If the straight line 2x + 3y + 1 = 0 bisects | vedclass

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(C) Let the two lines be $L_1: 3x + 2y + 4 = 0$ and $L_2: ax + by + c = 0$. The line $L: 2x + 3y + 1 = 0$ is the angle bisector.First,find the interse

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$9x + 46y - 28 = 0$

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(C) Let the two lines be $L_1: 3x + 2y + 4 = 0$ and $L_2: ax + by + c = 0$. The line $L: 2x + 3y + 1 = 0$ is the angle bisector.First,find the intersection point of $L_1$ and $L$:$3x + 2y = -4$ (multiplied by $3$ gives $9x + 6y = -12$)$2x + 3y = -1$ (multiplied by $2$ gives $4x + 6y = -2$)Subtracting the equations: $(9x - 4x) = -12 - (-2)$ $\Rightarrow 5x = -10$ $\Rightarrow x = -2$.Substituting $x = -2$ into $2x + 3y + 1 = 0$: $2(-2) + 3y + 1 = 0$ $\Rightarrow -4 + 3y + 1 = 0$ $\Rightarrow 3y = 3$ $\Rightarrow y = 1$.The intersection point is $(-2, 1)$.The angle bisector $L$ makes an angle $	heta$ with $L_1$. The other line $L_2$ must also make the same angle $	heta$ with $L$ on the other side.The slope of $L_1$ is $m_1 = -3/2$. The slope of $L$ is $m = -2/3$. Let the slope of $L_2$ be $m_2$.The angle between $L_1$ and $L$ is given by $	an 	heta = |(m - m_1) / (1 + m \cdot m_1)| = |(-2/3 - (-3/2)) / (1 + (-2/3)(-3/2))| = |(-4/6 + 9/6) / (1 + 1)| = |(5/6) / 2| = 5/12$.Since $L$ bisects the angle,$	an 	heta = |(m_2 - m) / (1 + m_2 \cdot m)| = 5/12$.$|(m_2 + 2/3) / (1 - 2m_2/3)| = 5/12 \Rightarrow |(3m_2 + 2) / (3 - 2m_2)| = 5/12$.Case $1$: $(3m_2 + 2) / (3 - 2m_2) = 5/12$ $\Rightarrow 36m_2 + 24 = 15 - 10m_2$ $\Rightarrow 46m_2 = -9$ $\Rightarrow m_2 = -9/46$.Using point-slope form with $(-2, 1)$: $y - 1 = (-9/46)(x + 2)$ $\Rightarrow 46y - 46 = -9x - 18$ $\Rightarrow 9x + 46y - 28 = 0$.

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

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