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

Vedclass · Accepted Answer

$11x - 3y + 9 = 0$

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(C) The equations of the bisectors of the angles between the lines 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 $\frac{3x - 4y + 7}{5} = \pm \frac{12x + 5y - 2}{13}$.Case $1$: $13(3x - 4y + 7) = 5(12x + 5y - 2) \implies 39x - 52y + 91 = 60x + 25y - 10 \implies 21x + 77y - 101 = 0$.Case $2$: $13(3x - 4y + 7) = -5(12x + 5y - 2) \implies 39x - 52y + 91 = -60x - 25y + 10 \implies 99x - 27y + 81 = 0 \implies 11x - 3y + 9 = 0$.To identify the acute angle bisector,we check the angle between the line $3x - 4y + 7 = 0$ (slope $m_1 = 3/4$) and the bisector $11x - 3y + 9 = 0$ (slope $m_2 = 11/3$).$	an 	heta = |\frac{m_2 - m_1}{1 + m_1 m_2}| = |\frac{11/3 - 3/4}{1 + (11/3)(3/4)}| = |\frac{44 - 9}{12 + 33}| = |\frac{35}{45}| = \frac{7}{9} Since $	an 	heta < 1$,the angle is acute. Thus,$11x - 3y + 9 = 0$ is the acute angle bisector.

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

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