} \\[1em] = \dfrac{-40}{8} = -5. ∴ x = m 1 + m 2 m 1 x 2 + m 2 x 1 = 3 + 5 3 × 0 + 5 × − 8 = 8 0 − 40 = 8 − 40 = − 5.
and,
y = m 1 y 2 + m 2 y 1 m 1 + m 2 = 3 × 16 + 5 × 0 3 + 5 = 48 + 0 8 = 6. y = \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2} \\[1em] = \dfrac{3 \times 16 + 5 \times 0}{3 + 5} \\[1em] = \dfrac{48 + 0}{8} \\[1em] = 6. y = m 1 + m 2 m 1 y 2 + m 2 y 1 = 3 + 5 3 × 16 + 5 × 0 = 8 48 + 0 = 6.
P = (x, y) = (-5, 6).
Let the co-ordinates of Q be (m, n)
∴ m = m 1 x 2 + m 2 x 1 m 1 + m 2 = 3 × 0 + 5 × − 8 3 + 5 = 0 − 40 8 = − 40 8 = − 5. \therefore m = \dfrac{m_1x_2 + m_2x_1}{m_1 + m_2} \\[1em] = \dfrac{3 \times 0 + 5 \times -8}{3 + 5} \\[1em] = \dfrac{0 - 40}{8} \\[1em] = \dfrac{-40}{8} = -5. ∴ m = m 1 + m 2 m 1 x 2 + m 2 x 1 = 3 + 5 3 × 0 + 5 × − 8 = 8 0 − 40 = 8 − 40 = − 5.
and,
n = m 1 y 2 + m 2 y 1 m 1 + m 2 = 3 × 0 + 5 × 0 3 + 5 = 0 + 0 8 = 0. n = \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2} \\[1em] = \dfrac{3 \times 0 + 5 \times 0}{3 + 5} \\[1em] = \dfrac{0 + 0}{8} \\[1em] = 0. n = m 1 + m 2 m 1 y 2 + m 2 y 1 = 3 + 5 3 × 0 + 5 × 0 = 8 0 + 0 = 0.
Q = (m, n) = (-5, 0).
Distance between two points = ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2
P Q = ( − 5 − ( − 5 ) ) 2 + ( 0 − 6 ) 2 = 0 2 + ( − 6 ) 2 = 36 = 6 units . B C = ( 0 − 0 ) 2 + ( 0 − 16 ) 2 = 0 2 + ( − 16 ) 2 = 256 = 16 units . PQ = \sqrt{(-5 - (-5))^2 + (0 - 6)^2} \\[1em] = \sqrt{0^2 + (-6)^2} \\[1em] = \sqrt{36} \\[1em] = 6 \text{ units}. \\[1em] BC = \sqrt{(0 - 0)^2 + (0 - 16)^2} \\[1em] = \sqrt{0^2 + (-16)^2} \\[1em] = \sqrt{256} \\[1em] = 16 \text{ units}. PQ = ( − 5 − ( − 5 ) ) 2 + ( 0 − 6 ) 2 = 0 2 + ( − 6 ) 2 = 36 = 6 units . BC = ( 0 − 0 ) 2 + ( 0 − 16 ) 2 = 0 2 + ( − 16 ) 2 = 256 = 16 units .
⇒ BC = 16 units
⇒ 3 8 B C = 16 × 3 8 \dfrac{3}{8}BC = 16 \times \dfrac{3}{8} 8 3 BC = 16 × 8 3
Hence, proved that PQ = 3 8 \dfrac{3}{8} 8 3
A line segment joining A ( − 1 , 5 3 ) \Big(-1, \dfrac{5}{3}\Big) ( − 1 , 3 5 )
(i) Calculate the value of 'a'.
(ii) Calculate the co-ordinates of 'P'.
Answer
(i) Since, P is the point where the line segment AB intersects the y-axis.
Let P = (0, y).
Since, P divides AB in the ratio 1 : 3.
∴ 0 = m 1 x 2 + m 2 x 1 m 1 + m 2 ⇒ 0 = 1 × a + 3 × − 1 1 + 3 ⇒ 0 = a − 3 4 ⇒ 0 = a − 3 ⇒ a = 3. \therefore 0 = \dfrac{m_1x_2 + m_2x_1}{m_1 + m_2} \\[1em] \Rightarrow 0 = \dfrac{1 \times a + 3 \times -1}{1 + 3} \\[1em] \Rightarrow 0 = \dfrac{a - 3}{4} \\[1em] \Rightarrow 0 = a - 3 \\[1em] \Rightarrow a = 3. \\[1em] ∴ 0 = m 1 + m 2 m 1 x 2 + m 2 x 1 ⇒ 0 = 1 + 3 1 × a + 3 × − 1 ⇒ 0 = 4 a − 3 ⇒ 0 = a − 3 ⇒ a = 3.
Hence, a = 3.
(ii) By section formula,
y = m 1 y 2 + m 2 y 1 m 1 + m 2 = 1 × 5 + 3 × 5 3 1 + 3 = 5 + 5 4 = 10 4 = 5 2 = 2 1 2 . y = \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2} \\[1em] = \dfrac{1 \times 5 + 3 \times \dfrac{5}{3}}{1 + 3} \\[1em] = \dfrac{5 + 5}{4} \\[1em] = \dfrac{10}{4} = \dfrac{5}{2} = 2\dfrac{1}{2}. y = m 1 + m 2 m 1 y 2 + m 2 y 1 = 1 + 3 1 × 5 + 3 × 3 5 = 4 5 + 5 = 4 10 = 2 5 = 2 2 1 .
P = (0, y) = ( 0 , 2 1 2 ) \Big(0, 2\dfrac{1}{2}\Big) ( 0 , 2 2 1 )
Hence, co-ordinates of P = ( 0 , 2 1 2 ) \Big(0, 2\dfrac{1}{2}\Big) ( 0 , 2 2 1 )
In what ratio is the line joining A(0, 3) and B (4, -1) divided by the x-axis ?
Write the co-ordinates of the point where AB intersects the x-axis.
Answer
Let AB intersect x-axis at P. So, co-ordinates of P = (x, 0).
Let ratio be m1 : m2 .
By section formula,
y = m 1 y 2 + m 2 y 1 m 1 + m 2 ⇒ 0 = m 1 × − 1 + m 2 × 3 m 1 + m 2 ⇒ 0 = − m 1 + 3 m 2 ⇒ m 1 = 3 m 2 ⇒ m 1 m 2 = 3 1 . y = \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2} \\[1em] \Rightarrow 0 = \dfrac{m_1 \times -1 + m_2 \times 3}{m_1 + m_2} \\[1em] \Rightarrow 0 = -m_1 + 3m_2 \\[1em] \Rightarrow m_1 = 3m_2 \\[1em] \Rightarrow \dfrac{m_1}{m_2} = \dfrac{3}{1}. y = m 1 + m 2 m 1 y 2 + m 2 y 1 ⇒ 0 = m 1 + m 2 m 1 × − 1 + m 2 × 3 ⇒ 0 = − m 1 + 3 m 2 ⇒ m 1 = 3 m 2 ⇒ m 2 m 1 = 1 3 .
m1 : m2 = 3 : 1.
x = m 1 x 2 + m 2 x 1 m 1 + m 2 = 3 × 4 + 1 × 0 3 + 1 = 12 + 0 4 = 12 4 = 3. x = \dfrac{m_1x_2 + m_2x_1}{m_1 + m_2} \\[1em] = \dfrac{3 \times 4 + 1 \times 0}{3 + 1} \\[1em] = \dfrac{12 + 0}{4} \\[1em] = \dfrac{12}{4} = 3. \\[1em] x = m 1 + m 2 m 1 x 2 + m 2 x 1 = 3 + 1 3 × 4 + 1 × 0 = 4 12 + 0 = 4 12 = 3.
P = (x, 0) = (3, 0).
Hence, co-ordinates of P = (3, 0) and ratio = 3 : 1.
The mid-point of the segment AB, as shown in diagram, is C(4, -3). Write down the co-ordinates of A and B.
Answer
Since, A lies on x-axis, its co-ordinates be (x, 0) and B lies on y-axis , its co-ordinates be (0, y).
By formula,
Mid-point (M) = ( x 1 + x 2 2 , y 1 + y 2 2 ) \Big(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\Big) ( 2 x 1 + x 2 , 2 y 1 + y 2 )
Substituting values we get,
⇒ ( 4 , − 3 ) = ( x + 0 2 , 0 + y 2 ) ∴ 4 = x 2 and − 3 = y 2 ⇒ x = 8 and y = − 6. \Rightarrow (4, -3) = \Big(\dfrac{x + 0}{2}, \dfrac{0 + y}{2}\Big) \\[1em] \therefore 4 = \dfrac{x}{2} \text{ and } -3 = \dfrac{y}{2} \\[1em] \Rightarrow x = 8 \text{ and } y = -6. ⇒ ( 4 , − 3 ) = ( 2 x + 0 , 2 0 + y ) ∴ 4 = 2 x and − 3 = 2 y ⇒ x = 8 and y = − 6.
Hence, A = (8, 0) and B = (0, -6).
AB is a diameter of a circle with center C = (-2, 5). If A = (3, -7), find
(i) the length of radius AC.
(ii) the co-ordinates of B.
Answer
Distance between two points = ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2
A C = [ 3 − ( − 2 ) ] 2 + [ − 7 − 5 ] 2 = ( 3 + 2 ) 2 + ( − 12 ) 2 = 25 + 144 = 169 = 13 units . AC = \sqrt{[3 - (-2)]^2 + [-7 - 5]^2} \\[1em] = \sqrt{(3 + 2)^2 + (-12)^2} \\[1em] = \sqrt{25 + 144} \\[1em] = \sqrt{169} \\[1em] = 13 \text{ units}. A C = [ 3 − ( − 2 ) ] 2 + [ − 7 − 5 ] 2 = ( 3 + 2 ) 2 + ( − 12 ) 2 = 25 + 144 = 169 = 13 units .
Hence, radius = 13 units.
(ii) Since, AB is diameter and C is center of circle so,
C is the mid-point of AB.
Let B = (x, y).
By formula,
Mid-point (M) = ( x 1 + x 2 2 , y 1 + y 2 2 ) \Big(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\Big) ( 2 x 1 + x 2 , 2 y 1 + y 2 )
Substituting values we get,
⇒ ( − 2 , 5 ) = ( 3 + x 2 , − 7 + y 2 ) ∴ − 2 = 3 + x 2 and 5 = − 7 + y 2 ⇒ x + 3 = − 4 and − 7 + y = 10 ⇒ x = − 7 and y = 17. \Rightarrow (-2, 5) = \Big(\dfrac{3 + x}{2}, \dfrac{-7 + y}{2}\Big) \\[1em] \therefore -2 = \dfrac{3 + x}{2} \text{ and } 5 = \dfrac{-7 + y}{2} \\[1em] \Rightarrow x + 3 = -4 \text{ and } -7 + y = 10 \\[1em] \Rightarrow x = -7 \text{ and } y = 17. ⇒ ( − 2 , 5 ) = ( 2 3 + x , 2 − 7 + y ) ∴ − 2 = 2 3 + x and 5 = 2 − 7 + y ⇒ x + 3 = − 4 and − 7 + y = 10 ⇒ x = − 7 and y = 17.
Hence, co-ordinates of B = (-7, 17).
The mid-point of the line segment joining (4a, 2b - 3) and (-4, 3b) is (2, -2a). Find the values of a and b.
Answer
By formula,
Mid-point (M) = ( x 1 + x 2 2 , y 1 + y 2 2 ) \Big(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\Big) ( 2 x 1 + x 2 , 2 y 1 + y 2 )
Substituting values we get,
⇒ ( 2 , − 2 a ) = ( 4 a + ( − 4 ) 2 , 2 b − 3 + 3 b 2 ) ∴ 2 = 4 a − 4 2 and − 2 a = 5 b − 3 2 ⇒ 4 a − 4 = 4 and 5 b − 3 = − 4 a ⇒ 4 a = 8 and 5 b = − 4 a + 3 ⇒ a = 2 and 5 b = − 4 ( 2 ) + 3 ⇒ a = 2 and 5 b = − 5 ⇒ a = 2 and b = − 1. \Rightarrow (2, -2a) = \Big(\dfrac{4a + (-4)}{2}, \dfrac{2b - 3 + 3b}{2}\Big) \\[1em] \therefore 2 = \dfrac{4a - 4}{2} \text{ and } -2a = \dfrac{5b - 3}{2} \\[1em] \Rightarrow 4a - 4 = 4 \text{ and } 5b - 3 = -4a \\[1em] \Rightarrow 4a = 8 \text{ and } 5b = -4a + 3 \\[1em] \Rightarrow a = 2 \text{ and } 5b = -4(2) + 3 \\[1em] \Rightarrow a = 2 \text{ and } 5b = -5 \\[1em] \Rightarrow a = 2 \text{ and } b = -1. ⇒ ( 2 , − 2 a ) = ( 2 4 a + ( − 4 ) , 2 2 b − 3 + 3 b ) ∴ 2 = 2 4 a − 4 and − 2 a = 2 5 b − 3 ⇒ 4 a − 4 = 4 and 5 b − 3 = − 4 a ⇒ 4 a = 8 and 5 b = − 4 a + 3 ⇒ a = 2 and 5 b = − 4 ( 2 ) + 3 ⇒ a = 2 and 5 b = − 5 ⇒ a = 2 and b = − 1.
Hence, a = 2 and b = -1.
(i) Write down the co-ordinates of the point P that divides the line joining A(-4, 1) and B(17, 10) in the ratio 1 : 2.
(ii) Calculate the distance OP, where O is the origin.
(iii) In what ratio does the y-axis divide the line AB ?
Answer
(i) Let the co-ordinates of P be (x, y)
∴ x = m 1 x 2 + m 2 x 1 m 1 + m 2 = 1 × 17 + 2 × − 4 1 + 2 = 17 + ( − 8 ) 3 = 9 3 = 3. \therefore x = \dfrac{m_1x_2 + m_2x_1}{m_1 + m_2} \\[1em] = \dfrac{1 \times 17 + 2 \times -4}{1 + 2} \\[1em] = \dfrac{17 + (-8)}{3} \\[1em] = \dfrac{9}{3} = 3. ∴ x = m 1 + m 2 m 1 x 2 + m 2 x 1 = 1 + 2 1 × 17 + 2 × − 4 = 3 17 + ( − 8 ) = 3 9 = 3.
and,
y = m 1 y 2 + m 2 y 1 m 1 + m 2 = 1 × 10 + 2 × 1 1 + 2 = 10 + 2 3 = 12 3 = 4. y = \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2} \\[1em] = \dfrac{1 \times 10 + 2 \times 1}{1 + 2} \\[1em] = \dfrac{10 + 2}{3} \\[1em] = \dfrac{12}{3} = 4. y = m 1 + m 2 m 1 y 2 + m 2 y 1 = 1 + 2 1 × 10 + 2 × 1 = 3 10 + 2 = 3 12 = 4.
P = (x, y) = (3, 4).
Hence, co-ordinates of point P = (3, 4).
(ii) Distance between two points = ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2
O P = ( 3 − 0 ) 2 + ( 4 − 0 ) 2 = ( 3 ) 2 + ( 4 ) 2 = 9 + 16 = 25 = 5 units . OP = \sqrt{(3 - 0)^2 + (4 - 0)^2} \\[1em] = \sqrt{(3)^2 + (4)^2} \\[1em] = \sqrt{9 + 16} \\[1em] = \sqrt{25} \\[1em] = 5 \text{ units}. OP = ( 3 − 0 ) 2 + ( 4 − 0 ) 2 = ( 3 ) 2 + ( 4 ) 2 = 9 + 16 = 25 = 5 units .
Hence, OP = 5 units.
(iii) Let point Q (0, z) on y-axis divide line AB in ratio m1 : m2 .
By section formula,
x = m 1 x 2 + m 2 x 1 m 1 + m 2 x = \dfrac{m_1x_2 + m_2x_1}{m_1 + m_2} x = m 1 + m 2 m 1 x 2 + m 2 x 1
Substituting values we get,
⇒ 0 = m 1 × 17 + m 2 × − 4 m 1 + m 2 ⇒ 0 = 17 m 1 − 4 m 2 ⇒ 4 m 2 = 17 m 1 ⇒ m 1 m 2 = 4 17 . \Rightarrow 0 = \dfrac{m_1 \times 17 + m_2 \times -4}{m_1 + m_2} \\[1em] \Rightarrow 0 = 17m_1 - 4m_2 \\[1em] \Rightarrow 4m_2 = 17m_1 \\[1em] \Rightarrow \dfrac{m_1}{m_2} = \dfrac{4}{17}. ⇒ 0 = m 1 + m 2 m 1 × 17 + m 2 × − 4 ⇒ 0 = 17 m 1 − 4 m 2 ⇒ 4 m 2 = 17 m 1 ⇒ m 2 m 1 = 17 4 .
m1 : m2 = 4 : 17.
Hence, ratio in which the y-axis divide the line AB = 4 : 17.
Prove that the points A(-5, 4); B(-1, -2) and C(5, 2) are the vertices of an isosceles right angled triangle. Find the co-ordinates of D so that ABCD is a square.
Answer
The points are shown in the figure below:
Distance between two points = ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2
A B = [ − 1 − ( − 5 ) ] 2 + [ − 2 − 4 ] 2 = [ − 1 + 5 ] 2 + [ − 6 ] 2 = [ 4 ] 2 + 36 = 16 + 36 = 52 . B C = [ 5 − ( − 1 ) ] 2 + [ 2 − ( − 2 ) ] 2 = [ 5 + 1 ] 2 + [ 4 ] 2 = [ 6 ] 2 + 16 = 36 + 16 = 52 . A C = [ 5 − ( − 5 ) ] 2 + [ 2 − 4 ] 2 = [ 5 + 5 ] 2 + [ − 2 ] 2 = [ 10 ] 2 + 4 = 100 + 4 = 104 . A B 2 + B C 2 = ( 52 ) 2 + ( 52 ) 2 = 52 + 52 = 104 = A C 2 . AB = \sqrt{[-1 - (-5)]^2 + [-2 - 4]^2} \\[1em] = \sqrt{[-1 + 5]^2 + [-6]^2} \\[1em] = \sqrt{[4]^2 + 36} \\[1em] = \sqrt{16 + 36} \\[1em] = \sqrt{52}. \\[1em] BC = \sqrt{[5 - (-1)]^2 + [2 - (-2)]^2} \\[1em] = \sqrt{[5 + 1]^2 + [4]^2} \\[1em] = \sqrt{[6]^2 + 16} \\[1em] = \sqrt{36 + 16} \\[1em] = \sqrt{52}. \\[1em] AC = \sqrt{[5 - (-5)]^2 + [2 - 4]^2} \\[1em] = \sqrt{[5 + 5]^2 + [-2]^2} \\[1em] = \sqrt{[10]^2 + 4} \\[1em] = \sqrt{100 + 4} \\[1em] = \sqrt{104}. \\[1em] AB^2 + BC^2 = (\sqrt{52})^2 + (\sqrt{52})^2 \\[1em] = 52 + 52 \\[1em] = 104 = AC^2. A B = [ − 1 − ( − 5 ) ] 2 + [ − 2 − 4 ] 2 = [ − 1 + 5 ] 2 + [ − 6 ] 2 = [ 4 ] 2 + 36 = 16 + 36 = 52 . BC = [ 5 − ( − 1 ) ] 2 + [ 2 − ( − 2 ) ] 2 = [ 5 + 1 ] 2 + [ 4 ] 2 = [ 6 ] 2 + 16 = 36 + 16 = 52 . A C = [ 5 − ( − 5 ) ] 2 + [ 2 − 4 ] 2 = [ 5 + 5 ] 2 + [ − 2 ] 2 = [ 10 ] 2 + 4 = 100 + 4 = 104 . A B 2 + B C 2 = ( 52 ) 2 + ( 52 ) 2 = 52 + 52 = 104 = A C 2 .
Since, AB = BC and AC2 = AB2 + BC2 .
Hence, proved that ABC is an isosceles right angled triangle.
Since, diagonals of square bisect each other so,
Mid-point of AC = Mid-point of BD = O.
O = ( − 5 + 5 2 , 4 + 2 2 ) = ( 0 2 , 6 2 ) = ( 0 , 3 ) . O = \Big(\dfrac{-5 + 5}{2}, \dfrac{4 + 2}{2}\Big) \\[1em] = \Big(\dfrac{0}{2}, \dfrac{6}{2}\Big) \\[1em] = (0, 3). O = ( 2 − 5 + 5 , 2 4 + 2 ) = ( 2 0 , 2 6 ) = ( 0 , 3 ) .
Let co-ordinates of D = (x, y).
∴ O = ( − 1 + x 2 , − 2 + y 2 ) ⇒ ( 0 , 3 ) = ( − 1 + x 2 , − 2 + y 2 ) ⇒ 0 = − 1 + x 2 and 3 = − 2 + y 2 ⇒ − 1 + x = 0 and − 2 + y = 6 ⇒ x = 1 and y = 8. \therefore O = \Big(\dfrac{-1 + x}{2}, \dfrac{-2 + y}{2}\Big) \\[1em] \Rightarrow (0, 3) = \Big(\dfrac{-1 + x}{2}, \dfrac{-2 + y}{2}\Big) \\[1em] \Rightarrow 0 = \dfrac{-1 + x}{2} \text{ and } 3 = \dfrac{-2 + y}{2} \\[1em] \Rightarrow -1 + x = 0 \text{ and } -2 + y = 6 \\[1em] \Rightarrow x = 1 \text{ and } y = 8. ∴ O = ( 2 − 1 + x , 2 − 2 + y ) ⇒ ( 0 , 3 ) = ( 2 − 1 + x , 2 − 2 + y ) ⇒ 0 = 2 − 1 + x and 3 = 2 − 2 + y ⇒ − 1 + x = 0 and − 2 + y = 6 ⇒ x = 1 and y = 8.
D = (x, y) = (1, 8).
Hence, co-ordinates of D = (1, 8).
M is the mid-point of the line segment joining the points A(-3, 7) and B(9, -1). Find the co-ordinates of point M. Further, if R(2, 2) divides the line segment joining M and the origin in the ratio p : q, find the ratio p : q.
Answer
By formula,
Mid-point (M) = ( x 1 + x 2 2 , y 1 + y 2 2 ) \Big(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\Big) ( 2 x 1 + x 2 , 2 y 1 + y 2 )
Substituting values we get,
M = ( − 3 + 9 2 , 7 + ( − 1 ) 2 ) = ( 6 2 , 6 2 ) = ( 3 , 3 ) . M = \Big(\dfrac{-3 + 9}{2}, \dfrac{7 + (-1)}{2}\Big) \\[1em] = \Big(\dfrac{6}{2}, \dfrac{6}{2}\Big) \\[1em] = (3, 3). M = ( 2 − 3 + 9 , 2 7 + ( − 1 ) ) = ( 2 6 , 2 6 ) = ( 3 , 3 ) .
Given, R(2, 2) divides the line segment joining M and the origin in the ratio p : q.
By section formula,
⇒ x = m 1 x 2 + m 2 x 1 m 1 + m 2 \Rightarrow x = \dfrac{m_1x_2 + m_2x_1}{m_1 + m_2} ⇒ x = m 1 + m 2 m 1 x 2 + m 2 x 1
Substituting values we get,
⇒ 2 = p × 0 + q × 3 p + q ⇒ 2 ( p + q ) = 0 + 3 q ⇒ 2 p + 2 q = 3 q ⇒ 2 p = q ⇒ p q = 1 2 . \Rightarrow 2 = \dfrac{p \times 0 + q \times 3}{p + q} \\[1em] \Rightarrow 2(p + q) = 0 + 3q \\[1em] \Rightarrow 2p + 2q = 3q \\[1em] \Rightarrow 2p = q \\[1em] \Rightarrow \dfrac{p}{q} = \dfrac{1}{2}. ⇒ 2 = p + q p × 0 + q × 3 ⇒ 2 ( p + q ) = 0 + 3 q ⇒ 2 p + 2 q = 3 q ⇒ 2 p = q ⇒ q p = 2 1 .
p : q = 1 : 2.
Hence, M = (3, 3) and p : q = 1 : 2.
Calculate the ratio in which the line joining A(-4, 2) and B(3, 6) is divided by point P(x, 3). Also, find (i) x (ii) length of AP.
Answer
Let ratio be m1 : m2 .
By section formula,
y = m 1 y 2 + m 2 y 1 m 1 + m 2 y = \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2} y = m 1 + m 2 m 1 y 2 + m 2 y 1
Substituting values we get,
3 = m 1 × 6 + m 2 × 2 m 1 + m 2 ⇒ 3 m 1 + 3 m 2 = 6 m 1 + 2 m 2 ⇒ m 2 = 3 m 1 ⇒ m 1 m 2 = 1 3 . 3 = \dfrac{m_1 \times 6 + m_2 \times 2}{m_1 + m_2} \\[1em] \Rightarrow 3m_1 + 3m_2 = 6m_1 + 2m_2 \\[1em] \Rightarrow m_2 = 3m_1 \\[1em] \Rightarrow \dfrac{m_1}{m_2} = \dfrac{1}{3}. 3 = m 1 + m 2 m 1 × 6 + m 2 × 2 ⇒ 3 m 1 + 3 m 2 = 6 m 1 + 2 m 2 ⇒ m 2 = 3 m 1 ⇒ m 2 m 1 = 3 1 .
m1 : m2 = 1 : 3.
(i) By section formula,
x = m 1 x 2 + m 2 x 1 m 1 + m 2 x = \dfrac{m_1x_2 + m_2x_1}{m_1 + m_2} x = m 1 + m 2 m 1 x 2 + m 2 x 1
Substituting values we get,
⇒ x = 1 × 3 + 3 × − 4 1 + 3 ⇒ x = 3 − 12 4 ⇒ x = − 9 4 . \Rightarrow x = \dfrac{1 \times 3 + 3 \times -4}{1 + 3} \\[1em] \Rightarrow x = \dfrac{3 - 12}{4} \\[1em] \Rightarrow x = -\dfrac{9}{4}. ⇒ x = 1 + 3 1 × 3 + 3 × − 4 ⇒ x = 4 3 − 12 ⇒ x = − 4 9 .
Hence, x = − 9 4 -\dfrac{9}{4} − 4 9
(ii) Distance between two points = ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2
A P = [ − 9 4 − ( − 4 ) ] 2 + [ 3 − 2 ] 2 = [ − 9 4 + 4 ] 2 + [ 1 ] 2 = [ − 9 + 16 4 ] 2 + 1 = [ 7 4 ] 2 + 1 = 49 16 + 1 = 49 + 16 16 = 65 16 = 65 4 . AP = \sqrt{\Big[-\dfrac{9}{4} - (-4)\Big]^2 + [3 - 2]^2} \\[1em] = \sqrt{\Big[-\dfrac{9}{4} + 4\Big]^2 + [1]^2} \\[1em] = \sqrt{\Big[\dfrac{-9 + 16}{4}\Big]^2 + 1} \\[1em] = \sqrt{\Big[\dfrac{7}{4}\Big]^2 + 1} \\[1em] = \sqrt{\dfrac{49}{16} + 1} \\[1em] = \sqrt{\dfrac{49 + 16}{16}} \\[1em] = \sqrt{\dfrac{65}{16}} \\[1em] = \dfrac{\sqrt{65}}{4}. A P = [ − 4 9 − ( − 4 ) ] 2 + [ 3 − 2 ] 2 = [ − 4 9 + 4 ] 2 + [ 1 ] 2 = [ 4 − 9 + 16 ] 2 + 1 = [ 4 7 ] 2 + 1 = 16 49 + 1 = 16 49 + 16 = 16 65 = 4 65 .
Hence, AP = 65 4 . \dfrac{\sqrt{65}}{4}. 4 65 .
Find the ratio in which the line 2x + y = 4 divides the line segment joining the points P(2, -2) and Q(3, 7).
Answer
Let ratio in which 2x + y = 4 divides the line segment joining the points P(2, -2) and Q(3, 7) be k : 1 at point (x, y).
By section formula,
x = m 1 x 2 + m 2 x 1 m 1 + m 2 x = \dfrac{m_1x_2 + m_2x_1}{m_1 + m_2} x = m 1 + m 2 m 1 x 2 + m 2 x 1
Substituting values we get,
⇒ x = k × 3 + 1 × 2 k + 1 ⇒ x ( k + 1 ) = 3 k + 2 ⇒ x = 3 k + 2 k + 1 . \Rightarrow x = \dfrac{k \times 3 + 1 \times 2}{k + 1} \\[1em] \Rightarrow x(k + 1) = 3k + 2 \\[1em] \Rightarrow x = \dfrac{3k + 2}{k + 1}. ⇒ x = k + 1 k × 3 + 1 × 2 ⇒ x ( k + 1 ) = 3 k + 2 ⇒ x = k + 1 3 k + 2 .
y = m 1 y 2 + m 2 y 1 m 1 + m 2 y = \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2} y = m 1 + m 2 m 1 y 2 + m 2 y 1
Substituting values we get,
⇒ y = k × 7 + 1 × − 2 k + 1 ⇒ y ( k + 1 ) = 7 k − 2 ⇒ y = 7 k − 2 k + 1 . \Rightarrow y = \dfrac{k \times 7 + 1 \times -2}{k + 1} \\[1em] \Rightarrow y(k + 1) = 7k - 2 \\[1em] \Rightarrow y = \dfrac{7k - 2}{k + 1}. ⇒ y = k + 1 k × 7 + 1 × − 2 ⇒ y ( k + 1 ) = 7 k − 2 ⇒ y = k + 1 7 k − 2 .
Substituting value of x and y in 2x + y = 4.
⇒ 2 ( 3 k + 2 k + 1 ) + 7 k − 2 k + 1 = 4 ⇒ 6 k + 4 k + 1 + 7 k − 2 k + 1 = 4 ⇒ 6 k + 4 + 7 k − 2 k + 1 = 4 ⇒ 13 k + 2 = 4 k + 4 ⇒ 9 k = 2 ⇒ k = 2 9 . ⇒ k : 1 = 2 9 : 1 = 2 : 9. \Rightarrow 2 \Big(\dfrac{3k + 2}{k + 1}\Big) + \dfrac{7k - 2}{k + 1} = 4 \\[1em] \Rightarrow \dfrac{6k + 4}{k + 1} + \dfrac{7k - 2}{k + 1} = 4 \\[1em] \Rightarrow \dfrac{6k + 4 + 7k - 2}{k + 1} = 4 \\[1em] \Rightarrow 13k + 2 = 4k + 4 \\[1em] \Rightarrow 9k = 2 \\[1em] \Rightarrow k = \dfrac{2}{9}. \\[1em] \Rightarrow k : 1 = \dfrac{2}{9} : 1 = 2 : 9. ⇒ 2 ( k + 1 3 k + 2 ) + k + 1 7 k − 2 = 4 ⇒ k + 1 6 k + 4 + k + 1 7 k − 2 = 4 ⇒ k + 1 6 k + 4 + 7 k − 2 = 4 ⇒ 13 k + 2 = 4 k + 4 ⇒ 9 k = 2 ⇒ k = 9 2 . ⇒ k : 1 = 9 2 : 1 = 2 : 9.
Hence, ratio in which the line 2x + y = 4 divides the line segment joining the points P(2, -2) and Q(3, 7) = 2 : 9.
If the abscissa of a point P is 2, find the ratio in which this point divides the line segment joining the points (-4, 3) and (6, 3). Also, find the co-ordinates of point P.
Answer
Let point P be (2, y) and ratio in which it divides line segment joining the points (-4, 3) and (6, 3) be m1 : m2 .
By section formula,
x = m 1 x 2 + m 2 x 1 m 1 + m 2 x = \dfrac{m_1x_2 + m_2x_1}{m_1 + m_2} x = m 1 + m 2 m 1 x 2 + m 2 x 1
Substituting values we get,
⇒ 2 = m 1 × 6 + m 2 × − 4 m 1 + m 2 ⇒ 2 m 1 + 2 m 2 = 6 m 1 − 4 m 2 ⇒ 2 m 2 + 4 m 2 = 6 m 1 − 2 m 1 ⇒ 6 m 2 = 4 m 1 ⇒ m 1 m 2 = 6 4 = 3 2 . \Rightarrow 2 = \dfrac{m_1 \times 6 + m_2 \times -4}{m_1 + m_2} \\[1em] \Rightarrow 2m_1 + 2m_2 = 6m_1 - 4m_2 \\[1em] \Rightarrow 2m_2 + 4m_2 = 6m_1 - 2m_1 \\[1em] \Rightarrow 6m_2 = 4m_1 \\[1em] \Rightarrow \dfrac{m_1}{m_2} = \dfrac{6}{4} = \dfrac{3}{2}. ⇒ 2 = m 1 + m 2 m 1 × 6 + m 2 × − 4 ⇒ 2 m 1 + 2 m 2 = 6 m 1 − 4 m 2 ⇒ 2 m 2 + 4 m 2 = 6 m 1 − 2 m 1 ⇒ 6 m 2 = 4 m 1 ⇒ m 2 m 1 = 4 6 = 2 3 .
m1 : m2 = 3 : 2.
y = m 1 y 2 + m 2 y 1 m 1 + m 2 y = \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2} y = m 1 + m 2 m 1 y 2 + m 2 y 1
Substituting values we get,
⇒ y = 3 × 3 + 2 × 3 3 + 2 ⇒ y = 9 + 6 5 ⇒ y = 15 5 = 3. \Rightarrow y = \dfrac{3 \times 3 + 2 \times 3}{3 + 2} \\[1em] \Rightarrow y = \dfrac{9 + 6}{5} \\[1em] \Rightarrow y = \dfrac{15}{5} = 3. ⇒ y = 3 + 2 3 × 3 + 2 × 3 ⇒ y = 5 9 + 6 ⇒ y = 5 15 = 3.
P = (2, y) = (2, 3).
Hence, ratio = 3 : 2 and co-ordinates of P = (2, 3).
Find the image of the point A(5, -3) under reflection in the point P(-1, 3).
Answer
Let image be B(x, y).
Since, A is reflected in P to become B. So, P is mid-point of AB.
By formula,
Mid-point (M) = ( x 1 + x 2 2 , y 1 + y 2 2 ) \Big(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\Big) ( 2 x 1 + x 2 , 2 y 1 + y 2 )
Substituting values we get,
⇒ P = ( 5 + x 2 , − 3 + y 2 ) ⇒ ( − 1 , 3 ) = ( 5 + x 2 , − 3 + y 2 ) ∴ − 1 = 5 + x 2 and 3 = − 3 + y 2 ⇒ x + 5 = − 2 and y − 3 = 6 ⇒ x = − 7 and y = 9. \Rightarrow P = \Big(\dfrac{5 + x}{2}, \dfrac{-3 + y}{2}\Big) \\[1em] \Rightarrow (-1, 3) = \Big(\dfrac{5 + x}{2}, \dfrac{-3 + y}{2}\Big) \\[1em] \therefore -1 = \dfrac{5 + x}{2} \text{ and } 3 = \dfrac{-3 + y}{2} \\[1em] \Rightarrow x + 5 = -2 \text{ and } y - 3 = 6 \\[1em] \Rightarrow x = -7 \text{ and } y = 9. ⇒ P = ( 2 5 + x , 2 − 3 + y ) ⇒ ( − 1 , 3 ) = ( 2 5 + x , 2 − 3 + y ) ∴ − 1 = 2 5 + x and 3 = 2 − 3 + y ⇒ x + 5 = − 2 and y − 3 = 6 ⇒ x = − 7 and y = 9.
B = (x, y) = (-7, 9).
Hence, image of the point A(5, -3) under reflection in the point P(-1, 3) is (-7, 9).
M is the mid-point of the line segment joining the points A(0, 4) and B(6, 0). M also divides the line segment OP in the ratio 1 : 3. Find :
(i) co-ordinates of M
(ii) co-ordinates of P
(iii) length of BP
Answer
(i) By formula,
Mid-point (M) = ( x 1 + x 2 2 , y 1 + y 2 2 ) \Big(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\Big) ( 2 x 1 + x 2 , 2 y 1 + y 2 )
Substituting values we get,
M = ( 0 + 6 2 , 4 + 0 2 ) = ( 6 2 , 4 2 ) = ( 3 , 2 ) . M = \Big(\dfrac{0 + 6}{2}, \dfrac{4 + 0}{2}\Big) \\[1em] = \Big(\dfrac{6}{2}, \dfrac{4}{2}\Big) \\[1em] = (3, 2). M = ( 2 0 + 6 , 2 4 + 0 ) = ( 2 6 , 2 4 ) = ( 3 , 2 ) .
Hence, M = (3, 2).
(ii) Let co-ordinates of P be (x, y).
Given, M divides the line segment OP in the ratio 1 : 3.
By section formula,
x = m 1 x 2 + m 2 x 1 m 1 + m 2 x = \dfrac{m_1x_2 + m_2x_1}{m_1 + m_2} x = m 1 + m 2 m 1 x 2 + m 2 x 1
Substituting values we get,
⇒ 3 = 1 × x + 3 × 0 1 + 3 ⇒ 3 = x 4 ⇒ x = 12. y = m 1 y 2 + m 2 y 1 m 1 + m 2 \Rightarrow 3 = \dfrac{1 \times x + 3 \times 0}{1 + 3} \\[1em] \Rightarrow 3 = \dfrac{x}{4} \\[1em] \Rightarrow x = 12. \\[1em] y = \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2} ⇒ 3 = 1 + 3 1 × x + 3 × 0 ⇒ 3 = 4 x ⇒ x = 12. y = m 1 + m 2 m 1 y 2 + m 2 y 1
Substituting values we get,
⇒ 2 = 1 × y + 3 × 0 1 + 3 ⇒ 2 = y 4 ⇒ y = 8. \Rightarrow 2 = \dfrac{1 \times y + 3 \times 0}{1 + 3} \\[1em] \Rightarrow 2 = \dfrac{y}{4} \\[1em] \Rightarrow y = 8. ⇒ 2 = 1 + 3 1 × y + 3 × 0 ⇒ 2 = 4 y ⇒ y = 8.
P = (x, y) = (12, 8).
Hence, co-ordinates of P = (12, 8).
(iii) Distance between two points = ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2
B P = ( 12 − 6 ) 2 + ( 8 − 0 ) 2 = ( 6 ) 2 + ( 8 ) 2 = 36 + 64 = 100 = 10 units . BP = \sqrt{(12 - 6)^2 + (8 - 0)^2} \\[1em] = \sqrt{(6)^2 + (8)^2} \\[1em] = \sqrt{36 + 64} \\[1em] = \sqrt{100} = 10 \text{ units}. \\[1em] BP = ( 12 − 6 ) 2 + ( 8 − 0 ) 2 = ( 6 ) 2 + ( 8 ) 2 = 36 + 64 = 100 = 10 units .
Hence, BP = 10 units.
A(3, 1), B(y, 4) and C(1, x) are vertices of triangle ABC and G(3, 4) is its centroid. Find the values of x and y. Also, find the length of side BC.
Answer
Centroid of the triangle is given by (G) = ( x 1 + x 2 + x 3 3 , y 1 + y 2 + y 3 3 ) \Big(\dfrac{x_1 + x_2 + x_3}{3}, \dfrac{y_1 + y_2 + y_3}{3}\Big) ( 3 x 1 + x 2 + x 3 , 3 y 1 + y 2 + y 3 )
Substituting values we get,
⇒ G 1 = ( 3 + y + 1 3 , 1 + 4 + x 3 ) ⇒ ( 3 , 4 ) = ( y + 4 3 , x + 5 3 ) ⇒ 3 = y + 4 3 and 4 = x + 5 3 ⇒ y + 4 = 9 and 12 = x + 5 ⇒ y = 5 and x = 7. \Rightarrow G_1 = \Big(\dfrac{3 + y + 1}{3}, \dfrac{1 + 4 + x}{3}\Big) \\[1em] \Rightarrow (3, 4) = \Big(\dfrac{y + 4}{3}, \dfrac{x + 5}{3}\Big) \\[1em] \Rightarrow 3 = \dfrac{y + 4}{3} \text{ and } 4 = \dfrac{x + 5}{3} \\[1em] \Rightarrow y + 4 = 9 \text{ and } 12 = x + 5 \\[1em] \Rightarrow y = 5 \text{ and } x = 7. ⇒ G 1 = ( 3 3 + y + 1 , 3 1 + 4 + x ) ⇒ ( 3 , 4 ) = ( 3 y + 4 , 3 x + 5 ) ⇒ 3 = 3 y + 4 and 4 = 3 x + 5 ⇒ y + 4 = 9 and 12 = x + 5 ⇒ y = 5 and x = 7.
B = (y, 4) = (5, 4) and C = (1, x) = (1, 7).
Distance between two points = ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2
B C = ( 1 − 5 ) 2 + ( 7 − 4 ) 2 = ( − 4 ) 2 + ( 3 ) 2 = 16 + 9 = 25 = 5 units . BC = \sqrt{(1 - 5)^2 + (7 - 4)^2} \\[1em] = \sqrt{(-4)^2 + (3)^2} \\[1em] = \sqrt{16 + 9} \\[1em] = \sqrt{25} = 5 \text{ units}. \\[1em] BC = ( 1 − 5 ) 2 + ( 7 − 4 ) 2 = ( − 4 ) 2 + ( 3 ) 2 = 16 + 9 = 25 = 5 units .
Hence, x = 7, y = 5 and BC = 5 units.
