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(A) The equation of the angle bisectors of the lines $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$ is given by $\frac{a_1x + b_1y + c_1}{\sqrt{a

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$99x - 77y + 51 = 0, 21x + 27y - 131 = 0$

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(A) The equation of the angle bisectors of the lines $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$ is given by $\frac{a_1x + b_1y + c_1}{\sqrt{a_1^2 + b_1^2}} = \pm \frac{a_2x + b_2y + c_2}{\sqrt{a_2^2 + b_2^2}}$.First,ensure the constant terms $c_1$ and $c_2$ are positive. Rewrite the lines as $-3x + 4y - 7 = 0$ and $12x - 5y - 8 = 0$. To make constants positive,use $3x - 4y + 7 = 0$ and $-12x + 5y + 8 = 0$.Now,$\frac{3x - 4y + 7}{\sqrt{3^2 + (-4)^2}} = \pm \frac{-12x + 5y + 8}{\sqrt{(-12)^2 + 5^2}}$.$\frac{3x - 4y + 7}{5} = \pm \frac{-12x + 5y + 8}{13}$.Taking the positive sign: $13(3x - 4y + 7) = 5(-12x + 5y + 8) \implies 39x - 52y + 91 = -60x + 25y + 40 \implies 99x - 77y + 51 = 0$.Taking the negative sign: $13(3x - 4y + 7) = -5(-12x + 5y + 8) \implies 39x - 52y + 91 = 60x - 25y - 40 \implies 21x + 27y - 131 = 0$.

The equations of the angle bisectors between the lines $3x - 4y + 7 = 0$ and $12x - 5y - 8 = 0$ are:

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