If \(A(2,-3)\) and \(B(-2,1)\) are two vertices of a \(\triangle A B C\) and if the centroid of \(\triangle…
If \(A(2,-3)\) and \(B(-2,1)\) are two vertices of a \(\triangle A B C\) and if the centroid of \(\triangle A B C\) lies on the line \(2 x+3 y=1\), then the locus of vertex \(C\) of \(\triangle A B C\) is equal to
\(2 x+3 y=5\)
\(2 x+3 y=9\)
\(3 x+2 y=5\)
\(3 x+2 y=9\)
Solution
Let third vertex be \(C=(h, k)\)
\(\begin{aligned}
A & \equiv(2,-3) \\
B & \equiv(-2,1)
\end{aligned}\)
Centroid \((G)=\left(\frac{h}{3}, \frac{-2+k}{3}\right)\)
Since, given \(G\) lies on \(2 x+3 y=1\)
\(\begin{aligned}
2\left(\frac{h}{3}\right)+3\left(\frac{-2+k}{3}\right) & =1 \\
2 h-6+3 k & =3 \\
2 h+3 k & =9
\end{aligned}\)
\(\therefore\) Required Locus is \(2 x+3 y=9\)
Hence, option (d) is correct.