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(C) Let the lines be $L_1: 2x + 3y + 4 = 0$ and $L_2: 3x - 2y + 4 = 0$.The length of the perpendicular from $(a, b)$ to $L_1$ is $d_1 = \frac{|2a + 3b

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$x - 5y = 0$ or $5x + y + 8 = 0$

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(C) Let the lines be $L_1: 2x + 3y + 4 = 0$ and $L_2: 3x - 2y + 4 = 0$.The length of the perpendicular from $(a, b)$ to $L_1$ is $d_1 = \frac{|2a + 3b + 4|}{\sqrt{2^2 + 3^2}} = \frac{|2a + 3b + 4|}{\sqrt{13}}$.The length of the perpendicular from $(a, b)$ to $L_2$ is $d_2 = \frac{|3a - 2b + 4|}{\sqrt{3^2 + (-2)^2}} = \frac{|3a - 2b + 4|}{\sqrt{13}}$.Given $d_1 = d_2$,we have $|2a + 3b + 4| = |3a - 2b + 4|$.This implies $2a + 3b + 4 = 3a - 2b + 4$ or $2a + 3b + 4 = -(3a - 2b + 4)$.Case $1$: $2a + 3b + 4 = 3a - 2b + 4 \Rightarrow a - 5b = 0$.Case $2$: $2a + 3b + 4 = -3a + 2b - 4 \Rightarrow 5a + b + 8 = 0$.Thus,the locus of $(a, b)$ is $x - 5y = 0$ or $5x + y + 8 = 0$.

If the lengths of the perpendiculars drawn from a point $(a, b)$ to the lines $2x + 3y + 4 = 0$ and $3x - 2y + 4 = 0$ are equal,then the point $(a, b)$ lies on the line

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