Vertex A of triangle ABC is (-3, 5) and mid-points of the sides AB and AC are (-4, 4) and (2, -2) respectively. Find :

(i) co-ordinates of vertices B and C.

(ii) equation of line through vertex C and parallel to side AB.

(i) By mid-point formula,

M = $\Big(\dfrac{x$1 + x2}{2}, \dfrac{y1 + y2}{2}\Big)

Let coordinates of B be (a, b) and C be (c, d).

Given,

Mid-point of AB is (-4, 4).

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

B = (a, b) = (-5, 3).

Given,

Mid-point of AC is (2, -2).

$\therefore (2, -2) = \Big(\dfrac{-3 + c}{2}, \dfrac{5 + d}{2}\Big) \\[1em] \Rightarrow \dfrac{-3 + c}{2} = 2 \text{ and } \dfrac{5 + d}{2} = -2 \\[1em] \Rightarrow -3 + c = 4 \text{ and } 5 + d = -4 \\[1em] \Rightarrow c = 4 + 3 \text{ and } d = -4 - 5 \\[1em] \Rightarrow c = 7 \text{ and } d = -9.$

C = (c, d) = (7, -9).

Hence, co-ordinates of B = (-5, 3) and C = (7, -9).

(ii) By formula,

Slope = $\dfrac{y$2 - y1}{x2 - x1}

Slope of AB = $\dfrac{3 - 5}{-5 - (-3)} = \dfrac{-2}{-5 + 3} = \dfrac{-2}{-2}$ = 1.

We know that,

Slope of parallel lines are equal.

∴ Slope of required line = 1.

By point-slope form,

Equation of line is : y - y₁ = m(x - x₁)

So, equation of line through vertex C and parallel to side AB is

⇒ y - (-9) = 1(x - 7)

⇒ y + 9 = x - 7

⇒ y = x - 7 - 9

⇒ y = x - 16.

Hence, equation of line through vertex C and parallel to side AB is y = x - 16.