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Locus of the points which are at equal distance from $3 x+4 y-11=0$ and $12 x+5 y+2=0$ and which is near the origin is
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The correct answer is:
$99 x+77 y-133=0$
Let point be $\left(x_1, y_1\right)$, then according to the condition $\frac{3 x_1+4 y_1-11}{5}=-\left(\frac{12 x_1+5 y_1+2}{13}\right)$ Since the given lines are on opposite sides with respect to origin, hence the required locus is $99 x+77 y-133=0$
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