Section & Mid-Point Formulae — Exercise 13

Find the co-ordinates of point P which divides the line segment joining A(-2, -7) and B(6, 1) in the ratio 5 : 3.

Answer

Let point P be (x, y).

Given,

m₁ : m₂ = 5 : 3

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 :

$\Rightarrow (x, y) = \Big(\dfrac{5 \times 6 + 3 \times -2}{5 + 3}, \dfrac{5 \times 1 + 3 \times -7}{5 + 3}\Big) \\[1em] \Rightarrow (x, y) = \Big(\dfrac{30 - 6}{8}, \dfrac{5 - 21}{8}\Big) \\[1em] \Rightarrow (x, y) = \Big(\dfrac{24}{8}, \dfrac{-16}{8}\Big) \\[1em] \Rightarrow (x, y) = (3, -2).$

Hence, the coordinates of P are (3, -2).

The line segment joining the points A(4, -3) and B(4, 2) is divided by the point P such that AP : AB = 2 : 5. Find the co-ordinates of P.

Answer

Let point P be (x, y).

$\Rightarrow \dfrac{AP}{AB} = \dfrac{2}{5} \\[1em] \Rightarrow \dfrac{AP}{AP + PB} = \dfrac{2}{5} \\[1em] \Rightarrow 5AP = 2AP + 2PB \\[1em] \Rightarrow 3AP = 2PB \\[1em] \Rightarrow \dfrac{AP}{PB} = \dfrac{2}{3} \\[1em] \Rightarrow AP : PB = 2 : 3.$

m₁ : m₂ = AP : PB = 2 : 3.

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 :

$\Rightarrow (x, y) = \Big(\dfrac{2 \times 4+ 3 \times 4}{2 + 3}, \dfrac{2 \times 2 + 3 \times -3}{2 + 3}\Big) \\[1em] \Rightarrow (x, y) = \Big(\dfrac{8 + 12}{5}, \dfrac{4 - 9}{5}\Big) \\[1em] \Rightarrow (x, y) = \Big(\dfrac{20}{5}, \dfrac{-5}{5}\Big) \\[1em] \Rightarrow (x, y) = (4, -1).$

Hence, the coordinates of P are (4, -1).

P(1, -2) is a point on the line segment A(3, -6) and B(x, y) such that AP : PB is equal to 2 : 3. Find the co-ordinates of B.

Answer

Given,

m₁ : m₂ = 2 : 3

By section-formula,

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

Substituting values we get :

$\Rightarrow 1 = \Big(\dfrac{2 \times x + 3 \times 3}{2 + 3}\Big) \\[1em] \Rightarrow 1 = \Big(\dfrac{2x + 9}{5}\Big) \\[1em] \Rightarrow 5 = 2x + 9 \\[1em] \Rightarrow 5 - 9 = 2x \\[1em] \Rightarrow -4 = 2x \\[1em] \Rightarrow x = \dfrac{-4}{2} \\[1em] \Rightarrow x = -2.$

By section-formula,

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

Substituting values we get :

$\Rightarrow -2 = \Big(\dfrac{2 \times y + 3 \times -6}{2 + 3}\Big) \\[1em] \Rightarrow -2 = \Big(\dfrac{2y - 18}{5}\Big) \\[1em] \Rightarrow -10 = 2y - 18 \\[1em] \Rightarrow -10 + 18 = 2y \\[1em] \Rightarrow 8 = 2y \\[1em] \Rightarrow y = \dfrac{8}{2} \\[1em] \Rightarrow y = 4.$

Hence, the coordinates of B are (-2, 4).

Find a point P on the line segment joining A(14, -5) and B(-4, 4), which is twice as far from A as from B.

Answer

Let point P be (x, y).

Since, point P is twice as far from A as from B.

⇒ AP = 2BP

⇒ $\dfrac{AP}{BP} = \dfrac{2}{1}$

⇒ AP : BP = 2 : 1.

⇒ m₁ : m₂ = AP : PB = 2 : 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 :

$\Rightarrow (x, y) = \Big(\dfrac{2 \times -4+ 1 \times 14}{2 + 1}, \dfrac{2 \times 4 + 1 \times -5}{2 + 1}\Big) \\[1em] \Rightarrow (x, y) = \Big(\dfrac{-8 + 14}{3}, \dfrac{8 - 5}{3}\Big) \\[1em] \Rightarrow (x, y) = \Big(\dfrac{6}{3}, \dfrac{3}{3}\Big) \\[1em] \Rightarrow (x, y) = (2, 1).$

Hence, the coordinates of P are (2, 1).

Find the co-ordinates of the points of trisection of the line segment joining the points A(5, -3) and B(2, -9).

Answer

Let point P is the first point of trisection, meaning it divides the segment AB internally in the ratio 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 :

$\Rightarrow P(x, y) = \Big(\dfrac{1 \times 2 + 2 \times 5}{1 + 2}, \dfrac{1 \times -9 + 2 \times -3}{1 + 2}\Big) \\[1em] \Rightarrow P(x, y) = \Big(\dfrac{2 + 10}{3}, \dfrac{-9 - 6}{3}\Big) \\[1em] \Rightarrow P(x, y) = \Big(\dfrac{12}{3}, \dfrac{-15}{3}\Big) \\[1em] \Rightarrow P(x, y) = (4, -5).$

Let point Q is the second point of trisection, meaning it divides the segment AB internally in the ratio m₁ : m₂ = 2 : 1

Substituting values we get :

$\Rightarrow Q(a, b) = \Big(\dfrac{2 \times 2 + 1 \times 5}{2 + 1}, \dfrac{2 \times -9 + 1 \times -3}{2 + 1}\Big) \\[1em] \Rightarrow Q(a, b) = \Big(\dfrac{4 + 5}{3}, \dfrac{-18 - 3}{3}\Big) \\[1em] \Rightarrow Q(a, b) = \Big(\dfrac{9}{3}, \dfrac{-21}{3}\Big) \\[1em] \Rightarrow Q(a, b) = (3, -7).$

Hence, the coordinates of trisection are P(4, -5) and Q(3, -7).

Find the co-ordinates of the mid-point of the line segment joining :

(i) A(5, 7) and B(-3, -1)

(ii) P(-5, -8) and Q(3, 4)

Answer

(i) Let mid-point of the line segment joining AB be P(x,y).

By using mid-point formula,

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

Substituting values we get :

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

Hence, the coordinates of mid-point are (1, 3).

(ii) Let mid-point of the line segment joining PQ be B(x,y).

By using mid-point formula,

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

Substituting values we get :

$\Rightarrow (x, y) = \Big(\dfrac{-5 + 3}{2}, \dfrac{-8 + 4}{2}\Big) \\[1em] \Rightarrow (x, y) = \Big(\dfrac{-2}{2}, \dfrac{-4}{2}\Big) \\[1em] \Rightarrow (x, y) = (-1, -2).$

Hence, the coordinates of mid-point are (-1, -2).

The line segment joining A(-3, 1) and B(7, -5) is a diameter of a circle whose centre is C. Find the co-ordinates of the centre C.

Answer

Since the line segment joining points A and B is the diameter of the circle, the centre of the circle must be the mid-point of the diameter AB.

By using mid-point formula,

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

Substituting values we get :

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

Hence, the coordinates of centre (C) are (2, -2).

A(10, 5), B(6, -3) and C(2, 1) are the vertices of a ΔABC. L is the mid-point of AB and M is the mid-point of AC. Write down the co-ordinates of L and M. Show that $LM = \dfrac{1}{2} BC.$

Answer

By using mid-point formula,

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

L is the mid-point of AB.

Substituting values we get :

$\Rightarrow L = \Big(\dfrac{10 + 6}{2}, \dfrac{5 + (-3)}{2}\Big) \\[1em] \Rightarrow L = \Big(\dfrac{16}{2}, \dfrac{2}{2}\Big) \\[1em] \Rightarrow L = (8, 1).$

Given,

M is the mid-point of AC.

Substituting values we get :

$\Rightarrow M = \Big(\dfrac{10 + 2}{2}, \dfrac{5 + 1}{2}\Big) \\[1em] \Rightarrow M = \Big(\dfrac{12}{2}, \dfrac{6}{2}\Big) \\[1em] \Rightarrow M = (6, 3).$

By distance formula,

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

Substituting values we get :

$LM = \sqrt{(6 - 8)^2 + (3 - 1)^2} \\[1em] = \sqrt{(-2)^2 + (2)^2} \\[1em] = \sqrt{4 + 4} \\[1em] = \sqrt{8} = 2\sqrt{2}. \\[1em] BC = \sqrt{(2 - 6)^2 + (1 - (-3))^2} \\[1em] = \sqrt{(-4)^2 + (4)^2} \\[1em] = \sqrt{16 + 16} \\[1em] = \sqrt{32} \\[1em] = \sqrt{16 \times 2} \\[1em] = 4\sqrt{2} \\[1em] = 2 \times 2\sqrt{2} \\[1em] = 2 \times LM.$

Thus, BC = 2LM or LM = $\dfrac{1}{2}BC.$

Hence, proved that $LM = \dfrac{1}{2}BC$.

The mid-point of the line segment joining A(p, 5) and B(3, q) is M(-1, 4). Find the values of p and q.

Answer

Given,

Mid-point of the line segment joining A(p, 5) and B(3, q) is M(-1, 4).

By using mid-point formula,

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

Substituting values we get :

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

Hence, p = -5 and q = 3.

The centre of a circle is C(-2, 3) and one end of a diameter PQ is P(2, -4). Find the co-ordinates of Q.

Answer

Since PQ is the diameter of the circle and C is the center, C must be the mid-point of the segment PQ.

Let coordinates of Q be (x₁, y₁).

By using mid-point formula,

x-coordinate = $\Big(\dfrac{x_1 + x_2}{2}\Big)$

Substituting values we get :

$\Rightarrow -2 = \Big(\dfrac{2 + x_1}{2}\Big) \\[1em] \Rightarrow -4 = 2 + x_1 \\[1em] \Rightarrow -4 - 2 = x_1 \\[1em] \Rightarrow x_1 = -6.$

By using mid-point formula,

y-coordinate = $\Big(\dfrac{y_1 + y_2}{2}\Big)$

Substituting values we get :

$\Rightarrow 3 = \Big(\dfrac{-4 + y_1}{2}\Big) \\[1em] \Rightarrow 6 = -4 + y_1 \\[1em] \Rightarrow 6 + 4 = y_1 \\[1em] \Rightarrow y_1 = 10.$

Q = (x₁, y₁) = (-6, 10).

Hence, coordinates of Q are (-6, 10).

The point P(-4, 1) divides the line segment joining the points A(2, -2) and B in the ratio 3 : 5. Find the co-ordinates of point B.

Answer

Let coordinates of B be (a, b).

Point P(-4, 1) divides the line segment joining A(2, -2) and B(a, b) in the ratio 3 : 5.

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 :

$\Rightarrow (-4, 1) = \Big(\dfrac{3a + 5(2)}{3 + 5}, \dfrac{3b + 5(-2)}{3 + 5}\Big) \\[1em] \Rightarrow (-4, 1) = \Big(\dfrac{3a + 10}{8}, \dfrac{3b - 10}{8}\Big) \\[1em] \Rightarrow -4 = \Big(\dfrac{3a + 10}{8}\Big) \text{ and } 1 = \Big(\dfrac{3b - 10}{8}\Big) \\[1em] \Rightarrow -32 = 3a + 10 \text{ and } 8 = 3b - 10 \\[1em] \Rightarrow -32 - 10 = 3a \text{ and } 8 + 10 = 3b \\[1em] \Rightarrow -42 = 3a \text{ and } 18 = 3b \\[1em] \Rightarrow a = \dfrac{-42}{3} \text{ and } b = \dfrac{18}{3} \\[1em] \Rightarrow a = -14 \text{ and }b = 6 .$

B = (a, b) = (-14, 6).

Hence, coordinates of B(-14, 6).

In what ratio does the point P(2, -5) divide the join of A(-3, 5) and B(4, -9)?

Answer

Let the point P(2, -5) divide the line segment AB in the ratio k : 1.

By section-formula,

x-coordinate = $\Big(\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}\Big)$

Substituting values we get :

$\Rightarrow 2 = \Big(\dfrac{k(4) + 1(-3)}{k + 1}\Big) \\[1em] \Rightarrow 2(k + 1) = 4k - 3 \\[1em] \Rightarrow 2k + 2 = 4k - 3 \\[1em] \Rightarrow 2 + 3 = 4k - 2k \\[1em] \Rightarrow 5 = 2k \\[1em] \Rightarrow k = \dfrac{5}{2} \\[1em] \Rightarrow k : 1 = \dfrac{5}{2} : 1 = 5 : 2.$

Hence, point P divide AB in the ratio 5 : 2.

In what ratio does the point P(a, -1) divide the join of A(1, -3) and B(6, 2)? Hence, find the value of a.

Answer

Let the point P(2, -5) divide the segment AB in the ratio k : 1.

By section-formula,

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

Substituting values we get :

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

By section-formula,

x-coordinate = $\Big(\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}\Big)$

Substitute values we get:

$\Rightarrow a = \Big(\dfrac{2 \times 6 + 3 \times 1}{2 + 3}\Big) \\[1em] = \Big(\dfrac{12 + 3}{5}\Big) \\[1em] = \dfrac{15}{5} \\[1em] = 3.$

Hence, point P divide AB in the ratio 2 : 3 and value of a = 3.

The line segment joining A(2, 3) and B(6, -5) is intercepted by the x-axis at the point k. Find the ratio in which k divides AB. Also, write the co-ordinates of the point k.

Answer

Since the point k lies on the x-axis, its y-coordinate must be 0. Let the coordinates of k be (x, 0).

Let k divide the line segment joining A(2, 3) and B(6, -5) in the ratio m₁:m₂

By section-formula,

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

Substituting values we get :

$\Rightarrow 0 = \Big(\dfrac{m_1(-5) + m_2(3)}{m_1 + m_2}\Big) \\[1em] \Rightarrow -5m_1 + 3m_2 = 0 \\[1em] \Rightarrow 5m_1 = 3m_2 \\[1em] \Rightarrow \dfrac{m_1}{m_2} = \dfrac{3}{5} \\[1em] \Rightarrow m_1 : m_2 = 3 : 5.$

By section-formula,

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

Substituting values we get :

$\Rightarrow x = \Big(\dfrac{3(6) + 5(2)}{3 + 5}\Big) \\[1em] \Rightarrow x = \Big(\dfrac{18 + 10}{8}\Big) \\[1em] \Rightarrow x = \Big(\dfrac{28}{8}\Big) \\[1em] \Rightarrow x = \dfrac{7}{2}$

Hence, ratio in which k divides AB = 3 : 5 and coordinates of the point k are $\Big(\dfrac{7}{2}, 0\Big)$ .

In what ratio is the segment joining the points A(6, 5) and B(-3, 2) divided by the y-axis? Find the point at which the y-axis cuts AB.

Answer

When a point lies on the y-axis, its x-coordinate is always 0. Let the point where the y-axis cuts AB be P(0, y).

Let ratio in which P divides AB be m₁ : m₂.

By section-formula,

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

Substituting values we get :

$\Rightarrow 0 = \Big(\dfrac{m_1(-3) + m_2(6)}{m_1 + m_2}\Big) \\[1em] \Rightarrow 0 = -3m_1 + 6m_2 \\[1em] \Rightarrow 3m_1 = 6m_2 \\[1em] \Rightarrow \dfrac{m_1}{m_2} = \dfrac{6}{3} = 2.$

Thus, m₁ : m₂ = 2 : 1.

By section-formula,

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

Substitute values we get:

$\Rightarrow y = \Big(\dfrac{2(2) + 1(5)}{2 + 1}\Big) \\[1em] = \Big(\dfrac{4 + 5}{3}\Big) \\[1em] = \Big(\dfrac{9}{3}\Big) \\[1em] = 3.$

P = (0, y) = (0, 3).

Hence, AB is divided in ratio 2 : 1 and point at which the y-axis cuts AB is (0, 3).

(i) Write down the co-ordinates of the point P that divides the line segment 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) The point P divides the line segment joining A(-4, 1) and B(17, 10) in the ratio m₁ : m₂ = 1 : 2.

Let coordinates of P 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 :

$\Rightarrow (x, y) = \Big(\dfrac{1(17) + 2(-4)}{1 + 2}, \dfrac{1(10) + 2(1)}{1 + 2}\Big) \\[1em] \Rightarrow (x, y) = \Big(\dfrac{17 - 8}{3}, \dfrac{10 + 2}{3}\Big) \\[1em] \Rightarrow (x, y) = \Big(\dfrac{9}{3}, \dfrac{12}{3}\Big) \\[1em] \Rightarrow (x, y) = (3, 4).$

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

Hence, coordinates of P = (3, 4).

(ii) Using distance formula,

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

Substitute values we get,

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

Hence, distance of OP is 5 units.

(iii) A point on the y-axis has an x-coordinate of 0. Let the y-axis cut AB at K(0, y). let the ratio be k : 1.

By section-formula,

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

Substitute values we get,

$\Rightarrow 0 = \Big(\dfrac{k(17) + 1(-4)}{k + 1}\Big) \\[1em] \Rightarrow 17k - 4 = 0 \\[1em] \Rightarrow 17k = 4 \\[1em] \Rightarrow k = \dfrac{4}{17} \\[1em] \Rightarrow k : 1 = \dfrac{4}{17} : 1 = 4 : 17.$

Hence, y-axis cut line AB in ratio 4 : 17.

The line segment joining P(-4, 5) and Q(3, 2) intersects the y-axis at R. PM and QN are perpendiculars from P and Q on the x-axis. Find :

(i) the ratio PR : RQ

(ii) the coordinates of R.

(iii) the area of quadrilateral PMNQ.

Answer

(i) The point R lies on the y-axis, so its x-coordinate is 0.

Let R(0, y) divide the line segment PQ in the ratio m₁ : m₂

By section-formula,

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

Substitute values we get,

$\Rightarrow 0 = \Big(\dfrac{m_1(3) + m_2(-4)}{m_1 + m_2}\Big) \\[1em] \Rightarrow 0 = 3m_1 - 4m_2 \\[1em] \Rightarrow 3m_1 = 4m_2 \\[1em] \Rightarrow \dfrac{m_1}{m_2} = \dfrac{4}{3}.$

Hence, the ratio PR : RQ = 4 : 3.

(ii) Given,

P(-4, 5) and Q(3, 2)

m₁ : m₂ = 4 : 3.

By section-formula,

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

Substitute values we get,

$\Rightarrow y = \Big(\dfrac{4(2) + 3(5)}{4 + 3}\Big) \\[1em] \Rightarrow y = \Big(\dfrac{8 + 15}{7}\Big) \\[1em] \Rightarrow y = \Big(\dfrac{23}{7}\Big).$

Hence, the coordinates of R = $\Big(0, \dfrac{23}{7}\Big)$.

(iii) From figure,

PQNM is a trapezium, with PM and QN being the parallel sides, as both are perpendicular to x-axis.

From figure,

PM = 5 units, MN = 7 units and QN = 2 units.

The area of a trapezoid is given by :

Area = $\dfrac{1}{2}$ × Sum of parallel sides × Distance between them

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

Hence, area of PQNM = 24.5 sq.units.

In the given figure, the line segment AB meets x-axis at A and y-axis at B. The point P(-3, 1) on AB divides it in ratio 2 : 3. Find the coordinates of A and B.

Answer

Since, point A and B lies on x-axis and y-axis respectively. Let their coordinates be A(a, 0) and B(0, b).

Given,

The line segment AB be divided by point P(-3, 1) in the ratio AP : PB = 2 : 3.

By section formula,

$\Rightarrow (x, y) = \Big(\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}, \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big) \\[1em] \Rightarrow (-3, 1) = \Big(\dfrac{2(0) + 3(a)}{2 + 3}, \dfrac{2(b) + 3(0)}{2 + 3}\Big) \\[1em] \Rightarrow (-3, 1) = \Big(\dfrac{3a}{5}, \dfrac{2b}{5}\Big) \\[1em] \Rightarrow -3 = \Big(\dfrac{3a}{5}\Big), 1 = \Big(\dfrac{2b}{5}\Big) \\[1em] \Rightarrow -15 = 3a, 5 = 2b \\[1em] \Rightarrow a = \dfrac{-15}{3}, b = \dfrac{5}{2} \\[1em] \Rightarrow a = -5, b = \dfrac{5}{2} \\[1em] \Rightarrow A = (a, 0) = (-5, 0) \\[1em] \Rightarrow B = (0, b) = \Big(0, \dfrac{5}{2}\Big).$

Hence, A(-5, 0) and $B\Big(0, \dfrac{5}{2}\Big)$.

Show that the line segment joining the points A(-5, 8) and B(10, -4) is trisected by the coordinate axes. Also, find the points of trisection of AB.

Answer

Let x-axis divide AB in the ratio k : 1 at the point P(x, 0).

By section-formula,

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

Substituting values we get,

$\Rightarrow 0 = \Big(\dfrac{k(-4) + 1(8)}{k + 1}\Big) \\[1em] \Rightarrow 0 = \Big(\dfrac{-4k + 8}{k + 1}\Big) \\[1em] \Rightarrow 0 = -4k + 8 \\[1em] \Rightarrow 4k = 8 \\[1em] \Rightarrow k = \dfrac{8}{4} = 2.$

The x-axis divides AB in the ratio k : 1 = 2 : 1.

Thus, m₁ : m₂ = 2 : 1.

By section-formula,

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

Substituting values we get,

$\Rightarrow x = \Big(\dfrac{2(10) + 1(-5)}{2 + 1}\Big) \\[1em] \Rightarrow x = \Big(\dfrac{20 - 5}{3}\Big) \\[1em] \Rightarrow x = \Big(\dfrac{15}{3}\Big) \\[1em] \Rightarrow x = 5.$

The coordinates of P = (x, 0) = (5, 0).

Let the y-axis divide AB in the ratio p : 1 at the point Q(0, y).

By section-formula,

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

Substituting values we get,

$\Rightarrow 0 = \Big(\dfrac{p(10) + 1(-5)}{p + 1}\Big) \\[1em] \Rightarrow 0 = \Big(\dfrac{10p - 5}{p + 1}\Big) \\[1em] \Rightarrow 10p - 5 = 0 \\[1em] \Rightarrow 10p = 5 \\[1em] \Rightarrow p = \dfrac{5}{10} \\[1em] \Rightarrow p = \dfrac{1}{2} \\[1em] \Rightarrow p : 1 = \dfrac{1}{2} : 1 = 1 : 2.$

By section-formula,

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

Substitute values we get,

$\Rightarrow y = \Big(\dfrac{1 \times -4 + 2 \times 8}{1 + 2}\Big) \\[1em] = \Big(\dfrac{-4 + 16}{3}\Big) \\[1em] = \dfrac{12}{3} \\[1em] = 4.$

The coordinates of Q = (0, y) = (0, 4).

Hence, points of trisection of AB are Q(5, 0) and P(0, 4).

The mid-points of the sides BC, CA and AB of ΔABC are D(2, 1), E(-1, -3) and F(4, 5) respectively. Find the co-ordinates of A, B and C.

Answer

Let the vertices of ΔABC be A(x₁, y₁), B(x₂, y₂), and C(x₃, y₃).

By mid-point formula,

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

Given,

D(2, 1) is the midpoint of BC.

Substitute values we get,

$\Rightarrow \dfrac{x_2 + x_3}{2} = 2 \\[1em] \Rightarrow x_2 + x_3 = 4 \text{ ....(1)} \\[1em] \Rightarrow \dfrac{y_2 + y_3}{2} = 1 \\[1em] \Rightarrow y_2 + y_3 = 2 \text{ ....(2)}$

Given,

E(-1, -3) is the midpoint of CA.

$\Rightarrow \dfrac{x_3 + x_1}{2} = -1 \\[1em] \Rightarrow x_3 + x_1 = -2 \text{ .....(3)} \\[1em] \Rightarrow \dfrac{y_3 + y_1}{2} = -3 \\[1em] \Rightarrow y_3 + y_1 = -6 \text{ ....(4)}$

Given,

F(4, 5) is the midpoint of CA.

$\Rightarrow \dfrac{x_1 + x_2}{2} = 4 \\[1em] \Rightarrow x_1 + x_2 = 8 \text{ ....(5)} \\[1em] \Rightarrow \dfrac{y_1 + y_2}{2} = 5 \\[1em] \Rightarrow y_1 + y_2 = 10 \text{ ....(6)}$

Adding the three equations (1), (3) and (5), we get :

⇒ (x₂ + x₃) + (x₃ + x₁) + (x₁ + x₂) = 4 + (-2) + 8

⇒ 2x₁ +2x₂ + 2x₃ = 10

⇒ 2(x₁ +x₂ + x₃) = 10

⇒ (x₁ +x₂ + x₃) = $\dfrac{10}{2}$

⇒ (x₁ +x₂ + x₃) = 5 .....(7)

Subtract (Eq. 3) from (Eq. 7) :

⇒ (x₁ +x₂ + x₃) - ( x₂ + x₃) = 5 - 4

⇒ (x₁ +x₂ + x₃ -x₂ - x₃) = 5 - 4

⇒ x₁ = 1.

Subtract (Eq. 2) from (Eq. 7) :

⇒ (x₁ +x₂ + x₃) - ( x₃ + x₁) = 5 - (-2)

⇒ (x₁ +x₂ + x₃ -x₃ - x₁) = 5 + 2

⇒ x₂ = 7.

Subtract (Eq. 5) from (Eq. 7):

⇒ (x₁ +x₂ + x₃) - ( x₁ + x₂) = 5 - 8

⇒ (x₁ +x₂ + x₃ -x₁ - x₂) = -3

⇒ x₃ = -3.

Adding equations (2), (4) and (5), we get :

⇒ (y₂ + y₃) + (y₃ + y₁) + (y₁ + y₂) = 2 + (-6) + 10

⇒ 2y₁ +2y₂ + 2y₃ = 6

⇒ 2(y₁ +y₂ + y₃) = 6

⇒ (y₁ +y₂ + y₃) = $\dfrac{6}{2}$

⇒ (y₁ +y₂ + y₃) = 3 .....(8)

Subtract (Eq. 2) from (Eq. 8):

⇒ (y₁ +y₂ + y₃) - ( y₂ + y₃) = 3 - 2

⇒ (y₁ +y₂ + y₃ -y₂ - y₃) = 1

⇒ y₁ = 1.

Subtract (Eq. 4) from (Eq. 8):

⇒ (y₁ + y₂ + y₃) - ( y₃ + y₁) = 3-(-6)

⇒ (y₁ +y₂ + y₃ -y₃ - y₁) = 3 + 6

⇒ y₂ = 9.

Subtract (Eq. 6) from (Eq. 8):

⇒ (y₁ + y₂ + y₃) - ( y₁ + y₂) = 3 - 10

⇒ (y₁ +y₂ + y₃ -y₁ - y₂) = -7

⇒ y₃ = -7.

⇒ A = (x₁, y₁) = (1, 1), B = (x₂, y₂) = (7, 9), C = (x₃, y₃) = (-3, -7).

Hence, A = (1, 1), B = (7, 9) and C = (-3, -7).

Prove that the points A(-2, -1), B(1, 0), C(4, 3) and D(1, 2) are the vertices of a parallelogram ABCD.

Answer

Given,

A(-2, -1), B(1, 0), C(4, 3) and D(1, 2).

By using mid-point formula,

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

The Midpoint of Diagonal AC :

$\Rightarrow M_{(AC)} = \Big(\dfrac{-2 + 4}{2}, \dfrac{-1 + 3}{2}\Big) \\[1em] \Rightarrow M_{(AC)} = \Big(\dfrac{2}{2}, \dfrac{2}{2}\Big) \\[1em] \Rightarrow M_{(AC)} = (1, 1).$

The Midpoint of Diagonal BD :

$\Rightarrow M_{(BD)} = \Big(\dfrac{1 + 1}{2}, \dfrac{0 + 2}{2}\Big) \\[1em] \Rightarrow M_{(BD)} = \Big(\dfrac{2}{2}, \dfrac{2}{2}\Big) \\[1em] \Rightarrow M_{(BD)} = (1, 1)$

Since the midpoint of diagonal AC, M_AC(1, 1), is the same as the midpoint of diagonal BD, M_BD (1, 1), the diagonals AC and BD bisect each other.

We know that,

A quadrilateral is a parallelogram if its diagonals bisect each other.

Hence, proved that ABCD is a parallelogram.

If the points A(-2, -1), B(1, 0), C(a, 3) and D(1, b) form a parallelogram, find the values of a and b.

Answer

We know that,

Diagonals of a parallelogram bisect each other.

Since, ABCD is a // gm.

Thus, Mid-point of AC = Mid-point of BD.

Given,

A(-2, -1), B(1, 0), C(a, 3) and D(1, b)

By using mid-point formula,

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

Mid-point of AC :

$\Rightarrow M_{AC} = \Big(\dfrac{-2 + a}{2}, \dfrac{-1 + 3}{2}\Big) \\[1em] \Rightarrow M_{AC} = \Big(\dfrac{-2 + a}{2}, \dfrac{2}{2}\Big) \\[1em] \Rightarrow M_{AC} = \Big(\dfrac{-2 + a}{2}, 1\Big).$

Midpoint of Diagonal BD :

$\Rightarrow M_{BD} = \Big(\dfrac{1 + 1}{2}, \dfrac{0 + b}{2}\Big) \\[1em] \Rightarrow M_{BD} = \Big(\dfrac{2}{2}, \dfrac{b}{2}\Big) \\[1em] \Rightarrow M_{BD} = \Big(1, \dfrac{b}{2}\Big) .$

Equating the x-coordinates of mid-points of AC and BD, we get :

⇒ $\dfrac{-2 + a}{2}$ = 1

⇒ -2 + a = 2

⇒ a = 2 + 2

⇒ a = 4.

Equating the y-coordinates:

⇒ 1 = $\dfrac{b}{2}$

⇒ b = 2.

Hence, a = 4 and b = 2.

The three vertices of a parallelogram ABCD, taken in order, are A(-1, 0), B(3, 1) and C(2, 2). Find the co-ordinates of the fourth vertex of the parallelogram.

Answer

Let coordinates of D be (x, y).

By using mid-point formula,

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

Mid-point of AC :

$\Rightarrow M_{AC} = \Big(\dfrac{-1 + 2}{2}, \dfrac{0 + 2}{2}\Big) \\[1em] \Rightarrow M_{AC} = \Big(\dfrac{1}{2}, 1\Big).$

Midpoint of Diagonal BD :

$\Rightarrow M_{BD} = \Big(\dfrac{3 + x}{2}, \dfrac{1 + y}{2}\Big)$

Since its a parallelogram, the midpoint of diagonal AC must be equal to the midpoint of diagonal BD, as diagonals bisect each other.

Equating the x-coordinates:

$\Rightarrow \dfrac{3 + x}{2} = \dfrac{1}{2}$

⇒ 3 + x = 1

⇒ x = 1 - 3

⇒ x = -2.

Equating the y-coordinates:

$\Rightarrow \dfrac{1 + y}{2} = 1$

⇒ 1 + y = 2

⇒ y = 2 - 1

⇒ y = 1.

Hence, coordinates of D = (-2, 1).

Find the lengths of the medians of a ΔABC whose vertices are A(-1, 3), B(1, -1) and C(5, 1). Also, find the co-ordinates of the centroid of ΔABC.

Answer

Using the Midpoint Formula,

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

Let D be the midpoint of BC.

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

Let E be the midpoint of AC.

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

Let F be the midpoint of AB.

$\Rightarrow F = \Big( \dfrac{-1 + 1}{2}, \dfrac{3 + (-1)}{2} \Big) \\[1em] \Rightarrow F = \Big( \dfrac{0}{2}, \dfrac{2}{2} \Big) \\[1em] \Rightarrow F = (0, 1).$

Length of Median AD : A(-1, 3) and D(3, 0)

Using Distance Formula,

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

Substituting values we get:

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

Length of Median BE:

B(1, -1) and E(2, 2)

$BE = \sqrt{(2 - 1)^2 + (2 - (-1))^2} \\[1em]= \sqrt{(1)^2 + (3)^2} \\[1em] = \sqrt{1 + 9} \\[1em] = \sqrt{10} \text{ units}.$

Length of Median CF:

C(5, 1) and F(0, 1)

$CF = \sqrt{(0 - 5)^2 + (1 - 1)^2} \\[1em] = \sqrt{(-5)^2 + (0)^2} \\[1em] = \sqrt{25 + 0} \\[1em] = \sqrt{25} \\[1em] = 5 \text{ units}.$

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

Substituting values we get :

$\Rightarrow \text{Centroid} = \Big(\dfrac{-1 + 1 + 5}{3}, \dfrac{3 + (-1) + 1}{3}\Big) \\[1em] = \Big(\dfrac{5}{3}, \dfrac{3}{3}\Big) \\[1em] = \Big(\dfrac{5}{3}, 1\Big).$

Hence, AD = 5 units , $BE = \sqrt{10}$ units, CF = 5 units, coordinates of centroid are $\Big(\dfrac{5}{3}, 1\Big)$.

Find the co-ordinates of the centroid of ΔPQR whose vertices are P(6, 3), Q(-2, 5) and R(-1, 7).

Answer

By using the centroid formula,

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

P(6, 3), Q(-2, 5) and R(-1, 7)

Substitute values we get:

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

Hence, coordinates of centroid of ΔPQR = (1, 5).

Find the co-ordinates of the point of intersection of the medians of the triangle whose vertices are A(-7, 5), B(-1, -3) and C(5, 7).

Answer

The medians of a triangle intersect at centroid of triangle.

By using the centroid formula,

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

A(-7, 5), B(-1, -3) and C(5, 7)

Substitute values we get:

$\Rightarrow (x, y) = \Big( \dfrac{-7 + (-1) + 5}{3}, \dfrac{5 + (-3) + 7}{3} \Big) \\[1em] \Rightarrow (x, y) = \Big( \dfrac{-8 + 5}{3}, \dfrac{5 + 4}{3} \Big) \\[1em] \Rightarrow (x, y) = \Big( \dfrac{-3}{3}, \dfrac{9}{3} \Big) \\[1em] \Rightarrow (x, y) = (-1, 3).$

Hence, co-ordinates of the point of intersection of the medians = (-1, 3).

If G(-2, 1) is the centroid of ΔABC, two of whose vertices are A(1, -6) and B(-5, 2), find the third vertex of the triangle.

Answer

Let coordinates of third vertex be C(x₃, y₃).

By using the centroid formula,

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

Given,

G(-2, 1) is the centroid.

Substituting values we get :

$\Rightarrow (-2, 1) = \Big( \dfrac{1 + (-5) + x_3}{3}, \dfrac{-6 + 2 + y_3}{3}\Big) \\[1em] \Rightarrow -2 = \Big( \dfrac{1 + (-5) + x_3}{3}\Big), 1 = \Big( \dfrac{-6 + 2 + y_3}{3}\Big) \\[1em] \Rightarrow -6 = 1 + (-5) + x_3, 3 = -6 + 2 + y_3 \\[1em] \Rightarrow -6 = -4 + x_3, 3 = -4 + y_3 \\[1em] \Rightarrow -6 + 4 = x_3, 3 + 4 = y_3 \\[1em] \Rightarrow x_3 = -2, y_3 = 7.$

Hence, coordinates of third vertex = (-2, 7).

A(6, y), B(-4, 4) and C(x, -1) are the vertices of ΔABC whose centroid is the origin. Calculate the values of x and y.

Answer

By using the centroid formula,

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

Origin (0,0) is the centroid of triangle ABC.

Solving for x-coordinate:

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

Solving for y-coordinate:

$\Rightarrow 0 = \Big(\dfrac{y + 4 + (-1)}{3}\Big) \\[1em] \Rightarrow 0 = y + 3 \\[1em] \Rightarrow y = -3.$

Hence, x = -2 and y = -3.

ABC is a triangle and G(4, 3) is its centroid. If A(1, 3), B(4, b) and C(a, 1) be the vertices, find the values of a and b and hence find the length of side BC.

Answer

By using the centroid formula,

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

Given,

A(1, 3), B(4, b) and C(a, 1).

G(4, 3) is the centroid of triangle ABC.

Solving for x-coordinate :

$\Rightarrow 4 = \dfrac{1 + 4 + a}{3} \\[1em] \Rightarrow 12 = 5 + a \\[1em] \Rightarrow a = 12 - 5 \\[1em] \Rightarrow a = 7.$

C = (a, 1) = (7, 1)

Solving for y-coordinate :

$\Rightarrow 3 = \dfrac{3 + b + 1}{3} \\[1em] \Rightarrow 9 = 4 + b \\[1em] \Rightarrow b = 9 - 4 \\[1em] \Rightarrow b = 5.$

B = (4, b) = (4, 5)

By using distance formula,

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

Substituting values we get:

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

Hence, a = 7 and b = 5, length of BC = 5 units.

Calculate the ratio in which the line segment joining A(-4, 2) and B(3, 6) is divided by the point P(x, 3). Also, find

(i) x

(ii) length of AP.

Answer

(i) Let the point P(x, 3) divide the line segment joining A(-4, 2) and B(3, 6) in the ratio m₁ : m₂.

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 :

$\Rightarrow 3 = \Big(\dfrac{m_1(6) + m_2(2)}{m_1 + m_2}\Big) \\[1em] \Rightarrow 3(m_1 + m_2) = {6m_1 + 2m_2} \\[1em] \Rightarrow 3m_1 + 3m_2 = 6m_1 + 2m_2 \\[1em] \Rightarrow 3m_2 - 2m_2 = 6m_1 - 3m_1 \\[1em] \Rightarrow m_2 = 3m_1 \\[1em] \Rightarrow \dfrac{m_1}{m_2} = \dfrac{1}{3}.$

Solving for x-coordinate with the ratio $m_1 = 1$ and $m_2 = 3$:

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

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

(ii) Solving length of AP

By using Distance Formula

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

A(-4, 2) and $P\Big(\dfrac{-9}{4}, 3\Big)$.

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

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

In what ratio is the line joining P(5, 3) and Q(-5, 3) divided by the x-axis? Also, find the co-ordinates of the point of intersection.

Answer

Let point on x-axis dividing PQ be (a, 0).

By section formula,

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

Substituting values we get :

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

The negative ratio shows that the line PQ does not intersects with x-axis.

Hence, the x-axis does not intersect the line PQ, so no ratio or point of intersection exists.

Find a point P which divides internally the line segment joining the points A(-3, 9) and B(1, -3) in the ratio 1 : 3.

Answer

Let Point P divides the line segment joining A(−3, 9) and B(1, −3) internally in the ratio m₁ : m₂ = 1 : 3.

By using section formula,

P(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:

$P = \Big(\dfrac{1 \times 1 + 3 \times -3}{1 + 3}, \dfrac{1 \times -3 + 3 \times 9}{1 + 3}\Big) \\[1em] = \Big(\dfrac{1 - 9}{4}, \dfrac{-3 + 27}{4}\Big) \\[1em] = \Big(\dfrac{-8}{4}, \dfrac{24}{4}\Big) \\[1em] = (-2, 6).$