The point (a, b) is the foot of the perpendic | vedclass

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(B) Since $(a, b)$ is the foot of the perpendicular drawn from $(3, 1)$ to the line $x + 3y + 4 = 0$,we have $a + 3b + 4 = 0$ $(i)$.The equation of th

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$-5$

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(B) Since $(a, b)$ is the foot of the perpendicular drawn from $(3, 1)$ to the line $x + 3y + 4 = 0$,we have $a + 3b + 4 = 0$ $(i)$.The equation of the line passing through $(3, 1)$ and perpendicular to $x + 3y + 4 = 0$ is $3x - y - 8 = 0$,so $3a - b - 8 = 0$ $(ii)$.Solving $(i)$ and $(ii)$,we get $(a, b) = (2, -2)$.Let $(p, q)$ be the image of $(2, -2)$ with respect to $3x - 4y + 11 = 0$. The midpoint $P = \left(\frac{2+p}{2}, \frac{-2+q}{2}\right)$ lies on the line $3x - 4y + 11 = 0$,so $3(\frac{2+p}{2}) - 4(\frac{-2+q}{2}) + 11 = 0$,which simplifies to $3p - 4q + 36 = 0$ $(iii)$.The line joining $(2, -2)$ and $(p, q)$ is perpendicular to $3x - 4y + 11 = 0$. The slope of the given line is $\frac{3}{4}$,so the slope of the line joining $(2, -2)$ and $(p, q)$ is $-\frac{4}{3}$.Thus,$\frac{q - (-2)}{p - 2} = -\frac{4}{3}$ $\Rightarrow 3(q + 2) = -4(p - 2)$ $\Rightarrow 4p + 3q - 2 = 0$ $(iv)$.Solving $(iii)$ and $(iv)$,we get $p = -4$ and $q = 6$.Finally,$\frac{p}{a} + \frac{q}{b} = \frac{-4}{2} + \frac{6}{-2} = -2 - 3 = -5$.

The point $(a, b)$ is the foot of the perpendicular drawn from the point $(3, 1)$ to the line $x + 3y + 4 = 0$. If $(p, q)$ is the image of $(a, b)$ with respect to the line $3x - 4y + 11 = 0$,then $\frac{p}{a} + \frac{q}{b} =$

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