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(D) Let the coordinates of the foot of the perpendicular be $(h, k)$.Given the line equation $ax + by + c = 0$ and point $(x_1, y_1)$,the formula for

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$(3, 1)$

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(D) Let the coordinates of the foot of the perpendicular be $(h, k)$.Given the line equation $ax + by + c = 0$ and point $(x_1, y_1)$,the formula for the foot of the perpendicular $(h, k)$ is given by:$\frac{h - x_1}{a} = \frac{k - y_1}{b} = -\frac{ax_1 + by_1 + c}{a^2 + b^2}$Here,$a = 3, b = -4, c = -5$ and $(x_1, y_1) = (0, 5)$.Substituting these values into the formula:$\frac{h - 0}{3} = \frac{k - 5}{-4} = -\frac{3(0) - 4(5) - 5}{3^2 + (-4)^2}$$\frac{h}{3} = \frac{k - 5}{-4} = -\frac{0 - 20 - 5}{9 + 16}$$\frac{h}{3} = \frac{k - 5}{-4} = -\frac{-25}{25}$$\frac{h}{3} = \frac{k - 5}{-4} = 1$From $\frac{h}{3} = 1$,we get $h = 3$.From $\frac{k - 5}{-4} = 1$,we get $k - 5 = -4$,so $k = 1$.Thus,the coordinates of the foot of the perpendicular are $(3, 1)$.

Find the coordinates of the foot of the perpendicular drawn from the point $(0, 5)$ to the line $3x - 4y - 5 = 0$.

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