Question

Let $(1,-2)$ and $(-3,4)$ be two vertices of a triangle whose centroid lies on the line $x+3y=2.$ Then locus of the third vertex of the triangle is

• A. $x+3y=2$
• B. $x+3y=4$
• C. $x+3y=8$
• D. $x+3y=16$

Answer 1

The correct option is A $x+3y=2$

Let $(\alpha ,\beta )$ be the third vertex.
$\therefore$ Centroid $\equiv (\frac{\alpha +1-3}{3},\frac{\beta -2+4}{3})\equiv (\frac{\alpha -2}{3},\frac{\beta +2}{3})$ lies on $x+3y=2.$
$\Rightarrow \left(\frac{\alpha -2}{3}\right)+3\left(\frac{\beta +2}{3}\right)=2$
$\Rightarrow \alpha +3\beta =2$
$\therefore$ Locus of $(\alpha ,\beta )$ is $x+3y=2.$