Section & Mid-Point Formulae

Let point P be (x, y).

Given,

m₁ : m₂ = 5 : 3

By section-formula,

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

Substituting values we get :

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

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

Let point P be (x, y).

$$\begin{aligned}\Rightarrow \frac{AP}{AB} = \frac{2}{5} \\[1em] \Rightarrow \frac{AP}{AP + PB} = \frac{2}{5} \\[1em] \Rightarrow 5AP = 2AP + 2PB \\[1em] \Rightarrow 3AP = 2PB \\[1em] \Rightarrow \frac{AP}{PB} = \frac{2}{3} \\[1em] \Rightarrow AP : PB = 2 : 3.\end{aligned}$$

m₁ : m₂ = AP : PB = 2 : 3.

By section-formula,

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

Substituting values we get :

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

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

Given,

m₁ : m₂ = 2 : 3

By section-formula,

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

Substituting values we get :

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

By section-formula,

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

Substituting values we get :

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

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

Let point P be (x, y).

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

⇒ AP = 2BP

⇒ $\frac{AP}{BP} = \frac{2}{1}$

⇒ AP : BP = 2 : 1.

⇒ m₁ : m₂ = AP : PB = 2 : 1

By section-formula,

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

Substituting values we get :

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

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

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

Substituting values we get :

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

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 :

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

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

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

By using mid-point formula,

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

Substituting values we get :

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

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

Substituting values we get :

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

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

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

Substituting values we get :

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

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

By using mid-point formula,

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

L is the mid-point of AB.

Substituting values we get :

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

Given,

M is the mid-point of AC.

Substituting values we get :

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

By distance formula,

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

Substituting values we get :

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

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

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

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

Substituting values we get :

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

Hence, p = -5 and q = 3.

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 = $\Big(\frac{x_1 + x_2}{2}\Big)$

Substituting values we get :

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

By using mid-point formula,

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

Substituting values we get :

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

Q = (x₁, y₁) = (-6, 10).

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

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

Substituting values we get :

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

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

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

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

By section-formula,

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

Substituting values we get :

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

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

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

By section-formula,

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

Substituting values we get :

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

By section-formula,

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

Substitute values we get:

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

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

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

Substituting values we get :

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

By section-formula,

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

Substituting values we get :

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

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

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

Substituting values we get :

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

Thus, m₁ : m₂ = 2 : 1.

By section-formula,

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

Substitute values we get:

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

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

Substituting values we get :

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

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,

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

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

Substitute values we get,

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

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

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

Substitute values we get,

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

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

(ii) Given,

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

m₁ : m₂ = 4 : 3.

By section-formula,

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

Substitute values we get,

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

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

(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 = $\frac{1}{2}$ × Sum of parallel sides × Distance between them

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

Hence, area of PQNM = 24.5 sq.units.

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,

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

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

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

By section-formula,

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

Substituting values we get,

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

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

Thus, m₁ : m₂ = 2 : 1.

By section-formula,

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

Substituting values we get,

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

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

Substituting values we get,

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

By section-formula,

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

Substitute values we get,

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

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

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

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

By mid-point formula,

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

Given,

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

Substitute values we get,

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

Given,

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

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

Given,

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

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

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) = $\frac{10}{2}$

⇒ (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) = $\frac{6}{2}$

⇒ (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).

Given,

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

By using mid-point formula,

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

The Midpoint of Diagonal AC :

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

The Midpoint of Diagonal BD :

$$\begin{aligned}\Rightarrow M_{(BD)} = \Big(\frac{1 + 1}{2}, \frac{0 + 2}{2}\Big) \\[1em] \Rightarrow M_{(BD)} = \Big(\frac{2}{2}, \frac{2}{2}\Big) \\[1em] \Rightarrow M_{(BD)} = (1, 1)\end{aligned}$$

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.

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

Mid-point of AC :

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

Midpoint of Diagonal BD :

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

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

⇒ $\frac{-2 + a}{2}$ = 1

⇒ -2 + a = 2

⇒ a = 2 + 2

⇒ a = 4.

Equating the y-coordinates:

⇒ 1 = $\frac{b}{2}$

⇒ b = 2.

Hence, a = 4 and b = 2.

Let coordinates of D be (x, y).

By using mid-point formula,

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

Mid-point of AC :

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

Midpoint of Diagonal BD :

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

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:

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

⇒ 3 + x = 1

⇒ x = 1 - 3

⇒ x = -2.

Equating the y-coordinates:

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

⇒ 1 + y = 2

⇒ y = 2 - 1

⇒ y = 1.

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

Using the Midpoint Formula,

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

Let D be the midpoint of BC.

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

Let E be the midpoint of AC.

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

Let F be the midpoint of AB.

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

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:

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

Length of Median BE:

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

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

Length of Median CF:

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

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

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

Substituting values we get :

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

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

By using the centroid formula,

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

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

Substitute values we get:

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

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

The medians of a triangle intersect at centroid of triangle.

By using the centroid formula,

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

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

Substitute values we get:

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

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

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

By using the centroid formula,

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

Given,

G(-2, 1) is the centroid.

Substituting values we get :

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

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

By using the centroid formula,

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

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

Solving for x-coordinate:

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

Solving for y-coordinate:

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

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

By using the centroid formula,

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

Given,

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

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

Solving for x-coordinate :

$$\begin{aligned}\Rightarrow 4 = \frac{1 + 4 + a}{3} \\[1em] \Rightarrow 12 = 5 + a \\[1em] \Rightarrow a = 12 - 5 \\[1em] \Rightarrow a = 7.\end{aligned}$$

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

Solving for y-coordinate :

$$\begin{aligned}\Rightarrow 3 = \frac{3 + b + 1}{3} \\[1em] \Rightarrow 9 = 4 + b \\[1em] \Rightarrow b = 9 - 4 \\[1em] \Rightarrow b = 5.\end{aligned}$$

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:

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

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

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

Substituting values we get :

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

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

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

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

(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 $P\Big(\frac{-9}{4}, 3\Big)$.

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

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

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

By section formula,

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

Substituting values we get :

$$\begin{aligned}\Rightarrow 0 = \frac{m_1 \times 3 + m_2 \times 3}{m_1 + m_2} \\[1em] \Rightarrow 0 = 3m_1 + 3m_2 \\[1em] \Rightarrow 3m_1 = -3m_2 \\[1em] \Rightarrow \frac{m_1}{m_2} = -\frac{3}{3} \\[1em] \Rightarrow \frac{m_1}{m_2} = -\frac{1}{1}\end{aligned}$$

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.

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

Substituting values we get:

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

Hence, coordinates of P = (-2, 6).

Let point P be (x, y).

Given,

m₁ : m₂ = 3 : 1

By section-formula,

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

Substituting values we get :

$$\begin{aligned}\Rightarrow (x, y) = \Big(\frac{3 \times 9 + 1 \times 5}{3 + 1}, \frac{3 \times 6 + 1 \times -2}{3 + 1}\Big) \\[1em] = \Big(\frac{27 + 5}{4}, \frac{18 - 2}{4}\Big) \\[1em] = \Big(\frac{32}{4}, \frac{16}{4}\Big) \\[1em] = (8, 4).\end{aligned}$$

Hence, Option 3 is the correct option.

Let point P be (x, y).

Given,

m₁ : m₂ = 1 : 2

By section-formula,

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

Substituting values we get :

$$\begin{aligned}\Rightarrow (x, y) = \Big(\frac{1 \times 5 + 2 \times 2}{1 + 2}, \frac{1 \times (-6) + 2 \times 3}{1 + 2}\Big) \\[1em] = \Big(\frac{5 + 4}{3}, \frac{-6 + 6}{3}\Big) \\[1em] = \Big(\frac{9}{3}, \frac{0}{3}\Big) \\[1em] = (3, 0).\end{aligned}$$

Hence, Option 3 is the correct option.

Let point P be (x, y).

Given,

m₁ : m₂ = 3 : 2

By section-formula,

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

Substituting values we get :

$$\begin{aligned}\Rightarrow (x, y) = \Big(\frac{3 \times 6 + 2 \times 3}{3 + 2}, \frac{3 \times 7 + 2 \times (-2)}{3 + 2}\Big) \\[1em] = \Big(\frac{18 + 6}{5}, \frac{21 - 4}{5}\Big) \\[1em] = \Big(\frac{24}{5}, \frac{17}{5}\Big).\end{aligned}$$

Here, both x and y are positive.

Therefore, the point lies in the 1st Quadrant.

Hence, Option 1 is the correct option.

Given,

R = (k, 4)

m₁ : m₂ = 2 : 3

R(k, 4) divides the line segment joining the points P(2, 6) and Q(5, 1) in the ratio 2 : 3.

By section-formula,

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

Substituting values we get :

$$\begin{aligned}\Rightarrow (k, 4) = \Big(\frac{2 \times 5 + 3 \times 2}{2 + 3}, \frac{2 \times 1 + 3 \times 6}{2 + 3}\Big) \\[1em] \Rightarrow (k, 4) = \Big(\frac{10 + 6}{5}, \frac{2 + 18}{5}\Big) \\[1em] \Rightarrow (k, 4) = \Big(\frac{16}{5}, \frac{20}{5}\Big) \\[1em] \Rightarrow (k, 4) = \Big(\frac{16}{5}, 4\Big).\end{aligned}$$

Thus, k = $\frac{16}{5}$.

Hence, Option 4 is the correct option.

Since, points A, B and C are collinear.

∴ Slope of AB = Slope of BC.

$$\begin{aligned}\Rightarrow \frac{-2 - y}{3 - x} = \frac{-5 - (-2)}{4 - 3} \\[1em] \Rightarrow \frac{-2 - y}{3 - x} = \frac{-5 + 2}{1} \\[1em] \Rightarrow \frac{-2 - y}{3 - x} = -3 \\[1em] \Rightarrow -2 - y = -3(3 - x) \\[1em] \Rightarrow -2 - y = -9 + 3x \\[1em] \Rightarrow y = -2 + 9 - 3x \\[1em] \Rightarrow y = 7 - 3x.\end{aligned}$$

Hence, Option 4 is the correct option.

Let point P be (6, 2).

Given,

m₁ : m₂ = 3 : 1

P(6, 2) divides the line segment joining A(6, 5) and B(4, y) in the ratio 3 : 1.

By section-formula,

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

We use the y–coordinate to find y.

$$\begin{aligned}\Rightarrow 2 = \frac{3 \times y + 1 \times 5}{3 + 1} \\[1em] \Rightarrow 2 = \frac{3y + 5}{4} \\[1em] \Rightarrow 8 = 3y + 5 \\[1em] \Rightarrow 3y = 3 \\[1em] \Rightarrow y = 1.\end{aligned}$$

Hence, Option 1 is the correct option.

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

By section-formula,

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

Using the y–coordinate to find ratio.

$$\begin{aligned}\Rightarrow 2 = \Big(\frac{k(4) + 1(1)}{k + 1}\Big) \\[1em] \Rightarrow 2 = \Big(\frac{4k + 1}{k + 1}\Big) \\[1em] \Rightarrow 2(k + 1) = 4k + 1 \\[1em] \Rightarrow 2k + 2 = 4k + 1 \\[1em] \Rightarrow 2 - 1 = 4k - 2k \\[1em] \Rightarrow 2k = 1 \\[1em] \Rightarrow k = \frac{1}{2}\\[1em] \Rightarrow k : 1 = \frac{1}{2} : 1 = 1 : 2.\end{aligned}$$

Hence, Option 1 is the correct option.

Given,

AB is divided by the x-axis, thus y-coordinate = 0 at point of division.

A(-7, 2) and B(3, -8)

Let ratio be m : n.

By section-formula,

y = $\frac{my_2 + ny_1}{m + n}$

Substituting values we get :

⇒ 0 = $\frac{m \times (-8) + n \times (2)}{m + n}$

⇒ 0 = -8m + 2n

⇒ 8m = 2n

⇒ $\frac{m}{n} = \frac{2}{8}$

⇒ $\frac{m}{n} = \frac{1}{4}$

⇒ m : n = 1 : 4.

Hence, option 1 is the correct option.

Let the point where y-axis divides the line segment be R(0, y).

Let the ratio be m₁ : m₂.

Using section-formula,

$$\begin{aligned}\Rightarrow x = \frac{m_1x_2 + m_2x_1}{m_1 + m_2} \\[1em] \Rightarrow 0 = \frac{m_1 \times 8 + m_2 \times (-4)}{m_1 + m_2} \\[1em] \Rightarrow 0 = 8m_1 - 4m_2 \\[1em] \Rightarrow 8m_1 = 4m_2 \\[1em] \Rightarrow \frac{m_1}{m_2} = \frac{4}{8} = \frac{1}{2}.\end{aligned}$$

Thus, the required ratio is 1 : 2.

Hence, Option 3 is the correct option.

Let point P be (x, y).

Given,

m₁ : m₂ = k : 1

By section-formula,

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

Substituting values we get :

$$\begin{aligned}\Rightarrow (x, y) = \Big(\frac{k \times 9 + 1 \times (-1)}{k + 1}, \frac{k \times 8 + 1 \times 3}{k + 1}\Big) \\[1em] = \Big(\frac{9k - 1}{k + 1}, \frac{8k + 3}{k + 1}\Big).\end{aligned}$$

Since P lies on the line x - y + 2 = 0, substituting values of x and y:

$$\begin{aligned}\Rightarrow \frac{9k - 1}{k + 1} - \frac{8k + 3}{k + 1} + 2 = 0 \\[1em] \Rightarrow \frac{9k - 1 - (8k + 3)}{k + 1} + 2 = 0 \\[1em] \Rightarrow \frac{9k - 1 - 8k - 3}{k + 1} + 2 = 0 \\[1em] \Rightarrow \frac{k - 4}{k + 1} + 2 = 0 \\[1em] \Rightarrow \frac{k - 4 + 2(k + 1)}{k + 1} = 0 \\[1em] \Rightarrow \frac{k - 4 + 2k + 2}{k + 1} = 0 \\[1em] \Rightarrow \frac{3k - 2}{k + 1} = 0 \\[1em] \Rightarrow 3k - 2 = 0 \\[1em] \Rightarrow 3k = 2 \\[1em] \Rightarrow k = \frac{2}{3}.\end{aligned}$$

Hence, Option 4 is the correct option.

Since P and Q trisect the line segment AB, the point Q(3, y) divides A(7, -2) and B(1, -5) in the ratio 2 : 1.

Let point Q be (3, y).

Given,

m₁ : m₂ = 2 : 1

By section-formula,

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

Substituting values we get :

$$\begin{aligned}\Rightarrow (3, y) = \Big(\frac{2 \times 1 + 1 \times 7}{2 + 1}, \frac{2 \times (-5) + 1 \times (-2)}{2 + 1}\Big) \\[1em] \Rightarrow (3, y) = \Big(\frac{2 + 7}{3}, \frac{-10 - 2}{3}\Big) \\[1em] \Rightarrow (3, y) = \Big(\frac{9}{3}, \frac{-12}{3}\Big) \\[1em] \Rightarrow (3, y) = (3, -4).\end{aligned}$$

Thus, y = -4.

Hence, Option 1 is the correct option.

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

By section-formula,

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

Substituting values we get :

$$\begin{aligned}\Rightarrow 6 = \Big(\frac{k(8) + 1(4)}{k + 1}\Big) \\[1em] \Rightarrow 6 = \Big(\frac{8k + 4}{k + 1}\Big) \\[1em] \Rightarrow 6(k + 1) = 8k + 4 \\[1em] \Rightarrow 6k + 6 = 8k + 4 \\[1em] \Rightarrow 6 - 4 = 8k - 6k \\[1em] \Rightarrow 2 = 2k \\[1em] \Rightarrow k = \frac{1}{1} = 1:1.\end{aligned}$$

This means that P is the midpoint of AB.

∴ AP = $\frac{1}{2}$ AB.

Hence, Option 4 is the correct option.

Let the mid-point be M(x, y).

By mid-point formula,

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

Substituting values we get :

$$\begin{aligned}\Rightarrow M(x, y) = \Big(\frac{-3 + 7}{2}, \frac{2 + 6}{2}\Big) \\[1em] \Rightarrow \Big(\frac{4}{2}, \frac{8}{2}\Big) \\[1em] \Rightarrow (2, 4).\end{aligned}$$

Hence, Option 3 is the correct option.

Since AD is a median, D is the mid-point of BC.

Let point D be (x, y).

By mid-point formula,

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

$$\begin{aligned}\Rightarrow (x, y) = \Big(\frac{6 + 1}{2}, \frac{5 + 4}{2}\Big) \\[1em] \Rightarrow \Big(\frac{7}{2}, \frac{9}{2}\Big).\end{aligned}$$

Hence, Option 3 is the correct option.

Given, (3, -6) is the mid-point of the line segment joining (0, 0) and (x, y).

$$\begin{aligned}\therefore (3, -6) = \Big(\frac{0 + x}{2}, \frac{0 + y}{2}\Big) \\[1em] \Rightarrow (3, -6) = \Big(\frac{x}{2}, \frac{y}{2}\Big) \\[1em] \Rightarrow 3 = \frac{x}{2} \text{ and } -6 = \frac{y}{2} \\[1em] \Rightarrow x = 6 \text{ and } y = -12\end{aligned}$$

(x, y) = (6, -12).

Hence, Option 3 is the correct option.

Let the coordinates of P be (x, y).

Given, the origin (0, 0) is the mid-point of PQ.

By mid-point formula,

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

Substituting values we get :

$$\begin{aligned}\Rightarrow (0, 0) = \Big(\frac{x + 3}{2}, \frac{y + (-4)}{2}\Big) \\[1em] \Rightarrow 0 = \frac{x + 3}{2} \text{ and } 0 = \frac{y - 4}{2} \\[1em] \Rightarrow x + 3 = 0 \text{ and } y - 4 = 0 \\[1em] \Rightarrow x = -3 \text{ and } y = 4 \\[1em] \Rightarrow P = (x, y) = (-3, 4).\end{aligned}$$

Hence, Option 4 is the correct option.

Given,

Mid-point of AB = (-1, 1).

By mid-point formula,

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

Substituting values we get :

$$\begin{aligned}\Rightarrow (-1, 1) = \Big(\frac{-3 + 1}{2}, \frac{b + (b + 4)}{2}\Big) \\[1em] \Rightarrow (-1, 1) = \Big(\frac{-2}{2}, \frac{2b + 4}{2}\Big) \\[1em] \Rightarrow (-1, 1) = (-1, b + 2)\end{aligned}$$

Comparing the y-coordinates, we get :

⇒ 1 = b + 2

⇒ b = -1.

Hence, Option 1 is the correct option.

Given,

R(5, 7) is the mid-point of the line segment joining the points P(3, y) and Q(x, 9).

By mid-point formula,

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

Substituting values we get :

$$\begin{aligned}\Rightarrow (5, 7) = \Big(\frac{3 + x}{2}, \frac{y + 9}{2}\Big) \\[1em] \Rightarrow 5 = \frac{3 + x}{2} \text{ and } 7 = \frac{y + 9}{2} \\[1em] \Rightarrow 10 = 3 + x \text{ and } 14 = y + 9 \\[1em] \Rightarrow x = 7 \text{ and } y = 5.\end{aligned}$$

x + y = 7 + 5 = 12.

Hence, Option 3 is the correct option.

Given,

Mid-point of the line segment joining (4p, 5) and (2, 3q) is (5, 5p - 1).

By mid-point formula,

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

Substituting values we get :

$$\begin{aligned}\Rightarrow (5, 5p - 1) = \Big(\frac{4p + 2}{2}, \frac{5 + 3q}{2}\Big) \\[1em] \Rightarrow (5, 5p - 1) = \Big(\frac{2(2p + 1)}{2}, \frac{5 + 3q}{2}\Big) \\[1em] \Rightarrow (5, 5p - 1) = \Big(2p + 1, \frac{5 + 3q}{2}\Big)\end{aligned}$$

Comparing the x coordinates, we get :

⇒ 2p + 1 = 5

⇒ 2p = 5 - 1

⇒ 2p = 4

⇒ p = $\frac{4}{2}$

⇒ p = 2.

Comparing y-coordinates we get :

$$\begin{aligned}\Rightarrow 5p - 1 = \frac{5 + 3q}{2} \\[1em] \Rightarrow 5 \times 2 - 1 = \frac{5 + 3q}{2} \\[1em] \Rightarrow 9 = \frac{5 + 3q}{2} \\[1em] \Rightarrow 18 = 5 + 3q \\[1em] \Rightarrow 3q = 18 - 5 \\[1em] \Rightarrow 3q = 13 \\[1em] \Rightarrow q = \frac{13}{3}.\end{aligned}$$

p = 2 and q = $\frac{13}{3}$.

Hence, Option 3 is the correct option.

Let P(0, a) be the point on y-axis and Q(b, 0) be the point on x-axis.

Given, (2, -5) is the mid-point of PQ.

By mid-point formula,

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

Substituting values we get :

$$\begin{aligned}\Rightarrow (2, -5) = \Big(\frac{0 + b}{2}, \frac{a + 0}{2}\Big) \\[1em] \Rightarrow (2, -5) = \Big(\frac{b}{2}, \frac{a}{2}\Big) \\[1em] \Rightarrow 2 = \frac{b}{2} \text{ and } -5 = \frac{a}{2} \\[1em] \Rightarrow b = 4 \text{ and } a = -10.\end{aligned}$$

P(0, -10) and Q(4, 0)

Hence, Option 4 is the correct option.

Given,

Point R = $\Big(\frac{23}{5}, \frac{33}{5}\Big)$ and P(3, 5), Q(x, y).

Given,

m₁ : m₂ = 2 : 3

By section-formula,

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

Substituting values we get :

$$\begin{aligned}\Rightarrow \Big(\frac{23}{5}, \frac{33}{5}\Big) = \Big(\frac{2x + 3 \times 3}{2 + 3}, \frac{2y + 3 \times 5}{2 + 3}\Big) \\[1em] \Rightarrow \frac{23}{5} = \frac{2x + 9}{5}, \frac{33}{5} = \frac{2y + 15}{5} \\[1em] \Rightarrow 23 = 2x + 9, 33 = 2y + 15 \\[1em] \Rightarrow 23 - 9 = 2x, 33 - 15 = 2y \\[1em] \Rightarrow 14 = 2x, 18 = 2y \\[1em] \Rightarrow x = \frac{14}{2}, y = \frac{18}{2} \\[1em] \Rightarrow x = 7, y = 9.\end{aligned}$$

Therefore, the values of x and y are 7 and 9 respectively.

Hence, Option 4 is the correct option.

By centroid formula,

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

Substituting values we get :

$$\begin{aligned}\Rightarrow G = \Big(\frac{-4 + 6 + 4}{3}, \frac{-2 + 2 + 6}{3}\Big) \\[1em] = \Big(\frac{6}{3}, \frac{6}{3}\Big) \\[1em] = (2, 2).\end{aligned}$$

Hence, Option 1 is the correct option.

Let the coordinates of C be (x, y).

By centroid formula,

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

Given,

G(0, -3) is the centroid of the triangle.

Substituting values we get :

$$\begin{aligned}\Rightarrow (0, -3) = \Big(\frac{-1 + 5 + x}{3}, \frac{4 + 2 + y}{3}\Big) \\[1em] \Rightarrow (0, -3) = \Big(\frac{4 + x}{3}, \frac{6 + y}{3}\Big) \\[1em] \Rightarrow 0 = \frac{4 + x}{3} \text{ and } -3 = \frac{6 + y}{3} \\[1em] \Rightarrow 4 + x = 0 \text{ and } 6 + y = -9 \\[1em] \Rightarrow x = -4 \text{ and } y = -9 - 6 \\[1em] \Rightarrow x = -4 \text{ and } y = -15.\end{aligned}$$

C = (x, y) = (-4, -15).

Hence, Option 3 is the correct option.

By formula,

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

$$\begin{aligned}\therefore (5, -5) = \Big(\frac{3 + 2 + x}{3}, \frac{(-3) + (-6) + y}{3}\Big) \\[1em] \Rightarrow (5, -5) = \Big(\frac{x + 5}{3}, \frac{y - 9}{3}\Big) \\[1em] \Rightarrow \frac{x + 5}{3} = 5 \text{ and } \frac{y - 9}{3} = -5 \\[1em] \Rightarrow x + 5 = 15 \text{ and } y - 9 = -15 \\[1em] \Rightarrow x = 15 - 5 = 10 \text{ and } y = -15 + 9 = -6.\end{aligned}$$

C(x, y) = (10, -6).

Since, centroid is the point of intersection of all the three medians of a triangle.

∴ AD is the median.

∴ D is mid-point of BC.

$$\begin{aligned}D = \Big(\frac{2 + 10}{2}, \frac{(-6) + (-6)}{2}\Big) \\[1em] = \Big(\frac{12}{2}, \frac{-12}{2}\Big) \\[1em] = (6, -6).\end{aligned}$$

Hence, Option 3 is the correct option.

Given,

The points dividing PQ into four equal parts be A, B and C such that A is nearest to P.

Since the line segment PQ is divided into four equal parts, the point A divides PQ in the ratio 1 : 3.

By section-formula,

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

Substituting values we get :

$$\begin{aligned}\Rightarrow (x, y) = \Big(\frac{1 \times 9 + 3 \times (-3)}{1 + 3}, \frac{1 \times (-4) + 3 \times 8}{1 + 3}\Big) \\[1em] \Rightarrow \Big(\frac{9 - 9}{4}, \frac{-4 + 24}{4}\Big) \\[1em] \Rightarrow \Big(\frac{0}{4}, \frac{20}{4}\Big) \\[1em] \Rightarrow (0, 5).\end{aligned}$$

Hence, Option 2 is the correct option.

Let the required point be P(x, y) which divides A(2, -2) and B(3, 7) in the ratio k : 1.

By section-formula,

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

Substituting values, we get :

$$\begin{aligned}\Rightarrow (x, y) = \Big(\frac{k \times 3 + 1 \times 2}{k + 1}, \frac{k \times 7 + 1 \times (-2)}{k + 1}\Big) \\[1em] \Rightarrow (x, y) = \Big(\frac{3k + 2}{k + 1}, \frac{7k - 2}{k + 1}\Big)\end{aligned}$$

Since P lies on the line 2x + y - 4 = 0, substituting the values of x and y:

$$\begin{aligned}\Rightarrow 2\Big(\frac{3k + 2}{k + 1}\Big) + \Big(\frac{7k - 2}{k + 1}\Big) - 4 = 0 \\[1em] \Rightarrow \frac{6k + 4 + 7k - 2}{k + 1} - 4 = 0 \\[1em] \Rightarrow \frac{13k + 2}{k + 1} = 4 \\[1em] \Rightarrow 13k + 2 = 4(k + 1) \\[1em] \Rightarrow 13k + 2 = 4k + 4 \\[1em] \Rightarrow 9k = 2 \\[1em] \Rightarrow k = \frac{2}{9} \\[1em] \Rightarrow k : 1 = \frac{2}{9} : 1 = 2 : 9.\end{aligned}$$

Hence, Option 4 is the correct option.

Let the end points of the diameter be A(-4, 2) and B(4, -3). The centre of the circle is the mid-point of the diameter AB.

By mid-point formula,

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

Substituting values we get :

$$\begin{aligned}\Rightarrow (x, y) = \Big(\frac{-4 + 4}{2}, \frac{2 + (-3)}{2}\Big) \\[1em] = \Big(\frac{0}{2}, \frac{-1}{2}\Big) \\[1em] = \Big(0, -\frac{1}{2}\Big).\end{aligned}$$

Hence, Option 3 is the correct option.

Let one end of the diameter be A(4, 0) and the other end be B(x, y). Given that the centre of the circle is (4, 4), which is the mid-point of AB.

By mid-point formula,

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

Substituting values we get :

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

The coordinates of the other end of the diameter are (4, 8).

Hence, Option 2 is the correct option.

In a parallelogram, the diagonals bisect each other. Therefore, the mid-point of AC = mid-point of BD.

By mid-point formula,

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

Substituting values, we get :

For diagonal AC :

$\text{Mid-point of AC} = \Big(\frac{1 + x}{2}, \frac{2 + 6}{2}\Big) = \Big(\frac{1 + x}{2}, 4\Big)$

For diagonal BD:

$\text{Mid-point of BD} = \Big(\frac{4 + 3}{2}, \frac{y + 5}{2}\Big) = \Big(\frac{7}{2}, \frac{y + 5}{2}\Big)$

Since both mid-points are equal, we equate their coordinates:

$$\begin{aligned}\Rightarrow \Big(\frac{1 + x}{2}, 4\Big) = \Big(\frac{7}{2}, \frac{y + 5}{2}\Big) \\[1em] \Rightarrow \frac{1 + x}{2} = \frac{7}{2}, 4 = \frac{y + 5}{2}\\[1em] \Rightarrow 1 + x = 7, y + 5 = 8 \\[1em] \Rightarrow x = 7 - 1, y = 8 - 5 \\[1em] \Rightarrow x = 6, y = 3.\end{aligned}$$

(x, y) = (6, 3).

Hence, Option 1 is the correct option.

In a parallelogram, the diagonals bisect each other. Therefore, the mid-point of AC = mid-point of BD.

By mid-point formula,

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

Substituting values, we get :

For diagonal AC:

$$\begin{aligned}\text{Mid-point of AC} = \Big(\frac{-2 + 8}{2}, \frac{3 + 3}{2}\Big) = \Big(\frac{6}{2}, \frac{6}{2}\Big) \\[1em] = (3, 3).\end{aligned}$$

Let point D be (x, y).

For diagonal BD:

$$\begin{aligned}\text{Mid-point of BD} = \Big(\frac{6 + x}{2}, \frac{7 + y}{2}\Big) \\[1em] \Rightarrow (3, 3) = \Big(\frac{6 + x}{2}, \frac{7 + y}{2}\Big) \\[1em] \Rightarrow 3 = \frac{6 + x}{2} \text{ and } 3 = \frac{7 + y}{2} \\[1em] \Rightarrow 6 = 6 + x \text{ and } 6= 7 + y \\[1em] \Rightarrow x = 0 \text{ and } y = 6 - 7 = -1.\end{aligned}$$

D = (x, y) = (0, -1).

Hence, Option 2 is the correct option.

Given,

The vertices of the triangle be A(1, 4), B(4, 1) and C(x, 4). The centroid G of a triangle is given by the formula:

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

Substituting the given values we get :

$$\begin{aligned}\Rightarrow (4, 3) = \Big(\frac{1 + 4 + x}{3}, \frac{4 + 1 + 4}{3}\Big) \\[1em] \Rightarrow (4, 3) = \Big(\frac{5 + x}{3}, \frac{9}{3}\Big) \\[1em] \Rightarrow (4, 3) = \Big(\frac{5 + x}{3}, 3\Big) \\[1em] \Rightarrow 4 = \frac{5 + x}{3} \\[1em] \Rightarrow 12 = 5 + x \\[1em] \Rightarrow x = 12 - 5 = 7.\end{aligned}$$

Hence, Option 3 is the correct option.

Let point P be (x, y), which divides the line segment joining the points (-3, 4) and (7, -6) in the ratio 1 : 2.

Given,

m₁ : m₂ = 1 : 2

By section-formula,

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

Substitute values we get:

$$\begin{aligned}\Rightarrow (x, y) = \Big(\frac{1(7) + 2(-3)}{1 + 2}, \frac{1(-6) + 2(4)}{1 + 2}\Big) \\[1em] = \Big(\frac{7 - 6}{3}, \frac{-6 + 8}{3}\Big) \\[1em] = \Big(\frac{1}{3}, \frac{2}{3}\Big).\end{aligned}$$

So, Assertion (A) is true.

The reason states the standard section formula, which is exactly the formula used to obtain the point in the Assertion.

So, Reason (R) is true and it is the correct explanation of Assertion (A).

Hence, Option 1 is the correct option.

By mid-point formula,

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

We know that the centre is the mid-point of a diameter of a circle.

Given,

Centre = (2, -3)

One end of diameter = (1, 4)

Let the other end of the diameter be (a, b).

Substituting values, we get :

$$\begin{aligned}\Rightarrow (2, -3) = \Big(\frac{1 + a}{2}, \frac{4 + b}{2}\Big) \\[1em] \Rightarrow 2 = \frac{1 + a}{2} \text{ and } -3 = \frac{4 + b}{2} \\[1em] \Rightarrow 4 = 1 + a \text{ and } -6 = 4 + b \\[1em] \Rightarrow a = 3 \text{ and } b = -10.\end{aligned}$$

The correct coordinates of the other end are (3, -10).

So, Assertion (A) is false.

Since the centre is the mid-point of a diameter, it is equidistant from each end of the diameter.

So, Reason (R) is true.

Hence, Option 4 is the correct option.

By mid-point formula,

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

Mid-point of AC :

$M_{AC} = \Big(\frac{x + 9}{2}, \frac{1 + 4}{2}\Big) = \Big(\frac{x + 9}{2}, \frac{5}{2}\Big)$

Mid-point of BD :

$M_{BD} = \Big(\frac{8 + 7}{2}, \frac{2 + 3}{2}\Big) = \Big(\frac{15}{2}, \frac{5}{2}\Big)$

Diagonals of a parallelogram bisect each other.

Thus,

Mid-point of AC = Mid-point of BD

$$\begin{aligned}\Rightarrow \Big(\frac{x + 9}{2}, \frac{5}{2}\Big) = \Big(\frac{15}{2}, \frac{5}{2}\Big) \\[1em] \Rightarrow \frac{x + 9}{2} = \frac{15}{2} \\[1em] \Rightarrow x + 9 = 15 \\[1em] \Rightarrow x = 15 - 9 = 6.\end{aligned}$$

So, Assertion (A) is true.

It is also true that the adjacent angles of a parallelogram are supplementary. However, the value of x is obtained using the property that the diagonals of a parallelogram bisect each other, not from the adjacent angles being supplementary.

So, Reason (R) is true, but it is not the correct explanation of Assertion (A).

Hence, Option 2 is the correct option.

By the centroid formula,

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

Substitute values we get,

$$\begin{aligned}\Rightarrow (x, y) = \Big( \frac{-1 + 4 + 6}{3}, \frac{-4 + 3 + (-2)}{3} \Big) \\[1em] \Rightarrow (x, y) = \Big( \frac{9}{3}, \frac{-3}{3} \Big) \\[1em] \Rightarrow (x, y) = (3, -1).\end{aligned}$$

So, Assertion (A) is true.

The correct centroid formula is $\Big(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\Big)$. The formula given in the Reason has the wrong denominator (2 instead of 3) and uses x-coordinates in place of the y-coordinates for the second term.

So, Reason (R) is false.

Hence, Option 3 is the correct option.

By two point form,

Equation of line :

⇒ y - y₁ = $\frac{y_2 - y_1}{x_2 - x_1}$(x - x₁)

Equation of AB :

$$\begin{aligned}\Rightarrow y - 9 = \frac{2 - 9}{4 - (-3)}[x - (-3)] \\[1em] \Rightarrow y - 9 = \frac{-7}{4 + 3}[x + 3] \\[1em] \Rightarrow y - 9 = \frac{-7}{7}[x + 3] \\[1em] \Rightarrow y - 9 = -1[x + 3] \\[1em] \Rightarrow y - 9 = -x - 3 \\[1em] \Rightarrow y + x = -3 + 9 \\[1em] \Rightarrow x + y = 6.\end{aligned}$$

Solving equation y = 2 + 3x and x + y = 6 simultaneously,

⇒ x + y = 6 .......(1)

⇒ y = 2 + 3x .......(2)

Substituting value of y from equation (2) in (1), we get :

⇒ x + (2 + 3x) = 6

⇒ 4x + 2 = 6

⇒ 4x = 6 - 2

⇒ 4x = 4

⇒ x = $\frac{4}{4}$

⇒ x = 1.

Substituting value of x in equation (2), we get :

⇒ y = 2 + 3(1) = 2 + 3 = 5.

Let (1, 5) divide the line AB in the ratio k : 1.

By section formula,

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

Substituting values we get :

$$\begin{aligned}\Rightarrow (1, 5) = \Big(\frac{k \times 4 + 1 \times -3}{k + 1}, \frac{k \times 2 + 1 \times 9}{k + 1}\Big) \\[1em] \Rightarrow (1, 5) = \Big(\frac{4k - 3}{k + 1}, \frac{2k + 9}{k + 1}\Big) \\[1em] \Rightarrow 1 = \frac{4k - 3}{k + 1} \text{ or } 5 = \frac{2k + 9}{k + 1} \\[1em] \Rightarrow 1(k + 1) = 4k - 3 \text{ or } 5(k + 1) = 2k + 9 \\[1em] \Rightarrow k + 1 = 4k - 3 \text{ or } 5k + 5 = 2k + 9 \\[1em] \Rightarrow 4k - k = 1 + 3 \text{ or } 5k - 2k = 9 - 5 \\[1em] \Rightarrow 3k = 4 \text{ or } 3k = 4 \\[1em] \Rightarrow k = \frac{4}{3} \\[1em] \Rightarrow k : 1 = \frac{4}{3} : 1 = 4 : 3.\end{aligned}$$

Hence, the line y = 2 + 3x divides the line segment AB in the ratio 4 : 3.

Let coordinates of C be (0, b) and D be (a, 0).

Given,

AC : AD = 1 : 4

Let AC = x and AD = 4x.

From figure,

⇒ AD = AC + CD

⇒ 4x = x + CD

⇒ CD = 4x - x = 3x.

AC : CD = 1 : 3.

By section formula,

$$\begin{aligned}\Rightarrow (x, y) = \Big(\frac{m_1x_2 + m_2x_1}{m_1 + m_2}, \frac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big) \\[1em] \Rightarrow (0, b) = \Big(\frac{1 \times a + 3 \times -2}{1 + 3}, \frac{1 \times 0 + 3 \times 6}{1 + 3}\Big) \\[1em] \Rightarrow (0, b) = \Big(\frac{a - 6}{4}, \frac{0 + 18}{4}\Big) \\[1em] \Rightarrow (0, b) = \Big(\frac{a - 6}{4}, \frac{18}{4}\Big) \\[1em] \Rightarrow \frac{a - 6}{4} = 0 \text{ and } b = \frac{18}{4} \\[1em] \Rightarrow a - 6 = 0 \text{ and } b = \frac{9}{2} \\[1em] \Rightarrow a = 6 \text{ and } b = \frac{9}{2}.\end{aligned}$$

C = (0 , b) = $\Big(0, \frac{9}{2}\Big)$ and D = (6, 0).

Given, D is the mid-point of CB.

$$\begin{aligned}\therefore (6, 0) = \Big(\frac{0 + p}{2}, \frac{\frac{9}{2} + q}{2}\Big) \\[1em] \Rightarrow (6, 0) = \Big(\frac{p}{2}, \frac{9 + 2q}{2 \times 2}\Big) \\[1em] \Rightarrow (6, 0) = \Big(\frac{p}{2}, \frac{9 + 2q}{4}\Big) \\[1em] \Rightarrow \frac{p}{2} = 6 \text{ and } \frac{9 + 2q}{4} = 0 \\[1em] \Rightarrow p = 12 \text{ and } 9 + 2q = 0 \\[1em] \Rightarrow p = 12 \text{ and } 2q = -9 \\[1em] \Rightarrow p = 12 \text{ and } q = -\frac{9}{2} \\[1em] \Rightarrow B = (p, q) = \Big(12, -\frac{9}{2}\Big).\end{aligned}$$

Hence, coordinates of B = $\Big(12, -\frac{9}{2}\Big), C = \Big(0, \frac{9}{2}\Big)$ and D = (6, 0).