Question

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

• A. $x+3y=4$
• B. $x+3y=8$
• C. $x+3y=12$
• D. $x+3y=16$

Answer 1

The correct option is B $x+3y=8$

Let $(x,y)$ is the third vertex
$\therefore$Centroid$(h,k)=(\frac{x+1-3}{3},\frac{y-2+4}{3})$
$\Rightarrow (h,k)=(\frac{x-2}{3},\frac{y+2}{3})$
$\because (h,k)$ lies on the line $x+3y=4\Rightarrow \frac{x-2}{3}+3\left(\frac{y+2}{3}\right)=4$
$\Rightarrow x+3y=8$