A(1, 4), B(3, 2) and C(7, 5) are the vertices of a ΔABC. Find :

(i) the co-ordinates of the centroid G of ΔABC

(ii) the equation of a line through G and parallel to AB

(i) By formula,

Centroid of triangle = $\Big(\dfrac{x$1 + x2 + x3}{3}, \dfrac{y1 + y2 + y3}{3}\Big)

Substituting values we get,

Centroid = $\Big(\dfrac{1 + 3 + 7}{3}, \dfrac{4 + 2 + 5}{3}\Big) = \Big(\dfrac{11}{3}, \dfrac{11}{3}\Big)$

Hence, centroid of triangle = $\Big(\dfrac{11}{3}, \dfrac{11}{3}\Big)$.

(ii) Calculating,

Slope of AB = $\dfrac{y$2 - y1}{x2 - x1} = \dfrac{2 - 4}{3 - 1} = \dfrac{-2}{2} = -1

Slope of line parallel to AB will also be equal to -1, as slope of parallel lies are equal.

By point-slope form,

Equation of a line, through the centroid and parallel to AB,

⇒ y - y₁ = m(x - x₁)

⇒ y - $\Big(\dfrac{11}{3}\Big) = -1 \Big[x - \Big(\dfrac{11}{3}\Big)\Big]$

⇒ $\Big(\dfrac{3y - 11}{3}\Big) = -1 \Big(\dfrac{3x - 11}{3}\Big)$

⇒ 3y − 11 = −1(3x − 11)

⇒ 3y − 11 = −3x + 11

⇒ 3y + 3x = 11 + 11

⇒ 3x + 3y = 22.

Hence, the equation of a line, through the centroid and parallel to AB is 3x + 3y = 22.