A(–1, 3), B(4, 2) and C(3, –2) are the vertices of a triangle.

(i) Find the co-ordinates of the centroid G of the triangle.

(ii) Find the equation of the line through G and parallel to AC.

(i) Centroid of the triangle is given by,

$G = \Big(\dfrac{x$1 + x2 + x3}{3}, \dfrac{y1 + y2 + y3}{3}\Big) \\[1em] = \Big(\dfrac{-1 + 4 + 3}{3}, \dfrac{3 + 2 - 2}{3}\Big) \\[1em] = \Big(\dfrac{6}{3}, \dfrac{3}{3}\Big) \\[1em] = (2, 1).

Hence, the coordinates of the centroid G of the triangle is (2, 1).

(ii) Slope of AC = $\dfrac{y$2 - y1}{x2 - x1} = \dfrac{-2 - 3}{3 - (-1)} = -\dfrac{5}{4}

So, the slope of the line parallel to AC is also $-\dfrac{5}{4}$. and it passes through (2, 1). Hence, its equation can be given by point-slope form i.e.,

⇒ y - y₁ = m(x - x₁)

⇒ y - 1 = $-\dfrac{5}{4}$(x - 2)

⇒ 4(y − 1) = −5(x − 2)

⇒ 4y − 4 = −5x + 10

⇒ 4y + 5x = 14

⇒ 5x + 4y − 14 = 0.

Hence, the equation of the required line is 5x + 4y - 14 = 0.