The equation of the bisector of the acute ang | vedclass

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

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$11x - 3y + 9 = 0$

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(A) The equations of the angle bisectors are given by $\frac{3x - 4y + 7}{\sqrt{3^2 + (-4)^2}} = \pm \frac{12x + 5y - 2}{\sqrt{12^2 + 5^2}}$.This simplifies to $\frac{3x - 4y + 7}{5} = \pm \frac{12x + 5y - 2}{13}$.To identify the acute angle bisector,we first make the constant terms positive: $3x - 4y + 7 = 0$ and $-12x - 5y + 2 = 0$.Calculate $a_1a_2 + b_1b_2 = (3)(-12) + (-4)(-5) = -36 + 20 = -16$.Since $a_1a_2 + b_1b_2 $13(3x - 4y + 7) = 5(-12x - 5y + 2)$$39x - 52y + 91 = -60x - 25y + 10$$99x - 27y + 81 = 0$Dividing by $9$,we get $11x - 3y + 9 = 0$.

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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