Section & Mid-Point Formula — Test Yourself

If P(4, 3) is mid-point of line segment AB, then

1. A = (6, 0), B = (0, 8)

2. A = (8, 6), B = (8, 0)

3. A = (6, 8), B = (0, 6)

4. 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 = $\Big(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\Big)$

Substituting values we get :

$\Rightarrow (4, 3) = \Big(\dfrac{x + 0}{2}, \dfrac{0 + y}{2}\Big) \\[1em] \Rightarrow (4, 3) = \Big(\dfrac{x}{2}, \dfrac{y}{2}\Big) \\[1em] \Rightarrow \dfrac{x}{2} = 4 \text{ and } \dfrac{y}{2} = 3 \\[1em] \Rightarrow x = 8 \text{ and } y = 6.$

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 :

1. x = -4, y = 8

2. x = 4, y = -8

3. x = -4, y = -8

4. x = 4, y = 8

Answer

By mid-point formula,

Mid-point = $\Big(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\Big)$

Substituting values we get :

$\Rightarrow (0, 0) = \Big(\dfrac{4 + x}{2}, \dfrac{y + 8}{2}\Big) \\[1em] \Rightarrow \dfrac{x + 4}{2} = 0 \text{ and } \dfrac{y + 8}{2} = 0 \\[1em] \Rightarrow x + 4 = 0 \text{ and } y + 8 = 0 \\[1em] \Rightarrow x = -4 \text{ and } y = -8.$

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 :

1. (-2, 7)

2. (-2, 10)

3. (-2, -10)

4. (0, -2)

Answer

Let third vertex be (x, y).

By formula,

Centroid of triangle = $\Big(\dfrac{x_1 + x_2 + x_3}{3}, \dfrac{y_1 + y_2 + y_3}{3}\Big)$

Substituting values we get :

$\Rightarrow (-1, 2) = \Big(\dfrac{(-5) + 4 + x}{3}, \dfrac{3 + (-7) + y}{3}\Big) \\[1em] \Rightarrow (-1, 2) = \Big(\dfrac{x - 1}{3}, \dfrac{y - 4}{3}\Big) \\[1em] \Rightarrow -1 = \dfrac{x - 1}{3} \text{ and } 2 = \dfrac{y - 4}{3} \\[1em] \Rightarrow x - 1 = -3 \text{ and } y - 4 = 6 \\[1em] \Rightarrow x = -3 + 1 \text{ and } y = 6 + 4 \\[1em] \Rightarrow x = -2 \text{ and } y = 10.$

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 :

1. (10, 8)

2. (-8, 0)

3. (-10, 0)

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

Substituting values we get :

$\Rightarrow (x, y) = \Big(\dfrac{4 \times 0 + 1 \times 0}{4 + 1}, \dfrac{4 \times 10 + 1\times 0}{4 + 1}\Big) \\[1em] = \Big(\dfrac{0}{5}, \dfrac{40}{5}\Big) \\[1em] = (0, 8).$

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 :

1. 2 : 3

2. 3 : 2

3. 3 : 4

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

Substituting values for y-coordinate :

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

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

⇒ $\dfrac{3}{2} : 1$

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

1. A is true, R is false.

2. A is false, R is true.

3. Both A and R are true and R is correct reason for A.

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

So, reason (R) is true.

Here, (x₁, y₁) = (1, 2) and (x₂, y₂) = (-1, 2)

m₁ : m₂ = 1 : 2

Substituting the values, we get :

$\Rightarrow \text{P} = \Big[\dfrac{1\times (-1) + 2\times 1}{1 + 2}, \dfrac{1 \times 2 + 2 \times 2}{1 + 2}\Big]\\[1em] = \Big(\dfrac{-1 + 2}{3}, \dfrac{2 + 4}{3}\Big)\\[1em] = \Big(\dfrac{1}{3}, \dfrac{6}{3}\Big)\\[1em] = \Big(\dfrac{1}{3}, 2\Big)\\[1em]$

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 = $\dfrac{7 + 3}{2}$ and y = $\dfrac{1 + 5}{2}$.

Statement 2: (x - 7)² + (y - 1)² = (x - 3)² + (y - 5)².

1. Both the statement are true.

2. Both the statement are false.

3. Statement 1 is true, and statement 2 is false.

4. 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 = $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

⇒ PA = PB

⇒ $\Rightarrow \sqrt{(x - 7)^2 + (y - 1)^2} = \sqrt{(x - 3)^2 + (y - 5)^2}$

⇒ (x - 7)² + (y - 1)² = (x - 3)² + (y - 5)²

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)

$\therefore x = \dfrac{m_1x_2 + m_2x_1}{m_1 + m_2} \\[1em] = \dfrac{3 \times 5 + 2 \times 0}{3 + 2} \\[1em] = \dfrac{15 + 0}{5} \\[1em] = \dfrac{15}{5} = 3.$

and,

$y = \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2} \\[1em] = \dfrac{3 \times 10 + 2 \times 5}{3 + 2} \\[1em] = \dfrac{30 + 10}{5} \\[1em] = \dfrac{40}{5} = 8.$

P = (x, y) = (3, 8).

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

$AP = \sqrt{(3 - 4)^2 + (8 - (-4))^2} \\[1em] = \sqrt{(-1)^2 + 12^2} \\[1em] = \sqrt{1 + 144} \\[1em] = \sqrt{145} \\[1em] = 12.04$

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

⇒ $\dfrac{AP}{PB} = \dfrac{2}{1}$

⇒ AP : PB = 2 : 1.

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

$\therefore x = \dfrac{m_1x_2 + m_2x_1}{m_1 + m_2} \\[1em] = \dfrac{2 \times 10 + 1 \times 20}{2 + 1} \\[1em] = \dfrac{20 + 20}{3} \\[1em] = \dfrac{40}{3}.$

and,

$y = \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2} \\[1em] = \dfrac{2 \times -20 + 1 \times 0}{2 + 1} \\[1em] = \dfrac{-40 + 0}{3} \\[1em] = \dfrac{-40}{3}.$

From figure,

⇒ AB = AQ + QB

⇒ 6AQ = AQ + QB

⇒ QB = 5AQ

⇒ $\dfrac{AQ}{QB} = \dfrac{1}{5}$

⇒ AQ : QB = 1 : 5.

Let the co-ordinates of Q be (p, q)

$\therefore p = \dfrac{m_1x_2 + m_2x_1}{m_1 + m_2} \\[1em] = \dfrac{1 \times 10 + 5 \times 20}{1 + 5} \\[1em] = \dfrac{10 + 100}{6} \\[1em] = \dfrac{110}{6} = \dfrac{55}{3}.$

and,

$q = \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2} \\[1em] = \dfrac{1 \times -20 + 5 \times 0}{1 + 5} \\[1em] = \dfrac{-20 + 0}{6} \\[1em] = \dfrac{-10}{3}.$

Hence, P = $\Big(\dfrac{40}{3}, -\dfrac{40}{3}\Big) \text{ and Q} = \Big(\dfrac{55}{3}, -\dfrac{10}{3}\Big)$.

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 = $\dfrac{3}{8}$BC.

Answer

The triangle is shown in the figure below:

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

$\therefore x = \dfrac{m_1x_2 + m_2x_1}{m_1 + m_2} \\[1em] = \dfrac{3 \times 0 + 5 \times -8}{3 + 5} \\[1em] = \dfrac{0 - 40}{8} \\[1em] = \dfrac{-40}{8} = -5.$

and,

$y = \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2} \\[1em] = \dfrac{3 \times 16 + 5 \times 0}{3 + 5} \\[1em] = \dfrac{48 + 0}{8} \\[1em] = 6.$

P = (x, y) = (-5, 6).

Let the co-ordinates of Q be (m, n)

$\therefore m = \dfrac{m_1x_2 + m_2x_1}{m_1 + m_2} \\[1em] = \dfrac{3 \times 0 + 5 \times -8}{3 + 5} \\[1em] = \dfrac{0 - 40}{8} \\[1em] = \dfrac{-40}{8} = -5.$

and,

$n = \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2} \\[1em] = \dfrac{3 \times 0 + 5 \times 0}{3 + 5} \\[1em] = \dfrac{0 + 0}{8} \\[1em] = 0.$

Q = (m, n) = (-5, 0).

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

$PQ = \sqrt{(-5 - (-5))^2 + (0 - 6)^2} \\[1em] = \sqrt{0^2 + (-6)^2} \\[1em] = \sqrt{36} \\[1em] = 6 \text{ units}. \\[1em] BC = \sqrt{(0 - 0)^2 + (0 - 16)^2} \\[1em] = \sqrt{0^2 + (-16)^2} \\[1em] = \sqrt{256} \\[1em] = 16 \text{ units}.$

⇒ BC = 16 units

⇒ $\dfrac{3}{8}BC = 16 \times \dfrac{3}{8}$ = 6 units = PQ.

Hence, proved that PQ = $\dfrac{3}{8}$BC.

A line segment joining A $\Big(-1, \dfrac{5}{3}\Big)$ and B(a, 5) is divided in the ratio 1 : 3 at P, the point where the line segment AB intersects the y-axis.

(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.

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

Hence, a = 3.

(ii) By section formula,

$y = \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2} \\[1em] = \dfrac{1 \times 5 + 3 \times \dfrac{5}{3}}{1 + 3} \\[1em] = \dfrac{5 + 5}{4} \\[1em] = \dfrac{10}{4} = \dfrac{5}{2} = 2\dfrac{1}{2}.$

P = (0, y) = $\Big(0, 2\dfrac{1}{2}\Big)$.

Hence, co-ordinates of P = $\Big(0, 2\dfrac{1}{2}\Big)$.

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,

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

m₁ : m₂ = 3 : 1.

$x = \dfrac{m_1x_2 + m_2x_1}{m_1 + m_2} \\[1em] = \dfrac{3 \times 4 + 1 \times 0}{3 + 1} \\[1em] = \dfrac{12 + 0}{4} \\[1em] = \dfrac{12}{4} = 3. \\[1em]$

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) = $\Big(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\Big)$

Substituting values we get,

$\Rightarrow (4, -3) = \Big(\dfrac{x + 0}{2}, \dfrac{0 + y}{2}\Big) \\[1em] \therefore 4 = \dfrac{x}{2} \text{ and } -3 = \dfrac{y}{2} \\[1em] \Rightarrow x = 8 \text{ and } y = -6.$

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

$AC = \sqrt{[3 - (-2)]^2 + [-7 - 5]^2} \\[1em] = \sqrt{(3 + 2)^2 + (-12)^2} \\[1em] = \sqrt{25 + 144} \\[1em] = \sqrt{169} \\[1em] = 13 \text{ units}.$

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) = $\Big(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\Big)$

Substituting values we get,

$\Rightarrow (-2, 5) = \Big(\dfrac{3 + x}{2}, \dfrac{-7 + y}{2}\Big) \\[1em] \therefore -2 = \dfrac{3 + x}{2} \text{ and } 5 = \dfrac{-7 + y}{2} \\[1em] \Rightarrow x + 3 = -4 \text{ and } -7 + y = 10 \\[1em] \Rightarrow x = -7 \text{ and } y = 17.$

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) = $\Big(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\Big)$

Substituting values we get,

$\Rightarrow (2, -2a) = \Big(\dfrac{4a + (-4)}{2}, \dfrac{2b - 3 + 3b}{2}\Big) \\[1em] \therefore 2 = \dfrac{4a - 4}{2} \text{ and } -2a = \dfrac{5b - 3}{2} \\[1em] \Rightarrow 4a - 4 = 4 \text{ and } 5b - 3 = -4a \\[1em] \Rightarrow 4a = 8 \text{ and } 5b = -4a + 3 \\[1em] \Rightarrow a = 2 \text{ and } 5b = -4(2) + 3 \\[1em] \Rightarrow a = 2 \text{ and } 5b = -5 \\[1em] \Rightarrow a = 2 \text{ and } b = -1.$

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)

$\therefore x = \dfrac{m_1x_2 + m_2x_1}{m_1 + m_2} \\[1em] = \dfrac{1 \times 17 + 2 \times -4}{1 + 2} \\[1em] = \dfrac{17 + (-8)}{3} \\[1em] = \dfrac{9}{3} = 3.$

and,

$y = \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2} \\[1em] = \dfrac{1 \times 10 + 2 \times 1}{1 + 2} \\[1em] = \dfrac{10 + 2}{3} \\[1em] = \dfrac{12}{3} = 4.$

P = (x, y) = (3, 4).

Hence, co-ordinates of point P = (3, 4).

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

$OP = \sqrt{(3 - 0)^2 + (4 - 0)^2} \\[1em] = \sqrt{(3)^2 + (4)^2} \\[1em] = \sqrt{9 + 16} \\[1em] = \sqrt{25} \\[1em] = 5 \text{ units}.$

Hence, OP = 5 units.

(iii) Let point Q (0, z) on y-axis divide line AB in ratio m₁ : m₂.

By section formula,

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

Substituting values we get,

$\Rightarrow 0 = \dfrac{m_1 \times 17 + m_2 \times -4}{m_1 + m_2} \\[1em] \Rightarrow 0 = 17m_1 - 4m_2 \\[1em] \Rightarrow 4m_2 = 17m_1 \\[1em] \Rightarrow \dfrac{m_1}{m_2} = \dfrac{4}{17}.$

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

$AB = \sqrt{[-1 - (-5)]^2 + [-2 - 4]^2} \\[1em] = \sqrt{[-1 + 5]^2 + [-6]^2} \\[1em] = \sqrt{[4]^2 + 36} \\[1em] = \sqrt{16 + 36} \\[1em] = \sqrt{52}. \\[1em] BC = \sqrt{[5 - (-1)]^2 + [2 - (-2)]^2} \\[1em] = \sqrt{[5 + 1]^2 + [4]^2} \\[1em] = \sqrt{[6]^2 + 16} \\[1em] = \sqrt{36 + 16} \\[1em] = \sqrt{52}. \\[1em] AC = \sqrt{[5 - (-5)]^2 + [2 - 4]^2} \\[1em] = \sqrt{[5 + 5]^2 + [-2]^2} \\[1em] = \sqrt{[10]^2 + 4} \\[1em] = \sqrt{100 + 4} \\[1em] = \sqrt{104}. \\[1em] AB^2 + BC^2 = (\sqrt{52})^2 + (\sqrt{52})^2 \\[1em] = 52 + 52 \\[1em] = 104 = AC^2.$

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.

$O = \Big(\dfrac{-5 + 5}{2}, \dfrac{4 + 2}{2}\Big) \\[1em] = \Big(\dfrac{0}{2}, \dfrac{6}{2}\Big) \\[1em] = (0, 3).$

Let co-ordinates of D = (x, y).

$\therefore O = \Big(\dfrac{-1 + x}{2}, \dfrac{-2 + y}{2}\Big) \\[1em] \Rightarrow (0, 3) = \Big(\dfrac{-1 + x}{2}, \dfrac{-2 + y}{2}\Big) \\[1em] \Rightarrow 0 = \dfrac{-1 + x}{2} \text{ and } 3 = \dfrac{-2 + y}{2} \\[1em] \Rightarrow -1 + x = 0 \text{ and } -2 + y = 6 \\[1em] \Rightarrow x = 1 \text{ and } y = 8.$

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) = $\Big(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\Big)$

Substituting values we get,

$M = \Big(\dfrac{-3 + 9}{2}, \dfrac{7 + (-1)}{2}\Big) \\[1em] = \Big(\dfrac{6}{2}, \dfrac{6}{2}\Big) \\[1em] = (3, 3).$

Given, R(2, 2) divides the line segment joining M and the origin in the ratio p : q.

By section formula,

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

Substituting values we get,

$\Rightarrow 2 = \dfrac{p \times 0 + q \times 3}{p + q} \\[1em] \Rightarrow 2(p + q) = 0 + 3q \\[1em] \Rightarrow 2p + 2q = 3q \\[1em] \Rightarrow 2p = q \\[1em] \Rightarrow \dfrac{p}{q} = \dfrac{1}{2}.$

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,

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

Substituting values we get,

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

m₁ : m₂ = 1 : 3.

(i) By section formula,

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

Substituting values we get,

$\Rightarrow x = \dfrac{1 \times 3 + 3 \times -4}{1 + 3} \\[1em] \Rightarrow x = \dfrac{3 - 12}{4} \\[1em] \Rightarrow x = -\dfrac{9}{4}.$

Hence, x = $-\dfrac{9}{4}$.

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

$AP = \sqrt{\Big[-\dfrac{9}{4} - (-4)\Big]^2 + [3 - 2]^2} \\[1em] = \sqrt{\Big[-\dfrac{9}{4} + 4\Big]^2 + [1]^2} \\[1em] = \sqrt{\Big[\dfrac{-9 + 16}{4}\Big]^2 + 1} \\[1em] = \sqrt{\Big[\dfrac{7}{4}\Big]^2 + 1} \\[1em] = \sqrt{\dfrac{49}{16} + 1} \\[1em] = \sqrt{\dfrac{49 + 16}{16}} \\[1em] = \sqrt{\dfrac{65}{16}} \\[1em] = \dfrac{\sqrt{65}}{4}.$

Hence, AP = $\dfrac{\sqrt{65}}{4}.$

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,

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

Substituting values we get,

$\Rightarrow x = \dfrac{k \times 3 + 1 \times 2}{k + 1} \\[1em] \Rightarrow x(k + 1) = 3k + 2 \\[1em] \Rightarrow x = \dfrac{3k + 2}{k + 1}.$

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

Substituting values we get,

$\Rightarrow y = \dfrac{k \times 7 + 1 \times -2}{k + 1} \\[1em] \Rightarrow y(k + 1) = 7k - 2 \\[1em] \Rightarrow y = \dfrac{7k - 2}{k + 1}.$

Substituting value of x and y in 2x + y = 4.

$\Rightarrow 2 \Big(\dfrac{3k + 2}{k + 1}\Big) + \dfrac{7k - 2}{k + 1} = 4 \\[1em] \Rightarrow \dfrac{6k + 4}{k + 1} + \dfrac{7k - 2}{k + 1} = 4 \\[1em] \Rightarrow \dfrac{6k + 4 + 7k - 2}{k + 1} = 4 \\[1em] \Rightarrow 13k + 2 = 4k + 4 \\[1em] \Rightarrow 9k = 2 \\[1em] \Rightarrow k = \dfrac{2}{9}. \\[1em] \Rightarrow k : 1 = \dfrac{2}{9} : 1 = 2 : 9.$

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,

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

Substituting values we get,

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

m₁ : m₂ = 3 : 2.

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

Substituting values we get,

$\Rightarrow y = \dfrac{3 \times 3 + 2 \times 3}{3 + 2} \\[1em] \Rightarrow y = \dfrac{9 + 6}{5} \\[1em] \Rightarrow y = \dfrac{15}{5} = 3.$

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) = $\Big(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\Big)$

Substituting values we get,

$\Rightarrow P = \Big(\dfrac{5 + x}{2}, \dfrac{-3 + y}{2}\Big) \\[1em] \Rightarrow (-1, 3) = \Big(\dfrac{5 + x}{2}, \dfrac{-3 + y}{2}\Big) \\[1em] \therefore -1 = \dfrac{5 + x}{2} \text{ and } 3 = \dfrac{-3 + y}{2} \\[1em] \Rightarrow x + 5 = -2 \text{ and } y - 3 = 6 \\[1em] \Rightarrow x = -7 \text{ and } y = 9.$

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) = $\Big(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\Big)$

Substituting values we get,

$M = \Big(\dfrac{0 + 6}{2}, \dfrac{4 + 0}{2}\Big) \\[1em] = \Big(\dfrac{6}{2}, \dfrac{4}{2}\Big) \\[1em] = (3, 2).$

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,

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

Substituting values we get,

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

Substituting values we get,

$\Rightarrow 2 = \dfrac{1 \times y + 3 \times 0}{1 + 3} \\[1em] \Rightarrow 2 = \dfrac{y}{4} \\[1em] \Rightarrow y = 8.$

P = (x, y) = (12, 8).

Hence, co-ordinates of P = (12, 8).

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

$BP = \sqrt{(12 - 6)^2 + (8 - 0)^2} \\[1em] = \sqrt{(6)^2 + (8)^2} \\[1em] = \sqrt{36 + 64} \\[1em] = \sqrt{100} = 10 \text{ units}. \\[1em]$

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) = $\Big(\dfrac{x_1 + x_2 + x_3}{3}, \dfrac{y_1 + y_2 + y_3}{3}\Big)$

Substituting values we get,

$\Rightarrow G_1 = \Big(\dfrac{3 + y + 1}{3}, \dfrac{1 + 4 + x}{3}\Big) \\[1em] \Rightarrow (3, 4) = \Big(\dfrac{y + 4}{3}, \dfrac{x + 5}{3}\Big) \\[1em] \Rightarrow 3 = \dfrac{y + 4}{3} \text{ and } 4 = \dfrac{x + 5}{3} \\[1em] \Rightarrow y + 4 = 9 \text{ and } 12 = x + 5 \\[1em] \Rightarrow y = 5 \text{ and } x = 7.$

B = (y, 4) = (5, 4) and C = (1, x) = (1, 7).

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

$BC = \sqrt{(1 - 5)^2 + (7 - 4)^2} \\[1em] = \sqrt{(-4)^2 + (3)^2} \\[1em] = \sqrt{16 + 9} \\[1em] = \sqrt{25} = 5 \text{ units}. \\[1em]$

Hence, x = 7, y = 5 and BC = 5 units.