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

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(B) Let the two lines be $L_1: 3x + 2y + 4 = 0$ and $L_2: ax + by + c = 0$. The bisector is $L_B: 2x + 3y + 1 = 0$. Since $L_B$ is the angle bisector,

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$9x + 46y - 28 = 0$

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(B) Let the two lines be $L_1: 3x + 2y + 4 = 0$ and $L_2: ax + by + c = 0$. The bisector is $L_B: 2x + 3y + 1 = 0$. Since $L_B$ is the angle bisector,any point $(x, y)$ on $L_B$ is equidistant from $L_1$ and $L_2$. However,a simpler property is that the bisector of the angle between two lines $L_1$ and $L_2$ is given by $\frac{L_1}{\sqrt{A_1^2 + B_1^2}} = \pm \frac{L_2}{\sqrt{A_2^2 + B_2^2}}$. Given $L_1: 3x + 2y + 4 = 0$ and $L_B: 2x + 3y + 1 = 0$,the angle bisector $L_B$ satisfies the condition of being the locus of points equidistant from $L_1$ and $L_2$. By calculating the reflection or using the property of the angle bisector,we find the equation of the other line $L_2$ to be $9x + 46y - 28 = 0$.

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

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