Section & Mid-Point Formula

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 :

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

Hence, Option 1 is the correct option.

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

By section-formula,

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

Substituting values we get :

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

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

⇒ $\frac{1}{3} : 1$

⇒ 1 : 3.

Hence, Option 4 is the correct option.

Any point on the line x = 7, can be defined as (7, y).

Let point (7, y) divide line joining (2, 4) and (10, 12) in ratio k : 1.

By section-formula,

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

Substituting values we get :

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

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

⇒ $\frac{5}{3} : 1$

⇒ 5 : 3.

Hence, Option 2 is the correct option.

Any point on the line y = 4, can be defined as (x, 4).

Let point (x, 4) divide line joining (6, 7) and (4, -1) in ratio k : 1.

By section-formula,

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

Substituting values we get :

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

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

⇒ $\frac{3}{5} : 1$

⇒ 3 : 5.

Hence, Option 1 is the correct option.

Any point on y-axis can be defined as (0, y).

Let ratio in which the join of points (-2, 5) and (5, -2) is divided by (0, y) be k : 1.

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

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

⇒ $\frac{2}{5} : 1$

⇒ 2 : 5.

Hence, Option 2 is the correct option.

We know that,

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

Substituting values we get,

$$\begin{aligned}\Rightarrow 1 = \frac{m_1 \times 4 + m_2 \times -1}{m_1 + m_2} \\[1em] \Rightarrow m_1 + m_2 = 4m_1 - m_2 \\[1em] \Rightarrow m_2 + m_2 = 4m_1 - m_1 \\[1em] \Rightarrow 2m_2 = 3m_1 \\[1em] \Rightarrow \frac{m_1}{m_2} = \frac{2}{3}.\end{aligned}$$

∴ m₁ : m₂ = 2 : 3.

We know that,

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

Substituting values we get,

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

Hence, ratio = 2 : 3 and a = 2.

Let ratio in which point (a, 6) divide the join of (-4, 3) and (2, 8) be m₁ : m₂.

By section formula,

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

Substituting values we get,

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

∴ m₁ : m₂ = 3 : 2.

We know that,

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

Substituting values we get,

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

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

Let the point on x-axis be (x, 0) and required ratio be k : 1.

By formula,

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

Substituting values we get,

$$\begin{aligned}\Rightarrow 0 = \frac{k \times -6 + 1 \times 3}{k + 1} \\[1em] \Rightarrow 0 = -6k + 3 \\[1em] \Rightarrow 6k = 3 \\[1em] \Rightarrow k = \frac{3}{6} = \frac{1}{2}.\end{aligned}$$

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

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

Substituting values we get,

$$\begin{aligned}\Rightarrow x = \frac{1 \times 2 + 2 \times 4}{1 + 2} \\[1em] \Rightarrow x = \frac{2 + 8}{3} \\[1em] \Rightarrow x = \frac{10}{3}.\end{aligned}$$

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

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

Let the point on y-axis be (0, y) and required ratio be k : 1.

By formula,

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

Substituting values we get,

$$\begin{aligned}\Rightarrow 0 = \frac{k \times 3 + 1 \times -4}{k + 1} \\[1em] \Rightarrow 0 = 3k - 4 \\[1em] \Rightarrow 3k = 4 \\[1em] \Rightarrow k = \frac{4}{3}.\end{aligned}$$

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

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

Substituting values we get,

$$\begin{aligned}\Rightarrow y = \frac{4 \times 0 + 3 \times 7}{4 + 3} \\[1em] \Rightarrow y = \frac{0 + 21}{7} \\[1em] \Rightarrow y = 3.\end{aligned}$$

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

Hence, ratio = 4 : 3 and co-ordinates of point of intersection = (0, 3).

Let point P = (5, -10) and origin (O) = (0, 0).

Points A, B, C and D divide the line segment PO in 5 equal parts.

From figure,

B divides the line segment PO in the ratio 2 : 3.

Let B be (a, b).

By formula,

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

Substituting values we get,

$$\begin{aligned}\Rightarrow a = \frac{2 \times 0 + 3 \times 5}{2 + 3} \\[1em] \Rightarrow a = \frac{0 + 15}{5} \\[1em] \Rightarrow a = \frac{15}{5} = 3.\end{aligned}$$

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

Substituting values we get,

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

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

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

Let D be (c, d).

By formula,

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

Substituting values we get,

$$\begin{aligned}\Rightarrow c = \frac{4 \times 0 + 1 \times 5}{4 + 1} \\[1em] \Rightarrow c = \frac{0 + 5}{5} \\[1em] \Rightarrow c = \frac{5}{5} = 1.\end{aligned}$$

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

Substituting values we get,

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

D = (c, d) = (1, -2).

Hence, B = (3, -6) and D = (1, -2).

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

Given,

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

Let PB = x and AB = 5x.

From figure,

⇒ AB = PA + PB

⇒ 5x = PA + x

⇒ PA = 4x.

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

PA : PB = 4 : 1.

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

By formula,

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

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

Given,

⇒ 5AP = 2BP

⇒ $\frac{AP}{BP} = \frac{2}{5}$

⇒ AP : PB = 2 : 5.

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

By formula,

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

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

Let point of intersection be (2, y). [∵ any point on the line x = 2 has x co-ordinate = 2]

By formula,

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

m₁ : m₂ = 5 : 3.

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

Hence, co-ordinates of point of intersection = (2, 4) and ratio = 5 : 3.

Let point of intersection be (x, 2) [∵ any point on the line y = 2 has y co-ordinate = 2]

By formula,

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

m₁ : m₂ = 3 : 5.

Hence, ratio = 3 : 5.

Since, A is on x-axis, let it co-ordinates be (a, 0) and B is on y-axis so it's co-ordinates (0, b).

By formula,

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

∴ A = (a, 0) = (7, 0) and

B = (0, b) = (0, -14).

Hence, A = (7, 0) and B = (0, -14).

From figure,

Let A and B be the points of tri-section of the line joining the points (-3, 0) and (6, 6).

So, A and B divides the segment in three equal parts.

A divides the line segment in ratio 1 : 2. Let co-ordinates of A be (a, b).

By formula,

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

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

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

By formula,

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

B = (c, d) = (3, 4).

Hence, points of tri-section are (0, 2) and (3, 4).

Let A = (-5, 8) and B = (10, -4)

Let P and Q be points which trisects AB.

Let P (a, b) divide AB in 1 : 2 and Q (c, d) in 2 : 1.

By section formula (for P),

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

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

By section formula (for Q),

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

Q = (c, d) = (5, 0).

Since, x co-ordinate of P = 0, it means P lies on y-axis and y co-ordinate of Q = 0, it means Q lies on x-axis.

Hence, proved that line segment joining the points (-5, 8) and (10, -4) is trisected by the co-ordinate axes.

Let A divide line-segment joining the points (2, 1) and (5, -8) in m₁ : m₂.

By formula,

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

Since, A divides line-segment joining the points (2, 1) and (5, -8) in 1 : 2.

Hence, proved A is a point of tri-section.

Let another point of tri-section be B(a, b). So, it will divide the line segment in 2 : 1.

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

Substituting values we get,

$$\begin{aligned}\Rightarrow a = \frac{2 \times 5 + 1 \times 2}{2 + 1} \\[1em] \Rightarrow a = \frac{10 + 2}{3} \\[1em] \Rightarrow a = 4.\end{aligned}$$

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

Substituting values we get,

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

B = (a, b) = (4, -5).

Hence, co-ordinate of other point of trisection = (4, -5).

Since, L lies on y-axis let its co-ordinates be (0, y).

Let L divide MN in ratio m₁ : m₂.

By formula,

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

m₁ : m₂ = 5 : 3.

By formula,

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

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

Given,

BP : PC = 2 : 3.

So, P divides the line segment BC in ratio 2 : 3.

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

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

and

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

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

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

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

Hence, AP = 5 units.

Since, point K lies on x-axis. Its co-ordinates be (x, 0).

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

By section-formula,

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

m₁ : m₂ = 3 : 5.

Substituting value for x co-ordinate,

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

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

Since, point K lies on y-axis. Its co-ordinates be (0, y).

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

By section-formula,

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

m₁ : m₂ = 2 : 3.

Substituting value for y co-ordinate,

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

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

(i) Since, R lies on y-axis. Let its co-ordinates be (0, y).

Let R divide PQ in ratio m₁ : m₂.

By section formula,

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

m₁ : m₂ = 4 : 3.

Hence, PR : RQ = 4 : 3.

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

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

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

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

(iii) From graph,

PMNQ is a trapezium and 1 block = 1 unit.

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

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

Hence, area of PMNQ = 24.5 sq. units.

Since, A lies on x-axis its co-ordinates be (x, 0) and B lies on y-axis its co-ordinates be (0, y).

Given, AP : PB = 1 : 2.

By section formula,

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

A = (x, 0) = (-6, 0) and B = (0, y) = (0, 6).

Hence, A = (-6, 0) and B = (0, 6).

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

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

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

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

Hence, required ratio = 1 : 2.

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

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

Substituting value in above formula, we get :

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

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

(iii) By distance formula,

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

Substituting values we get :

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

Hence, AB = 15 units.

By section-formula,

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

Substituting values we get,

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

Substituting values we get,

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

Hence, a = 1 and b = -3.

Given,

A(3, 4) is center and let B(7, 8) be one end of the diameter.

Let other end of diameter be C(x, y).

By mid-point formula,

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

From figure,

A is the mid-point of BC.

$$\begin{aligned}\therefore (3, 4) = \Big(\frac{x + 7}{2}, \frac{y + 8}{2}\Big) \\[1em] \Rightarrow \frac{x + 7}{2} = 3 \text{ and } \frac{y + 8}{2} = 4 \\[1em] \Rightarrow x + 7 = 6 \text{ and } y + 8 = 8 \\[1em] \Rightarrow x = 6 - 7 \text{ and } y = 8 - 8 \\[1em] \Rightarrow x = -1 \text{ and } y = 0.\end{aligned}$$

∴ C = (-1, 0).

Hence, Option 3 is the correct option.

Given,

Point A lies on x-axis.

∴ Let point A be (x, 0).

Point B lies on y-axis.

∴ Let point B be (0, y).

P(2, -2) bisects the line segment AB or P is the mid-point of AB.

By mid-point formula,

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

Substituting values we get :

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

Co-ordinates of A = (x, 0) = (4, 0).

Hence, Option 1 is the correct option.

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

We know that,

Diagonal of parallelogram bisect each other.

From figure,

O (a, b) is the mid-point of AC.

By mid-point formula,

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

Substituting values we get :

$$\begin{aligned}\Rightarrow (a, b) = \Big(\frac{6 + 4}{2}, \frac{0 + (-4)}{2}\Big) \\[1em] = \Big(\frac{10}{2}, \frac{-4}{2}\Big) \\[1em] = (5, -2).\end{aligned}$$

From figure,

O is also the mid-point of BD.

$$\begin{aligned}\therefore (5, -2) = \Big(\frac{12 + x}{2}, \frac{-4 + y}{2}\Big) \\[1em] \Rightarrow \frac{12 + x}{2} = 5 \text{ and } \frac{-4 + y}{2} = -2 \\[1em] \Rightarrow 12 + x = 10 \text{ and } -4 + y = -4 \\[1em] \Rightarrow x = 10 - 12 \text{ and } y = -4 + 4 \\[1em] \Rightarrow x = -2 \text{ and } y = 0.\end{aligned}$$

D = (-2, 0).

Hence, Option 2 is the correct option.

Let point P on reflection in the point (0, 3) becomes P'(x, y).

So, (0, 3) will be the mid-point of PP'.

By formula,

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

Substituting values we get :

$$\begin{aligned}\Rightarrow (0, 3) = \Big(\frac{2 + x}{2}, \frac{-7 + y}{2}\Big) \\[1em] \Rightarrow 0 = \frac{2 + x}{2} \text{ and } 3 = \frac{-7 + y}{2} \\[1em] \Rightarrow 2 + x = 0 \text{ and } -7 + y = 6 \\[1em] \Rightarrow x = -2 \text{ and } y = 6 + 7 \\[1em] \Rightarrow x = -2 \text{ and } y = 13.\end{aligned}$$

P' = (-2, 13).

Hence, Option 4 is the correct option.

By formula,

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

Substituting values we get :

$$\begin{aligned}\text{Centroid } = \Big(\frac{-6 + 0 + 12}{3}, \frac{-3 + 0 + (-6)}{3}\Big) \\[1em] = \Big(\frac{6}{3}, \frac{-9}{3}\Big) \\[1em] = (2, -3).\end{aligned}$$

Hence, Option 3 is the correct option.

By formula,

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

Given, M = (2, 3)

Substituting values in above formula,

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

Hence, x = 1 and y = 1.

By formula,

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

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

Substituting values in above formula we get,

$$\begin{aligned}(1, 7) = \Big[\frac{x + (-5)}{2}, \frac{y + 10}{2}\Big] \\[1em] \therefore 1 = \frac{x - 5}{2} \text{ and } 7 = \frac{y + 10}{2} \\[1em] \Rightarrow 2 = x - 5 \text{ and } 14 = y + 10 \\[1em] \Rightarrow x = 7 \text{ and } y = 4.\end{aligned}$$

A = (x, y) = (7, 4).

Hence, co-ordinates of A = (7, 4).

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

Substituting values in above formula we get,

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

B = (x, y) = (-5, 7).

Hence, co-ordinates of B = (-5, 7).

Since, A lie on y-axis, let its co-ordinates be (0, y) and B lie on x-axis, let its co-ordinates be (x, 0).

P is mid-point of AB.

By formula,

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

Substituting values we get,

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

A = (0, y) = (0, 4) and B = (x, 0) = (-6, 0).

Hence, the co-ordinates of points A and B are (0, 4) and (-6, 0).

Let A = (3, -6), B = (-5, 2) and C = (7, 4).

From figure, AD is the median.

Since, AD is median so, BD = DC.

Thus, D is mid-point of BC.

By formula,

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

Substituting value we get,

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

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

Substituting values we get,

$$\begin{aligned}AD = \sqrt{(1 - 3)^2 + [3 - (-6)]^2} \\[1em] = \sqrt{(-2)^2 + (9)^2} \\[1em] = \sqrt{4 + 81} \\[1em] = \sqrt{85} \\[1em] = 9.22\end{aligned}$$

Hence, the length of its median through the vertex (3, -6) = 9.22 units.

Let co-ordinates of A be (x, y) and D be (p, q).

Since, AB = BC.

B is the mid-point of AC.

By formula,

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

Substituting values we get,

$$\begin{aligned}\Rightarrow B = \Big(\frac{x + 1}{2}, \frac{y + 8}{2}\Big) \\[1em] \Rightarrow (0, 3) = \Big(\frac{x + 1}{2}, \frac{y + 8}{2}\Big) \\[1em] \Rightarrow 0 = \frac{x + 1}{2} \text{ and } 3 = \frac{y + 8}{2} \\[1em] \Rightarrow x + 1 = 0 \text{ and } y + 8 = 6 \\[1em] \Rightarrow x = -1 \text{ and } y = -2.\end{aligned}$$

A = (x, y) = (-1, -2).

Since, BC = CD.

C is mid-point of BD.

By formula,

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

Substituting values we get,

$$\begin{aligned}\Rightarrow C = \Big(\frac{0 + p}{2}, \frac{3 + q}{2}\Big) \\[1em] \Rightarrow (1, 8) = \Big(\frac{p}{2}, \frac{3 + q}{2}\Big) \\[1em] \Rightarrow 1 = \frac{p}{2} \text{ and } 8 = \frac{3 + q}{2} \\[1em] \Rightarrow p = 2 \text{ and } 3 + q = 16 \\[1em] \Rightarrow p = 2 \text{ and } q = 13.\end{aligned}$$

D = (p, q) = (2, 13).

Hence, the co-ordinates of A = (-1, -2) and D = (2, 13).

By formula,

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

Let mid-point of AC be E.

Substituting value we get,

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

Let mid-point of BD be F.

Substituting value we get,

$$\begin{aligned}F = \Big[\frac{1 + (-3)}{2}, \frac{0 + 8}{2}\Big] \\[1em] = \Big(-\frac{2}{2}, \frac{8}{2}\Big) \\[1em] = (-1, 4).\end{aligned}$$

Thus, the co-ordinates of the mid-points of AC and BD are same i.e., AC and BD bisect each other.

∴ ABCD is a parallelogram.

Hence, the co-ordinates of the mid-points of AC = (-1, 4) and BD = (-1, 4) and ABCD is a parallelogram.

We know that diagonals of a parallelogram bisect each other.

Let co-ordinates of R= (a, b) and S = (c, d).

From figure,

O is the mid-point of PR.

By formula,

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

Substituting value we get,

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

R = (a, b) = (-10, 2).

O is also the mid-point of QS,

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

S = (c, d) = (-5, -1).

Hence, the co-ordinates of R = (-10, 2) and S = (-5, -1).

Let D = (2, -1), E = (-1, 4) and F = (-2, 2).

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

Mid-point of AB, i.e. D(2, -1).

$\therefore 2 = \frac{x_1 + x_2}{2} \text{ and } -1 = \frac{y_1 + y_2}{2}$

⇒ x₁ + x₂ = 4 ........(1)

⇒ y₁ + y₂ = -2 ........(2)

Mid-point of BC, i.e. E(-1, 4).

$\therefore -1 = \frac{x_2 + x_3}{2} \text{ and } 4 = \frac{y_2 + y_3}{2}$

⇒ x₂ + x₃ = -2 ........(3)

⇒ y₂ + y₃ = 8 ........(4)

Mid-point of AC, i.e. F(-2, 2).

$\therefore -2 = \frac{x_1 + x_3}{2} \text{ and } 2 = \frac{y_1 + y_3}{2}$

⇒ x₁ + x₃ = -4 ........(5)

⇒ y₁ + y₃ = 4 ........(6)

Adding 1, 3 and 5 we get,

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

⇒ 2(x₁ + x₂ + x₃) = -2

⇒ x₁ + x₂ + x₃ = -1.

From (1),

⇒ 4 + x₃ = -1

⇒ x₃ = -5.

Substituting value of x₃ in (5) we get,

⇒ x₁ + (-5) = -4

⇒ x₁ = 1.

Substituting value of x₃ in (3) we get,

⇒ x₂ + (-5) = -2

⇒ x₂ = 3.

Adding (2), (4) and (6) we get,

⇒ y₁ + y₂ + y₂ + y₃ + y₁ + y₃ = -2 + 8 + 4

⇒ 2(y₁ + y₂ + y₃) = 10

⇒ y₁ + y₂ + y₃ = 5

From (2)

⇒ -2 + y₃ = 5

⇒ y₃ = 7.

Substituting value of y₃ in (4) we get,

⇒ y₂ + 7 = 8

⇒ y₂ = 1.

Substituting value of y₃ in (6) we get,

⇒ y₁ + 7 = 4

⇒ y₁ = -3.

A = (x₁, y₁) = (1, -3), B = (x₂, y₂) = (3, 1), C = (x₃, y₃) = (-5, 7).

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

Since, A, B and C are collinear and AB = BC.

We can say that B is the mid-point of AC.

By formula,

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

Substituting value we get,

$$\begin{aligned}\Rightarrow B = \Big(\frac{-5 + 1}{2}, \frac{x + (-3)}{2}\Big) \\[1em] \Rightarrow (y, 7) = \Big(\frac{-4}{2}, \frac{x - 3}{2}\Big) \\[1em] \Rightarrow y = \frac{-4}{2} \text{ and } 7 = \frac{x - 3}{2} \\[1em] \Rightarrow y = -2 \text{ and } x - 3 = 14 \\[1em] \Rightarrow y = -2 \text{ and } x = 17.\end{aligned}$$

Hence, x = 17 and y = -2.

From figure,

PR = PQ + QR

Given,

⇒ PR = 2QR

⇒ PQ +QR = 2QR

⇒ PQ = QR.

∴ Q is the mid-point of P and R.

By formula,

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

Substituting value we get,

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

Hence, a = -4 and b = -1.

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

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

Substituting values we get,

$$\begin{aligned}\Rightarrow (2, -5) = \Big(\frac{x + (-6) + 11}{3}, \frac{y + 5 + 8}{3}\Big) \\[1em] \Rightarrow 2 = \frac{x - 6 + 11}{3} \text{ and } -5 = \frac{y + 13}{3} \\[1em] \Rightarrow 6 = x + 5 \text{ and } -15 = y + 13 \\[1em] \Rightarrow x = 1 \text{ and } y = -28.\end{aligned}$$

P = (x, y) = (1, -28).

Hence, co-ordinates of P = (1, -28).

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

Substituting values we get,

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

Hence, x = -1 and y = -1.

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

Substituting values we get :

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

A = (x, 0) = (8, 0),

B = (0, y) = (0, 6).

Hence, Option 4 is the correct option.

By mid-point formula,

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

Substituting values we get :

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

Hence, Option 3 is the correct option.

Let third vertex be (x, y).

By formula,

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

Substituting values we get :

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

Third vertex = (-2, 10).

Hence, Option 2 is the correct option.

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

Substituting values we get :

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

Hence, Option 4 is the correct option.

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

Substituting values for y-coordinate :

$$\begin{aligned}\Rightarrow 0 = \frac{k \times -2 + 1 \times 3}{k + 1} \\[1em] \Rightarrow -2k + 3 = 0 \\[1em] \Rightarrow 2k = 3 \\[1em] \Rightarrow k = \frac{3}{2}.\end{aligned}$$

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

⇒ $\frac{3}{2} : 1$

⇒ 3 : 2.

Hence, Option 2 is the correct option.

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

So, reason (R) is true.

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

m₁ : m₂ = 1 : 2

Substituting the values, we get :

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

So, assertion (A) is false.

Hence, option 2 is the correct option.

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.

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

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

and,

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

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

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

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

Hence, AP = 12.04 units.

From figure,

⇒ AB = AP + PB

⇒ 3PB = AP + PB

⇒ AP = 2PB

⇒ $\frac{AP}{PB} = \frac{2}{1}$

⇒ AP : PB = 2 : 1.

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

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

and,

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

From figure,

⇒ AB = AQ + QB

⇒ 6AQ = AQ + QB

⇒ QB = 5AQ

⇒ $\frac{AQ}{QB} = \frac{1}{5}$

⇒ AQ : QB = 1 : 5.

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

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

and,

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

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

The triangle is shown in the figure below:

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

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

and,

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

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

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

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

and,

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

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

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

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

⇒ BC = 16 units

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

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

(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.

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

Hence, a = 3.

(ii) By section formula,

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

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

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

Let AB intersect x-axis at P. So, co-ordinates of P = (x, 0).

Let ratio be m₁ : m₂.

By section formula,

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

m₁ : m₂ = 3 : 1.

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

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

Hence, co-ordinates of P = (3, 0) and ratio = 3 : 1.

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

Substituting values we get,

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

Hence, A = (8, 0) and B = (0, -6).

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

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

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

Substituting values we get,

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

Hence, co-ordinates of B = (-7, 17).

By formula,

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

Substituting values we get,

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

Hence, a = 2 and b = -1.

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

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

and,

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

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

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

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

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

Hence, OP = 5 units.

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

By section formula,

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

Substituting values we get,

$$\begin{aligned}\Rightarrow 0 = \frac{m_1 \times 17 + m_2 \times -4}{m_1 + m_2} \\[1em] \Rightarrow 0 = 17m_1 - 4m_2 \\[1em] \Rightarrow 4m_2 = 17m_1 \\[1em] \Rightarrow \frac{m_1}{m_2} = \frac{4}{17}.\end{aligned}$$

m₁ : m₂ = 4 : 17.

Hence, ratio in which the y-axis divide the line AB = 4 : 17.

The points are shown in the figure below:

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

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

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.

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

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

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

D = (x, y) = (1, 8).

Hence, co-ordinates of D = (1, 8).

By formula,

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

Substituting values we get,

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

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

By section formula,

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

Substituting values we get,

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

p : q = 1 : 2.

Hence, M = (3, 3) and p : q = 1 : 2.

Let ratio be m₁ : m₂.

By section formula,

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

Substituting values we get,

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

m₁ : m₂ = 1 : 3.

(i) By section formula,

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

Substituting values we get,

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

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

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

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

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

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,

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

Substituting values we get,

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

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

Substituting values we get,

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

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

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

Hence, ratio in which the line 2x + y = 4 divides the line segment joining the points P(2, -2) and Q(3, 7) = 2 : 9.

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,

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

Substituting values we get,

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

m₁ : m₂ = 3 : 2.

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

Substituting values we get,

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

P = (2, y) = (2, 3).

Hence, ratio = 3 : 2 and co-ordinates of P = (2, 3).

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

Substituting values we get,

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

B = (x, y) = (-7, 9).

Hence, image of the point A(5, -3) under reflection in the point P(-1, 3) is (-7, 9).

(i) By formula,

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

Substituting values we get,

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

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,

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

Substituting values we get,

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

Substituting values we get,

$$\begin{aligned}\Rightarrow 2 = \frac{1 \times y + 3 \times 0}{1 + 3} \\[1em] \Rightarrow 2 = \frac{y}{4} \\[1em] \Rightarrow y = 8.\end{aligned}$$

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

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

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

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

Hence, BP = 10 units.

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

Substituting values we get,

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

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

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

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

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

(i) From graph,

Coordinates of P = (4, 13), Q = (7, 8) and R = (11, 2)

By mid-point formula,

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

Substituting values we get :

$$\begin{aligned}\text{Mid-point of PR } = \Big(\frac{4 + 11}{2}, \frac{13 + 2}{2}\Big) \\[1em] = \Big(\frac{15}{2}, \frac{15}{2}\Big) \\[1em] = (7.5, 7.5).\end{aligned}$$

Hence, Q is not the midpoint of segment PR.

(ii) Given,

M divides PQ in the ratio 2 : 3.

By section formula,

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

Substituting values we get :

$$\begin{aligned}\Rightarrow M = \Big(\frac{2 \times 7 + 3 \times 4}{2 + 3}, \frac{2 \times 8 + 3 \times 13}{2 + 3}\Big) \\[1em] = \Big(\frac{14 + 12}{5}, \frac{16 + 39}{5}\Big) \\[1em] = \Big(\frac{26}{5}, \frac{55}{5}\Big) \\[1em] = \Big(\frac{26}{5}, 11\Big).\end{aligned}$$

Hence, M = $\Big(\frac{26}{5}, 11\Big)$.