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Let $A=(2,3), B=(3,-5)$ be two vertices of $\triangle A B C$ such that $C$ is a point on the line $L \equiv 3 x+4 y-5=0$. Then the locus of the centroid of $\triangle A B C$ is a line parallel to
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The correct answer is:
$L=0$
Given, $A=(2,3), B(3,-5), C=(x, y)$
Let centroid of $\triangle A B C$ is $(h, k)$
$\begin{array}{ll}
\therefore & h=\frac{2+3+x}{3}, k=\frac{3-5+y}{3} \\
\Rightarrow & x=3 h-5 \Rightarrow y=3 k+2
\end{array}$
$C$ lie on line $3 x+4 y-5=0$
$\begin{array}{ll}
\therefore & 3(3 h-5)+4(3 k+2)-5=0 \\
\Rightarrow & 9 h-15+12 k+8-5=0 \\
\Rightarrow & 9 h+12 k-12=0 \Rightarrow 3 h+4 k-4=0
\end{array}$
Locus of centroid is $3 x+4 y-4=0$
Which is parallel to line $L=0$
Let centroid of $\triangle A B C$ is $(h, k)$
$\begin{array}{ll}
\therefore & h=\frac{2+3+x}{3}, k=\frac{3-5+y}{3} \\
\Rightarrow & x=3 h-5 \Rightarrow y=3 k+2
\end{array}$
$C$ lie on line $3 x+4 y-5=0$
$\begin{array}{ll}
\therefore & 3(3 h-5)+4(3 k+2)-5=0 \\
\Rightarrow & 9 h-15+12 k+8-5=0 \\
\Rightarrow & 9 h+12 k-12=0 \Rightarrow 3 h+4 k-4=0
\end{array}$
Locus of centroid is $3 x+4 y-4=0$
Which is parallel to line $L=0$
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