Section & Mid-Point Formula — Exercise 13(A)

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 :

1. $\Big(\dfrac{7}{3}, 5\Big)$

2. $\Big(7, \dfrac{5}{3}\Big)$

3. $\Big(\dfrac{7}{3}, \dfrac{5}{3}\Big)$

4. $\Big(\dfrac{7}{3}, \dfrac{13}{3}\Big)$

Answer

Let point P be (x, y).

Given,

m₁ : m₂ = 1 : 2

By section-formula,

(x, y) = $\Big(\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}, \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big)$

Substituting values we get :

$\Rightarrow (x, y) = \Big(\dfrac{1 \times 5 + 2 \times 1}{1 + 2}, \dfrac{1 \times 9 + 2 \times 3}{1 + 2}\Big) \\[1em] = \Big(\dfrac{5 + 2}{3}, \dfrac{9 + 6}{3}\Big) \\[1em] = \Big(\dfrac{7}{3}, \dfrac{15}{3}\Big) \\[1em] = \Big(\dfrac{7}{3}, 5\Big).$

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 :

1. 3 : 2

2. 2 : 3

3. 3 : 1

4. 1 : 3

Answer

Let ratio in which P divides AB be k : 1.

By section-formula,

(x, y) = $\Big(\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}, \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big)$

Substituting values we get :

$\Rightarrow (3, x) = \Big(\dfrac{k \times 6 + 1 \times 2}{k + 1}, \dfrac{k \times 12 + 1 \times 4}{k + 1}\Big) \\[1em] \Rightarrow (3, x) = \Big(\dfrac{6k + 2}{k + 1}, \dfrac{12k + 4}{k + 1}\Big) \\[1em] \Rightarrow \dfrac{6k + 2}{k + 1} = 3 \\[1em] \Rightarrow 6k + 2 = 3(k + 1) \\[1em] \Rightarrow 6k + 2 = 3k + 3 \\[1em] \Rightarrow 6k - 3k = 3 - 2 \\[1em] \Rightarrow 3k = 1 \\[1em] \Rightarrow k = \dfrac{1}{3}.$

Substituting value of k in k : 1, we get :

⇒ $\dfrac{1}{3} : 1$

⇒ 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 :

1. 3 : 5

2. 5 : 3

3. 1 : 5

4. 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) = $\Big(\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}, \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big)$

Substituting values we get :

$\Rightarrow (7, y) = \Big(\dfrac{k \times 10 + 1 \times 2}{k + 1}, \dfrac{k \times 12 + 1 \times 4}{k + 1}\Big) \\[1em] \Rightarrow (7, y) = \Big(\dfrac{10k + 2}{k + 1}, \dfrac{12k + 4}{k + 1}\Big) \\[1em] \Rightarrow 7 = \dfrac{10k + 2}{k + 1} \\[1em] \Rightarrow 7(k + 1) = 10k + 2 \\[1em] \Rightarrow 7k + 7 = 10k + 2 \\[1em] \Rightarrow 10k - 7k = 7 - 2 \\[1em] \Rightarrow 3k = 5 \\[1em] \Rightarrow k = \dfrac{5}{3}.$

Substituting value of k in k : 1, we get :

⇒ $\dfrac{5}{3} : 1$

⇒ 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 :

1. 3 : 5

2. 5 : 3

3. 1 : 5

4. 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) = $\Big(\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}, \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big)$

Substituting values we get :

$\Rightarrow (x, 4) = \Big(\dfrac{k \times 4 + 1 \times 6}{k + 1}, \dfrac{k \times -1 + 1 \times 7}{k + 1}\Big) \\[1em] \Rightarrow (x, 4) = \Big(\dfrac{4k + 6}{k + 1}, \dfrac{-k + 7}{k + 1}\Big) \\[1em] \Rightarrow 4 = \dfrac{-k + 7}{k + 1} \\[1em] \Rightarrow 4(k + 1) = -k + 7 \\[1em] \Rightarrow 4k + 4 = -k + 7 \\[1em] \Rightarrow 4k + k = 7 - 4 \\[1em] \Rightarrow 5k = 3 \\[1em] \Rightarrow k = \dfrac{3}{5}.$

Substituting value of k in k : 1, we get :

⇒ $\dfrac{3}{5} : 1$

⇒ 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 :

1. 3 : 5

2. 2 : 5

3. 5 : 3

4. 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.

$\Rightarrow (0, y) = \Big(\dfrac{k \times 5 + 1 \times -2}{k + 1}, \dfrac{k \times -2 + 1 \times 5}{k + 1}\Big) \\[1em] \Rightarrow (0, y) = \Big(\dfrac{5k - 2}{k + 1}, \dfrac{-2k + 5}{k + 1}\Big) \\[1em] \Rightarrow 0 = \dfrac{5k - 2}{k + 1} \\[1em] \Rightarrow 5k - 2 = 0 \\[1em] \Rightarrow 5k = 2 \\[1em] \Rightarrow k = \dfrac{2}{5}.$

Substituting value of k in k : 1, we get :

⇒ $\dfrac{2}{5} : 1$

⇒ 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,

$x = \dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}$

Substituting values we get,

$\Rightarrow 1 = \dfrac{m_1 \times 4 + m_2 \times -1}{m_1 + m_2} \\[1em] \Rightarrow m_1 + m_2 = 4m_1 - m_2 \\[1em] \Rightarrow m_2 + m_2 = 4m_1 - m_1 \\[1em] \Rightarrow 2m_2 = 3m_1 \\[1em] \Rightarrow \dfrac{m_1}{m_2} = \dfrac{2}{3}.$

∴ m₁ : m₂ = 2 : 3.

We know that,

$y = \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}$

Substituting values we get,

$\Rightarrow a = \dfrac{2 \times -1 + 3 \times 4}{2 + 3} \\[1em] \Rightarrow a = \dfrac{-2 + 12}{5} \\[1em] \Rightarrow a = \dfrac{10}{5} \\[1em] \Rightarrow a = 2.$

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,

$y = \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}$

Substituting values we get,

$\Rightarrow 6 = \dfrac{m_1 \times 8 + m_2 \times 3}{m_1 + m_2} \\[1em] \Rightarrow 6m_1 + 6m_2 = 8m_1 + 3m_2 \\[1em] \Rightarrow 8m_1 - 6m_1 = 6m_2 - 3m_2 \\[1em] \Rightarrow 2m_1 = 3m_2 \\[1em] \Rightarrow \dfrac{m_1}{m_2} = \dfrac{3}{2}.$

∴ m₁ : m₂ = 3 : 2.

We know that,

$x = \dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}$

Substituting values we get,

$\Rightarrow a = \dfrac{3 \times 2 + 2 \times -4}{3 + 2} \\[1em] \Rightarrow a = \dfrac{6 - 8}{5} \\[1em] \Rightarrow a = -\dfrac{2}{5}.$

Hence, ratio = 3 : 2 and a = $-\dfrac{2}{5}$.

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,

$y = \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}$

Substituting values we get,

$\Rightarrow 0 = \dfrac{k \times -6 + 1 \times 3}{k + 1} \\[1em] \Rightarrow 0 = -6k + 3 \\[1em] \Rightarrow 6k = 3 \\[1em] \Rightarrow k = \dfrac{3}{6} = \dfrac{1}{2}.$

k : 1 = $\dfrac{1}{2} : 1 = 1 : 2.$

$x = \dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}$

Substituting values we get,

$\Rightarrow x = \dfrac{1 \times 2 + 2 \times 4}{1 + 2} \\[1em] \Rightarrow x = \dfrac{2 + 8}{3} \\[1em] \Rightarrow x = \dfrac{10}{3}.$

P = (x, 0) = $\Big(\dfrac{10}{3}, 0\Big).$

Hence, ratio = 1 : 2 and co-ordinates of point of intersection = $\Big(\dfrac{10}{3}, 0\Big).$

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,

$x = \dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}$

Substituting values we get,

$\Rightarrow 0 = \dfrac{k \times 3 + 1 \times -4}{k + 1} \\[1em] \Rightarrow 0 = 3k - 4 \\[1em] \Rightarrow 3k = 4 \\[1em] \Rightarrow k = \dfrac{4}{3}.$

k : 1 = $\dfrac{4}{3} : 1 = 4 : 3$.

$y = \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}$

Substituting values we get,

$\Rightarrow y = \dfrac{4 \times 0 + 3 \times 7}{4 + 3} \\[1em] \Rightarrow y = \dfrac{0 + 21}{7} \\[1em] \Rightarrow y = 3.$

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,

$x = \dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}$

Substituting values we get,

$\Rightarrow a = \dfrac{2 \times 0 + 3 \times 5}{2 + 3} \\[1em] \Rightarrow a = \dfrac{0 + 15}{5} \\[1em] \Rightarrow a = \dfrac{15}{5} = 3.$

$y = \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}$

Substituting values we get,

$\Rightarrow b = \dfrac{2 \times 0 + 3 \times -10}{2 + 3} \\[1em] \Rightarrow b = \dfrac{0 - 30}{5} \\[1em] \Rightarrow b = -\dfrac{30}{5} = -6.$

B = (a, b) = (3, -6).

D divides the line segment PO in the ratio 4 : 1.

Let D be (c, d).

By formula,

$x = \dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}$

Substituting values we get,

$\Rightarrow c = \dfrac{4 \times 0 + 1 \times 5}{4 + 1} \\[1em] \Rightarrow c = \dfrac{0 + 5}{5} \\[1em] \Rightarrow c = \dfrac{5}{5} = 1.$

$y = \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}$

Substituting values we get,

$\Rightarrow d = \dfrac{4 \times 0 + 1 \times -10}{4 + 1} \\[1em] \Rightarrow d = -\dfrac{10}{5} \\[1em] \Rightarrow d = -2.$

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 $\dfrac{PB}{AB} = \dfrac{1}{5}$. Find the co-ordinates of P.

Answer

Let co-ordinates of P be (a, b).

Given,

$\dfrac{PB}{AB} = \dfrac{1}{5}$

Let PB = x and AB = 5x.

From figure,

⇒ AB = PA + PB

⇒ 5x = PA + x

⇒ PA = 4x.

$\dfrac{PA}{PB} = \dfrac{4x}{x} = \dfrac{4}{1}$.

PA : PB = 4 : 1.

∴ P divides the line segment joining A and B in ratio = 4 : 1.

By formula,

$x = \dfrac{m_1x_2 + m_2x_1}{m_1 + m_2} \\[1em] = \dfrac{4 \times -2 + 1 \times -3}{4 + 1} \\[1em] = \dfrac{-8 - 3}{5} \\[1em] = -\dfrac{11}{5} \\[1em] y = \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2} \\[1em] = \dfrac{4 \times 6 + 1 \times -10}{4 + 1} \\[1em] = \dfrac{24 - 10}{5} \\[1em] = \dfrac{14}{5}.$

Hence, P = $\Big(-\dfrac{11}{5}, \dfrac{14}{5}\Big)$.

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

⇒ $\dfrac{AP}{BP} = \dfrac{2}{5}$

⇒ AP : PB = 2 : 5.

Let co-ordinates of P be (x, y).

By formula,

$x = \dfrac{m_1x_2 + m_2x_1}{m_1 + m_2} \\[1em] = \dfrac{2 \times -2 + 5 \times 4}{2 + 5} \\[1em] = \dfrac{-4 + 20}{7} \\[1em] = \dfrac{16}{7} \\[1em] y = \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2} \\[1em] = \dfrac{2 \times 6 + 5 \times 3}{2 + 5} \\[1em] = \dfrac{12 + 15}{7} \\[1em] = \dfrac{27}{7}.$

Hence, co-ordinates of P = $\Big(\dfrac{16}{7}, \dfrac{27}{7}\Big).$

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,

$x = \dfrac{m_1x_2 + m_2x_1}{m_1 + m_2} \\[1em] \Rightarrow 2 = \dfrac{m_1 \times 5 + m_2 \times -3}{m_1 + m_2} \\[1em] \Rightarrow 2m_1 + 2m_2 = 5m_1 - 3m_2 \\[1em] \Rightarrow 5m_1 - 2m_1 = 2m_2 + 3m_2 \\[1em] \Rightarrow 3m_1 = 5m_2 \\[1em] \Rightarrow \dfrac{m_1}{m_2} = \dfrac{5}{3}.$

m₁ : m₂ = 5 : 3.

$y = \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2} \\[1em] = \dfrac{5 \times 7 + 3 \times -1}{5 + 3} \\[1em] = \dfrac{35 - 3}{8} \\[1em] = \dfrac{32}{8} \\[1em] = 4.$

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,

$y = \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2} \\[1em] \Rightarrow 2 = \dfrac{m_1 \times -3 + m_2 \times 5}{m_1 + m_2} \\[1em] \Rightarrow 2m_1 + 2m_2 = -3m_1 + 5m_2 \\[1em] \Rightarrow 2m_1 + 3m_1 = 5m_2 - 2m_2 \\[1em] \Rightarrow 5m_1 = 3m_2 \\[1em] \Rightarrow \dfrac{m_1}{m_2} = \dfrac{3}{5}.$

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,

$x = \dfrac{m_1x_2 + m_2x_1}{m_1 + m_2} \\[1em] \Rightarrow 5 = \dfrac{2 \times 0 + 5 \times a}{2 + 5} \\[1em] \Rightarrow 5 \times 7 = 0 + 5a \\[1em] \Rightarrow 5a = 35 \\[1em] \Rightarrow a = \dfrac{35}{5} = 7 \\[1em] y = \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2} \\[1em] \Rightarrow -4 = \dfrac{2 \times b + 5 \times 0}{2 + 5} \\[1em] \Rightarrow -4 \times 7 = 2b + 0 \\[1em] \Rightarrow 2b = -28 \\[1em] \Rightarrow b = -14.$

∴ 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,

$x = \dfrac{m_1x_2 + m_2x_1}{m_1 + m_2} \\[1em] \Rightarrow a = \dfrac{1 \times 6 + 2 \times -3}{1 + 2} \\[1em] \Rightarrow a = \dfrac{6 - 6}{3} \\[1em] \Rightarrow a = 0 \\[1em] y = \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2} \\[1em] \Rightarrow b = \dfrac{1 \times 6 + 2 \times 0}{1 + 2} \\[1em] \Rightarrow b = \dfrac{6}{3} = 2.$

A = (a, b) = (0, 2).

B divides the line segment in ratio 2 : 1. Let co-ordinates of B be (c, d).

By formula,

$x = \dfrac{m_1x_2 + m_2x_1}{m_1 + m_2} \\[1em] \Rightarrow c = \dfrac{2 \times 6 + 1 \times -3}{1 + 2} \\[1em] \Rightarrow c = \dfrac{12 - 3}{3} \\[1em] \Rightarrow c = \dfrac{9}{3} = 3. \\[1em] y = \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2} \\[1em] \Rightarrow d = \dfrac{2 \times 6 + 1 \times 0}{2 + 1} \\[1em] \Rightarrow d = \dfrac{12}{3} = 4.$

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),

$x = \dfrac{m_1x_2 + m_2x_1}{m_1 + m_2} \\[1em] \Rightarrow a = \dfrac{1 \times 10 + 2 \times -5}{1 + 2} \\[1em] \Rightarrow a = \dfrac{10 - 10}{3} \\[1em] \Rightarrow a = \dfrac{0}{3} = 0. \\[1em] y = \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2} \\[1em] \Rightarrow b = \dfrac{1 \times -4 + 2 \times 8}{1 + 2} \\[1em] \Rightarrow b = \dfrac{-4 + 16}{3} \\[1em] \Rightarrow b = \dfrac{12}{3} = 4.$

P = (a, b) = (0, 4).

By section formula (for Q),

$x = \dfrac{m_1x_2 + m_2x_1}{m_1 + m_2} \\[1em] \Rightarrow c = \dfrac{2 \times 10 + 1 \times -5}{2 + 1} \\[1em] \Rightarrow c = \dfrac{20 - 5}{3} \\[1em] \Rightarrow c = \dfrac{15}{3} = 5. \\[1em] y = \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2} \\[1em] \Rightarrow d = \dfrac{2 \times -4 + 1 \times 8}{2 + 1} \\[1em] \Rightarrow d = \dfrac{-8 + 8}{3} \\[1em] \Rightarrow d = \dfrac{0}{3} = 0.$

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,

$x = \dfrac{m_1x_2 + m_2x_1}{m_1 + m_2} \\[1em] \Rightarrow 3 = \dfrac{m_1 \times 5 + m_2 \times 2}{m_1 + m_2} \\[1em] \Rightarrow 3m_1 + 3m_2 = 5m_1 + 2m_2 \\[1em] \Rightarrow 5m_1 - 3m_1 = 3m_2 - 2m_2 \\[1em] \Rightarrow 2m_1 = m_2 \\[1em] \Rightarrow \dfrac{m_1}{m_2} = \dfrac{1}{2} \\[1em] \Rightarrow m_1 : m_2 = 1 : 2.$

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.

$x = \dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}$

Substituting values we get,

$\Rightarrow a = \dfrac{2 \times 5 + 1 \times 2}{2 + 1} \\[1em] \Rightarrow a = \dfrac{10 + 2}{3} \\[1em] \Rightarrow a = 4.$

$y = \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}$

Substituting values we get,

$\Rightarrow b = \dfrac{2 \times -8 + 1 \times 1}{2 + 1} \\[1em] \Rightarrow b = \dfrac{-16 + 1}{3} \\[1em] \Rightarrow b = -\dfrac{15}{3} = -5.$

B = (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,

$\Rightarrow x = \dfrac{m_1x_2 + m_2x_1}{m_1 + m_2} \\[1em] \Rightarrow 0 = \dfrac{m_1 \times -3 + m_2 \times 5}{m_1 + m_2} \\[1em] \Rightarrow 0 = -3m_1 + 5m_2 \\[1em] \Rightarrow 3m_1 = 5m_2 \\[1em] \Rightarrow \dfrac{m_1}{m_2} = \dfrac{5}{3}.$

m₁ : m₂ = 5 : 3.

By formula,

$\Rightarrow y = \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2} \\[1em] \Rightarrow y = \dfrac{5 \times 2 + 3 \times 7}{5 + 3} \\[1em] \Rightarrow y = \dfrac{10 + 21}{8} \\[1em] \Rightarrow y = \dfrac{31}{8}.$

Hence, abscissa of L = 0, m₁ : m₂ = 5 : 3 and co-ordinates of L = $\Big(0, \dfrac{31}{8}\Big)$.

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).

$\therefore x = \dfrac{m_1x_2 + m_2x_1}{m_1 + m_2} \\[1em] = \dfrac{2 \times -2 + 3 \times 3}{2 + 3} \\[1em] = \dfrac{-4 + 9}{5} \\[1em] = \dfrac{5}{5} = 1.$

and

$y = \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2} \\[1em] = \dfrac{2 \times 4 + 3 \times -1}{2 + 3} \\[1em] = \dfrac{8 - 3}{5} \\[1em] = \dfrac{5}{5} = 1.$

P = (x, y) = (1, 1).

Distance between two points = $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

$AP = \sqrt{[1 - (-3)]^2 + (1 - 4)^2} \\[1em] = \sqrt{(4)^2 + (-3)^2} \\[1em] = \sqrt{16 + 9} \\[1em] = \sqrt{25} \\[1em] = 5 \text{ units}.$

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,

$y = \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2} \\[1em] \Rightarrow 0 = \dfrac{m_1 \times -5 + m_2 \times 3}{m_1 + m_2} \\[1em] \Rightarrow 0 = -5m_1 + 3m_2 \\[1em] \Rightarrow 5m_1 = 3m_2 \\[1em] \Rightarrow \dfrac{m_1}{m_2} = \dfrac{3}{5}.$

m₁ : m₂ = 3 : 5.

Substituting value for x co-ordinate,

$\Rightarrow x = \dfrac{m_1x_2 + m_2x_1}{m_1 + m_2} \\[1em] = \dfrac{3 \times 6 + 5 \times 2}{3 + 5} \\[1em] = \dfrac{18 + 10}{8} \\[1em] = \dfrac{28}{8} \\[1em] = \dfrac{7}{2} = 3\dfrac{1}{2}.$

Hence, ordinate of K = 0, ratio in which K divides AB = 3 : 5 and K = $\Big(3\dfrac{1}{2}, 0\Big).$

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,

$x = \dfrac{m_1x_2 + m_2x_1}{m_1 + m_2} \\[1em] \Rightarrow 0 = \dfrac{m_1 \times -6 + m_2 \times 4}{m_1 + m_2} \\[1em] \Rightarrow 0 = -6m_1 + 4m_2 \\[1em] \Rightarrow 6m_1 = 4m_2 \\[1em] \Rightarrow \dfrac{m_1}{m_2} = \dfrac{4}{6} = \dfrac{2}{3}.$

m₁ : m₂ = 2 : 3.

Substituting value for y co-ordinate,

$\Rightarrow y = \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2} \\[1em] = \dfrac{2 \times -2 + 3 \times 7}{2 + 3} \\[1em] = \dfrac{-4 + 21}{5} \\[1em] = \dfrac{17}{5}.$

Hence, abscissa of K = 0, ratio in which K divides AB = 2 : 3 and K = $\Big(0, \dfrac{17}{5}\Big).$

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,

$x = \dfrac{m_1x_2 + m_2x_1}{m_1 + m_2} \\[1em] \Rightarrow 0 = \dfrac{m_1 \times 3 + m_2 \times -4}{m_1 + m_2} \\[1em] \Rightarrow 3m_1 - 4m_2 = 0 \\[1em] \Rightarrow 3m_1 = 4m_2 \\[1em] \Rightarrow \dfrac{m_1}{m_2} = \dfrac{4}{3}.$

m₁ : m₂ = 4 : 3.

Hence, PR : RQ = 4 : 3.

(ii) Substituting m₁ : m₂ = 4 : 3 in section formula we get,

$y = \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2} \\[1em] = \dfrac{4 \times 2 + 3 \times 5}{4 + 3} \\[1em] = \dfrac{8 + 15}{7} \\[1em] = \dfrac{23}{7} \\[1em] = 3\dfrac{2}{7}$

R = (0, y) = $\Big(0, 3\dfrac{2}{7}\Big)$.

Hence, co-ordinates of R = $\Big(0, 3\dfrac{2}{7}\Big)$.

(iii) From graph,

PMNQ is a trapezium and 1 block = 1 unit.

Area of trapezium = $\dfrac{1}{2} \times$ (Sum of || sides) × Distance between them

$= \dfrac{1}{2} \times (PM + QN) \times MN \\[1em] = \dfrac{1}{2} \times (5 + 2) \times 7 \\[1em] = \dfrac{1}{2} \times 7 \times 7 \\[1em] = \dfrac{1}{2} \times 49 \\[1em] = 24.5$

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,

$x = \dfrac{m_1x_2 + m_2x_1}{m_1 + m_2} \\[1em] \Rightarrow -4 = \dfrac{1 \times 0 + 2 \times x}{1 + 2} \\[1em] \Rightarrow -4 \times 3 = 2x \\[1em] \Rightarrow 2x = -12 \\[1em] \Rightarrow x = -6. \\[1em] y = \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2} \\[1em] \Rightarrow 2 = \dfrac{1 \times y + 2 \times 0}{1 + 2} \\[1em] \Rightarrow 2 \times 3 = y \\[1em] \Rightarrow y = 6.$

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 AB is divided by the y-axis.

(ii) find the co-ordinates of the point of intersection.

(iii) the length of AB.

Answer

(i) Let the y-axis divide AB in the ratio m₁ : m₂.

By section-formula, the x-coordinate = $\Big(\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}\Big)$

Since, the x-coordinate on y-axis is 0. Putting value in above formula we get :

$\Rightarrow 0 = \dfrac{m_1 \times 8 + m_2 \times -4}{m_1 + m_2} \\[1em] \Rightarrow 8m_1 - 4m_2 = 0 \\[1em] \Rightarrow 8m_1 = 4m_2 \\[1em] \Rightarrow \dfrac{m_1}{m_2} = \dfrac{4}{8} = \dfrac{1}{2} \\[1em] \Rightarrow m_1 : m_2 = 1 : 2.$

Hence, required ratio = 1 : 2.

(ii) The x-coordinate equals to zero on y-axis.

By section formula, the y-coordinate = $\Big(\dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big)$

Substituting value in above formula, we get :

$\Rightarrow y = \dfrac{1 \times -3 + 2 \times 6}{1 + 2} \\[1em] = \dfrac{-3 + 12}{3} \\[1em] = \dfrac{9}{3} \\[1em] = 3.$

Hence, the coordinates of the point of intersection are (0, 3).

(iii) By distance formula,

Distance between two points = $\sqrt{(y_2 - y_1)^2 + (x_2 - x_1)^2}$

Substituting values we get :

$AB = \sqrt{(-3 - 6)^2 + (8 - (-4))^2} \\[1em] = \sqrt{(-9)^2 + (12)^2} \\[1em] = \sqrt{81 + 144} \\[1em] = \sqrt{225} \\[1em] = 15 \text{ units}.$

Hence, AB = 15 units.

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,

$x = \dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}$

Substituting values we get,

$\Rightarrow -b = \dfrac{3 \times 5 + 1 \times -3}{3 + 1} \\[1em] \Rightarrow -b = \dfrac{15 - 3}{4} \\[1em] \Rightarrow -b = \dfrac{12}{4} = 3 \\[1em] \Rightarrow b = -3 \\[1em] y = \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}$

Substituting values we get,

$\Rightarrow 9a - 2 = \dfrac{3 \times 8a + 1 \times (3a + 1)}{3 + 1} \\[1em] \Rightarrow 9a - 2 = \dfrac{24a + 3a + 1}{4} \\[1em] \Rightarrow 9a - 2 = \dfrac{27a + 1}{4} \\[1em] \Rightarrow 36a - 8 = 27a + 1 \\[1em] \Rightarrow 9a = 9 \\[1em] \Rightarrow a = 1.$

Hence, a = 1 and b = -3.