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Let $A(2,-3)$ and $B(-2,1)$ be two angular points of $\Delta A B C$. If the centroid of the triangle moves on the line $2 x+3 y=1$, then the locus of the angular point $C$ is given by
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Verified Answer
The correct answer is:
$2 x+3 y=9$
Let the coordinates of $C$ be $(\alpha, \beta)$.
$\therefore$ Coordinates of centroid
$$
\begin{array}{l}
=\left(\frac{2-2+\alpha}{3}, \frac{-3+1+\beta}{3}\right) \\
=\left(\frac{\alpha}{3}, \frac{\beta-2}{3}\right)
\end{array}
$$
Since, centroid lie on $2 x+3 y=1$
$\therefore \quad \frac{2 \alpha}{3}+3\left(\frac{\beta-2}{3}\right)=1$
$\Rightarrow \quad \frac{2 \alpha}{3}+\frac{3 \beta-6}{3}=1$
$\Rightarrow \quad 2 \alpha+3 \beta-6=3 \Rightarrow 2 \alpha+3 \beta=9$
$\therefore$ Locus of point $C$ will be $2 x+3 y=9$
$\therefore$ Coordinates of centroid
$$
\begin{array}{l}
=\left(\frac{2-2+\alpha}{3}, \frac{-3+1+\beta}{3}\right) \\
=\left(\frac{\alpha}{3}, \frac{\beta-2}{3}\right)
\end{array}
$$
Since, centroid lie on $2 x+3 y=1$
$\therefore \quad \frac{2 \alpha}{3}+3\left(\frac{\beta-2}{3}\right)=1$
$\Rightarrow \quad \frac{2 \alpha}{3}+\frac{3 \beta-6}{3}=1$
$\Rightarrow \quad 2 \alpha+3 \beta-6=3 \Rightarrow 2 \alpha+3 \beta=9$
$\therefore$ Locus of point $C$ will be $2 x+3 y=9$
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