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The point on the line $3 x+4 y=5$ which is equidistant from $(1,2)$ and $(3,4)$ is
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Verified Answer
The correct answer is:
$(15,-10)$
Let point $\left(x_1, y_1\right)$ be on the line $3 x+4 y=5$.
Also, $\quad\left(x_1-1\right)^2+\left(y_1-2\right)^2=\left(x_1-3\right)^2$
$\begin{aligned}
&+\left(y_1-4\right)^2 \\
\Rightarrow x_1^2+y_1^2-2 x_1-4 y_1+5=x_1^2+ & y_1^2-6 x_1 \\
& -8 y_1+25
\end{aligned}$
On solving Eqs. (i) and (ii), we get
$x_1=15, y_1=-10$
Also, $\quad\left(x_1-1\right)^2+\left(y_1-2\right)^2=\left(x_1-3\right)^2$
$\begin{aligned}
&+\left(y_1-4\right)^2 \\
\Rightarrow x_1^2+y_1^2-2 x_1-4 y_1+5=x_1^2+ & y_1^2-6 x_1 \\
& -8 y_1+25
\end{aligned}$
On solving Eqs. (i) and (ii), we get
$x_1=15, y_1=-10$
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