The family of lines, forming an isosceles triangle with the lines $3 x-4 y-2=0$ and $12 x-5 y+6=0$, is

The family of lines, forming an isosceles triangle with the lines $3 x-4 y-2=0$ and $12 x-5 y+6=0$, is
  1. $9 \mathrm{x}+7 \mathrm{y}+\mathrm{c}=0$
  2. $7 \mathrm{x}-9 \mathrm{y}+\mathrm{c}=0$
  3. $9 x-7 y+c=0$
  4. $x \pm y+c=0$

Solution

We know that obtuse bisector of equal sides of a triangle is parallel to third side. $ \begin{aligned} & \therefore \frac{3 x-4 y-2}{\sqrt{9+16}}=-\frac{12 x-5 y+6}{\sqrt{144+25}} \\ & \frac{3 x-4 y-2}{5}=\frac{-12 x+5 y-6}{13} \\ & 39 x-52 y-26=-60 x+25 y-30 \\ & 39 x-77 y+4=0 \\ & 9 x-7 y+(4 / 11)=0 \end{aligned} $ Equation of line parallel to above line is $ 9 x-7 y+c=0 $

Asked in: AP EAMCET 2022 (06 Jul Shift 1)

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