The equation of the bisectors of the angles between the straight line $3 x - 4 y + 7 = 0$ and $12 x - 5 y - 8 = 0$ , are:

21 x + 27 y + 131 = 0

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x + 27 y + 131 = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

21 x - 27 y + 131 = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

21 x + 27 y - 131 = 0

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Solution

The correct option is A $21 x + 27 y - 131 = 0$
As we know that,
The equation of the bisectors of the angles between the lines $a_{1} x + b_{2} y + c_{1} = 0$ and $a_{2} x + b_{2} y + c_{2} = 0$ are given by
$\frac{a_{1} x + b_{1} y + c_{1}}{\sqrt{a_{1}^{2} + b_{1}^{2}}} = \pm \frac{a_{2} x + b_{2} y + c_{2}}{\sqrt{a_{2}^{2} + b_{2}^{2}}}$ ...(1)
So, here we have
the lines $3 x - 4 y + 7 = 0$ ....(2)
and $12 x - 5 y - 8 = 0$ ...(3)
Therefore,
The equation of the bisectors of the angles between (2) and (3) are
$\frac{3 x - 4 y + 7}{\sqrt{(3)^{2} + (- 4)^{2}}} = \pm \frac{12 x + 5 y - 8}{\sqrt{(12)^{2} + (- 5)^{2}}}$ (from 1)

$\Rightarrow \frac{3 x - 4 y + 7}{\sqrt{9 + 16}} = \pm \frac{12 x - 5 y - 8}{\sqrt{144 + 25}}$

$\Rightarrow \frac{3 x - 4 y + 7}{\sqrt{25}} = \pm \frac{12 x - 5 y - 8}{\sqrt{169}}$

$\Rightarrow \frac{3 x - 4 y + 7}{5} = \pm \frac{12 x - 5 y - 8}{13}$

$\Rightarrow 39 x - 52 y + 91 = \pm (60 x - 25 y - 40)$

Now,
Case - I
$\Rightarrow 39 x - 52 y + 91 = 60 x - 25 y - 40$
$\Rightarrow 21 x + 27 y - 131 = 0$
$\Rightarrow$ Taking the positive sign $21 x + 27 y - 131 = 0$ as one bisector

Case - II
$\Rightarrow 39 x - 52 y + 91 = - 60 x + 25 y + 40$
$\Rightarrow 99 x - 77 y + 51 = 0$
$\Rightarrow$ Taking the negative sign $99 x - 77 y + 51 = 0$ is the equation of other bisector.

$\Rightarrow$ option D. $21 x + 27 y - 131 = 0$ is correct answer.

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