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If the lengths of the perpendiculars drawn from a point $(a, b)$ to the lines $2 x+3 y+4=0$ and $3 x-3 y+4=0$ are same, then the point $(a, b)$ lies on the line
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The correct answer is:
$x-5 y=0$ or $5 x+y+8=0$
$1_1 \equiv 2 \mathrm{x}+3 \mathrm{y}+4=0$
$1_2 \equiv 3 \mathrm{x}-2 \mathrm{y}+4=0$
$$
\because \mathrm{a}_1 \mathrm{a}_2+\mathrm{b}_1 \mathrm{~b}_2=6-6=0
$$
$\therefore 1_1$ and $l_2$ are perpendicular.
Since point $(\mathrm{a}, \mathrm{b})$ is equidistance from $\mathrm{l}_1$ and $\mathrm{l}_2$ so, locus of $(a, b)$ passes through intersection of $1_1$ and $l_2^2$.
Therefore locus of $(a, b)$ are
$$
\begin{array}{ll}
2 x+3 y+4=0 & 2 x+3 y+4=0 \\
\frac{3 x-2 y+4=0}{-x+5 y=0} & \frac{3 x-2 y+4=0}{5 x+y+8=0}
\end{array} \Rightarrow x-5 y=0
$$
$1_2 \equiv 3 \mathrm{x}-2 \mathrm{y}+4=0$
$$
\because \mathrm{a}_1 \mathrm{a}_2+\mathrm{b}_1 \mathrm{~b}_2=6-6=0
$$
$\therefore 1_1$ and $l_2$ are perpendicular.
Since point $(\mathrm{a}, \mathrm{b})$ is equidistance from $\mathrm{l}_1$ and $\mathrm{l}_2$ so, locus of $(a, b)$ passes through intersection of $1_1$ and $l_2^2$.
Therefore locus of $(a, b)$ are
$$
\begin{array}{ll}
2 x+3 y+4=0 & 2 x+3 y+4=0 \\
\frac{3 x-2 y+4=0}{-x+5 y=0} & \frac{3 x-2 y+4=0}{5 x+y+8=0}
\end{array} \Rightarrow x-5 y=0
$$
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