Let B_1: 3x + 4y - 7 = 0 and B_2: 4x - 3y - 1 | vedclass

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(A) Let $d_1$ and $d_2$ be the perpendicular distances of the point $(1, 2)$ from the bisectors $B_1$ and $B_2$ respectively.The distance $d_1$ from $

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$B_1$ is the acute angle bisector

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(A) Let $d_1$ and $d_2$ be the perpendicular distances of the point $(1, 2)$ from the bisectors $B_1$ and $B_2$ respectively.The distance $d_1$ from $(1, 2)$ to $3x + 4y - 7 = 0$ is $d_1 = \frac{|3(1) + 4(2) - 7|}{\sqrt{3^2 + 4^2}} = \frac{|3 + 8 - 7|}{5} = \frac{4}{5}$.The distance $d_2$ from $(1, 2)$ to $4x - 3y - 14 = 0$ is $d_2 = \frac{|4(1) - 3(2) - 14|}{\sqrt{4^2 + (-3)^2}} = \frac{|4 - 6 - 14|}{5} = \frac{|-16|}{5} = \frac{16}{5}$.Since $d_1 In the context of angle bisectors,the bisector closer to a point on one of the lines is the acute angle bisector. Therefore,$B_1$ is the acute angle bisector.

Let $B_1: 3x + 4y - 7 = 0$ and $B_2: 4x - 3y - 14 = 0$ be the angle bisectors of the angle between the lines $L_1 = 0$ and $L_2 = 0$. If $L_1$ passes through the point $(1, 2)$,then which of the following is true?

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