The equation of the bisector of the acute ang | vedclass

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(B) The equations of the angle bisectors are given by $\frac{3x - 4y + 7}{\sqrt{3^2 + (-4)^2}} = \pm \frac{12x + 5y - 2}{\sqrt{12^2 + 5^2}}$$\frac{3x

Vedclass · Accepted Answer

$11x - 3y + 9 = 0$

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(B) The equations of the angle bisectors are given by $\frac{3x - 4y + 7}{\sqrt{3^2 + (-4)^2}} = \pm \frac{12x + 5y - 2}{\sqrt{12^2 + 5^2}}$$\frac{3x - 4y + 7}{5} = \pm \frac{12x + 5y - 2}{13}$Case $1$: $13(3x - 4y + 7) = 5(12x + 5y - 2)$ $\Rightarrow 39x - 52y + 91 = 60x + 25y - 10 $ $\Rightarrow 21x + 77y - 101 = 0$Case $2$: $13(3x - 4y + 7) = -5(12x + 5y - 2)$ $\Rightarrow 39x - 52y + 91 = -60x - 25y + 10 $ $\Rightarrow 99x - 27y + 81 = 0$ $\Rightarrow 11x - 3y + 9 = 0$To identify the acute angle bisector,check the sign of $a_1a_2 + b_1b_2$. Here $a_1=3, b_1=-4, a_2=12, b_2=5$. $a_1a_2 + b_1b_2 = (3)(12) + (-4)(5) = 36 - 20 = 16 > 0$.Since the expression is positive,the equation $\frac{a_1x + b_1y + c_1}{\sqrt{a_1^2 + b_1^2}} = -\frac{a_2x + b_2y + c_2}{\sqrt{a_2^2 + b_2^2}}$ gives the acute angle bisector.This corresponds to $13(3x - 4y + 7) = -5(12x + 5y - 2)$,which simplifies to $11x - 3y + 9 = 0$.

The equation of the bisector of the acute angle between the lines $3x - 4y + 7 = 0$ and $12x + 5y - 2 = 0$ is

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