The straight line x+y+1=0 bisects an angle be | vedclass

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(D) The point of intersection of lines $x+y+1=0$ and $2x-3y+4=0$ is $A\left(-\frac{7}{5}, \frac{2}{5}\right)$.Let $P(-2, 0)$ be a point on the line $2

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$3x-2y+5=0$

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(D) The point of intersection of lines $x+y+1=0$ and $2x-3y+4=0$ is $A\left(-\frac{7}{5}, \frac{2}{5}\right)$.Let $P(-2, 0)$ be a point on the line $2x-3y+4=0$. The image of point $P$ with respect to the bisector line $x+y+1=0$ must lie on the other line.Let the image of $P(-2, 0)$ be $(h, k)$. Using the reflection formula $\frac{h-x_1}{a} = \frac{k-y_1}{b} = -2 \frac{ax_1+by_1+c}{a^2+b^2}$:$\frac{h+2}{1} = \frac{k-0}{1} = -2 \frac{-2+0+1}{1^2+1^2} = -2 \frac{-1}{2} = 1$.Thus,$h+2=1 \Rightarrow h=-1$ and $k=1$.The other line passes through $A\left(-\frac{7}{5}, \frac{2}{5}\right)$ and $(-1, 1)$.The slope $m = \frac{1 - 2/5}{-1 - (-7/5)} = \frac{3/5}{2/5} = \frac{3}{2}$.The equation is $y - 1 = \frac{3}{2}(x + 1)$ $\Rightarrow 2y - 2 = 3x + 3$ $\Rightarrow 3x - 2y + 5 = 0$.Hence,option $D$ is correct.

The straight line $x+y+1=0$ bisects an angle between a pair of lines,of which one is $2x-3y+4=0$. Then the equation of the other line in that pair is

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