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Q.
The straight lines $3x-4y+7=0$ and $12x+5y-2=0$ has equation of the acute angle bisector as:
NTA AbhyasNTA Abhyas 2022
Solution:
Equation of the straight lines are
$3x-4y+7=0..........\left(\right.i\left.\right)$
and $-12x-5y+2=0.............\left(\right.ii\left.\right)$
Here $\left(\right.3\left.\right)\left(\right.-12\left.\right)+\left(\right.-4\left.\right)\left(\right.-5\left.\right)+\left(\right.7\left.\right)\left(\right.2\left.\right)$
$=-36+20+14=-2 < 0$
$\therefore $ it makes acute angle.
$\therefore $ equation of the bisector of an acute angle between these lines is
$\frac{3 x - 4 y + 7}{\sqrt{3^{2} + 4^{2}}}=+\frac{\left( - 12 x - 5 y + 2 \right)}{\sqrt{\left( 12 \right)^{2} + \left( 5 \right)^{2}}}$ .
$\Rightarrow 99x-27y+81=0$
$\Rightarrow 11x-3y+9=0$