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.
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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:
⇒ Centroid of △ ABC = $\Big(\frac{-1 + 3 + 0}{3}, \frac{3 + (-1) + 0}{3}\Big)$
⇒ P = $\Big(\frac{2}{3}, \frac{2}{3}\Big)$
By formula,
Slope = $\frac{y_2 - y_1}{x_2 - x_1}$
$$\begin{aligned}\text{Slope of AB} = \frac{-1 - 3}{3 - (-1)} \\[1em] = \frac{-4}{4} \\[1em] = -1.\end{aligned}$$
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 :
$$\begin{aligned}\Rightarrow y - \frac{2}{3} = -1 \Big(x - \frac{2}{3}\Big) \\[1em] \Rightarrow \frac{3y - 2}{3} = -1 \times \Big(\frac{3x - 2}{3}\Big) \\[1em] \Rightarrow 3y - 2 = -1 \times (3x - 2) \\[1em] \Rightarrow 3y - 2 = -3x + 2 \\[1em] \Rightarrow 3x + 3y = 2 + 2 \\[1em] \Rightarrow 3x + 3y = 4.\end{aligned}$$
Hence, equation of a line passing through P and parallel to AB is 3x + 3y = 4.