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(A) The equation of the angle bisector between two lines $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$ is given by the formula: $\frac{a_1x + b_

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$\frac{3x + 4y - 7}{5} = \pm \frac{12x + 5y + 17}{13}$

Vedclass · Answer

(A) The equation of the angle bisector between two lines $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$ is given by the formula: $\frac{a_1x + b_1y + c_1}{\sqrt{a_1^2 + b_1^2}} = \pm \frac{a_2x + b_2y + c_2}{\sqrt{a_2^2 + b_2^2}}$ For the given lines $3x + 4y - 7 = 0$ and $12x + 5y + 17 = 0$: $a_1 = 3, b_1 = 4, c_1 = -7$ $a_2 = 12, b_2 = 5, c_2 = 17$ Calculating the denominators: $\sqrt{a_1^2 + b_1^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$ $\sqrt{a_2^2 + b_2^2} = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13$ Substituting these into the formula: $\frac{3x + 4y - 7}{5} = \pm \frac{12x + 5y + 17}{13}$ Thus,option $A$ is correct.

The equation of the angle bisectors between the lines $3x + 4y - 7 = 0$ and $12x + 5y + 17 = 0$ is:

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