The equation of the bisector of the acute angle between 4 x -3 y -6-0 and 5 x -12 y -9-0 isA- 9 x -7 y -3-0B- 9 x -7 y -41-0C- 9 x-7 y-41-0D- 7 x -9 y -3-0

The correct option is D 7x+9y 3=0Given lines are nbsp;4x+3y 6=0 and 5x+12y+9=0 Making the constant terms positive, we get 4x 3y+6=0 and 5x+12y+9=0 Now, a_1a ...

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Standard XII

Mathematics

Bisectors of Angle between Two Lines

The equation ...

Question

The equation of the bisector of the acute angle between $4 x + 3 y - 6 = 0$ and $5 x + 12 y + 9 = 0$ is

9 x - 7 y - 41 = 0

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7 x + 9 y - 3 = 0

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9 x + 7 y - 3 = 0

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9 x - 7 y + 41 = 0

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Solution

The correct option is B $7 x + 9 y - 3 = 0$
Given lines are $4 x + 3 y - 6 = 0$ and $5 x + 12 y + 9 = 0$
Making the constant terms positive, we get
$- 4 x - 3 y + 6 = 0$ and $5 x + 12 y + 9 = 0$
Now,
$a_{1} a_{2} + b_{1} b_{2} = - 20 - 36 = - 56 < 0$
So, the equation of acute angle bisector is
$\frac{- 4 x - 3 y + 6}{\sqrt{(- 4)^{2} + (- 3)^{2}}} = \frac{5 x + 12 y + 9}{\sqrt{(5)^{2} + (12)^{2}}} \Rightarrow \frac{- 4 x - 3 y + 6}{5} = \frac{5 x + 12 y + 9}{13}$
$\Rightarrow 13 (- 4 x - 3 y + 6) = 5 (5 x + 12 y + 9) \Rightarrow - 52 x - 39 y + 78 = 25 x + 60 y + 45 \Rightarrow 77 x + 99 y - 33 = 0 ∴ 7 x + 9 y - 3 = 0$

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