Question

Let $A\left(2,-3\right)$, $B\left(-2,1\right)$, be the vertices of a $∆ABC$. If the centroid of this triangle moves on the line $2x+3y=1$, then the locus of the vertex is the line.

• A. $3x-2y=3$
• B. $2x+3y=9$
• C. $2x-3y=7$
• D. $3x+2y=5$

Answer 1

The correct option is B

$2x+3y=9$

Explanation for the correct answer:

Step 1: Finding the locus of the vertex:

Given $A\left(2,-3\right)$, $B\left(-2,1\right)$, are the vertices of a $∆ABC$.

The centroid of the triangle $\left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}\right)$

Let the third vertex of the triangle is $C=(x_3,y_3)$

Coordinate of the centroid

$$\begin{gathered} \left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}\right)=\left(\frac{2-2+x_3}{3},\frac{-3+1+y_3}{3}\right)\left[\text{given}{x}_1=2,x_2=-2,y_1=-3,y_2=1\right] \\ =\left(\frac{x_3}{3},\frac{-2+y_3}{3}\right) \end{gathered}$$

Step 2: Find the locus of the vertex $C$

Centroid lies on the line $2x+3y=1$ .

Substitute Coordinate of the centroid in the line $2x+3y=1$

$$\begin{gathered} 2\left(\frac{x_3}{3}\right)+3\left(\frac{-2+y_3}{3}\right)=1\left[\because \left(\frac{x_3}{3},\frac{-2+y_3}{3}\right)\right] \\ \Rightarrow 2x_3+3y_3-6=3 \\ \Rightarrow 2x_3+3y_3=9 \end{gathered}$$

Therefore, $2x+3y=9$ is the locus.

Hence, option (B) is the correct answer.