Find the coordinates of the centroid P of the △ ABC, whose vertices are A(-1, 3), B(3, -1) and C(0, 0). Hence, find the equation of a line passing through P and parallel to AB.

By formula,

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

Substituting values we get :

$\Rightarrow \text{Centroid of △ ABC} = \Big(\dfrac{-1 + 3 + 0}{3}, \dfrac{3 + (-1) + 0}{3}\Big) \\[1em] \Rightarrow P = \Big(\dfrac{2}{3}, \dfrac{2}{3}\Big).$

By formula,

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

Substituting values we get :

$\text{Slope of AB} = \dfrac{-1 - 3}{3 - (-1)} \\[1em] = \dfrac{-4}{4} \\[1em] = -1.$

We know that,

Slope of parallel lines are equal.

By point-slope form,

Equation of line : y - y₁ = m(x - x₁)

Substituting values we get :

Equation of line passing through P and parallel to AB :

$\Rightarrow y - \dfrac{2}{3} = -1\Big(x - \dfrac{2}{3}\Big) \\[1em] \Rightarrow \dfrac{3y - 2}{3} = -1 \times \dfrac{3x - 2}{3} \\[1em] \Rightarrow 3y - 2 = -1(3x - 2) \\[1em] \Rightarrow 3y - 2 = -3x + 2 \\[1em] \Rightarrow 3y + 3x = 2 + 2 \\[1em] \Rightarrow 3y + 3x = 4.$

Hence, required equation is 3x + 3y = 4.