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(C) Let the slope of the required line be $m$. Since the line $x+y+1=0$ is the angle bisector,the angle between the bisector and the given line $2x+3y

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$3x+2y+9=0$

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(C) Let the slope of the required line be $m$. Since the line $x+y+1=0$ is the angle bisector,the angle between the bisector and the given line $2x+3y-4=0$ must be equal to the angle between the bisector and the required line. The slope of the bisector is $m_1 = -1$. The slope of the given line is $m_2 = -2/3$. The tangent of the angle $	heta$ between the bisector and the given line is: $	an 	heta = \left| \frac{-2/3 - (-1)}{1 + (-2/3)(-1)} ight| = \left| \frac{1/3}{1 + 2/3} ight| = \left| \frac{1/3}{5/3} ight| = \frac{1}{5}$. Now,let the slope of the required line be $m$. The tangent of the angle between the bisector and the required line is: $	an 	heta = \left| \frac{m - (-1)}{1 + m(-1)} ight| = \left| \frac{m+1}{1-m} ight|$. Equating the two: $\left| \frac{m+1}{1-m} ight| = \frac{1}{5} \implies \frac{m+1}{1-m} = \frac{1}{5}$ or $\frac{m+1}{1-m} = -\frac{1}{5}$. Case $1$: $5m+5 = 1-m \implies 6m = -4 \implies m = -2/3$ (This is the given line). Case $2$: $5m+5 = -1+m \implies 4m = -6 \implies m = -3/2$ (This is the required line). The point of intersection of $x+y+1=0$ and $2x+3y-4=0$ is found by solving the system: $2x+2y = -2$ and $2x+3y = 4$. Subtracting gives $y = 6$,and substituting back gives $x = -7$. The equation of the line with slope $m = -3/2$ passing through $(-7, 6)$ is: $y - 6 = -\frac{3}{2}(x + 7) \implies 2y - 12 = -3x - 21 \implies 3x + 2y + 9 = 0$.

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

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