Question

Let $A(-3,2)$ and $B(-2,1)$ be the vertices of a triangle $ABC$. If the centroid of this triangle lies on the line $3x+4y+2=0$, then the vertex $C$ lies on the line :

• A. $4x+3y+5=0$
• B. $3x+4y+3=0$
• C. $4x+3y+3=0$
• D. $3x+4y+5=0$

Answer 1

Answer: (B)

$LetC=\left(x_1,y_1\right)$
Centroid, E=$\left(\frac{x_1-5}{3},\frac{y_1+3}{3}\right)$
Since centroid lies on the line
3x + 4y + 2 = 0
$\therefore 3\left(\frac{x_1-5}{3}\right)+4\left(\frac{y_1+3}{3}\right)+2=0$
$\Rightarrow 3x_1+4y_1+3=0$
Hence vertex ($x_1,y_1$) lies on the line
$3x+4y+3=0$