Find the equation of the bisector of the acut | vedclass

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(C) Given lines are $L_1: 3x - 4y + 7 = 0$ and $L_2: 12x + 5y - 2 = 0$. First,make the constant terms positive: $L_1: 3x - 4y + 7 = 0$ $L_2: -12x - 5y

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$11x - 3y + 9 = 0$

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(C) Given lines are $L_1: 3x - 4y + 7 = 0$ and $L_2: 12x + 5y - 2 = 0$. First,make the constant terms positive: $L_1: 3x - 4y + 7 = 0$ $L_2: -12x - 5y + 2 = 0$ Now,check the sign of $a_1a_2 + b_1b_2$: $a_1a_2 + b_1b_2 = (3)(-12) + (-4)(-5) = -36 + 20 = -16 Since the value is negative,the positive sign in the bisector formula $\frac{a_1x + b_1y + c_1}{\sqrt{a_1^2 + b_1^2}} = \frac{a_2x + b_2y + c_2}{\sqrt{a_2^2 + b_2^2}}$ gives the acute angle bisector. $\frac{3x - 4y + 7}{5} = \frac{-12x - 5y + 2}{13}$ $13(3x - 4y + 7) = 5(-12x - 5y + 2)$ $39x - 52y + 91 = -60x - 25y + 10$ $99x - 27y + 81 = 0$ Dividing by $9$,we get $11x - 3y + 9 = 0$.

Find 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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