The foot of the perpendicular drawn from (2, | vedclass

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(B) Let the foot of the perpendicular from $P(x_1, y_1) = (2, 4)$ to the line $ax + by + c = 0$ be $(h, k)$.Here,$a = 1, b = 1, c = -1$.The formula fo

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$\left( -\frac{1}{2}, \frac{3}{2} \right)$

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(B) Let the foot of the perpendicular from $P(x_1, y_1) = (2, 4)$ to the line $ax + by + c = 0$ be $(h, k)$.Here,$a = 1, b = 1, c = -1$.The formula for the foot of the perpendicular is $\frac{h - x_1}{a} = \frac{k - y_1}{b} = -\frac{ax_1 + by_1 + c}{a^2 + b^2}$.Substituting the values: $\frac{h - 2}{1} = \frac{k - 4}{1} = -\frac{1(2) + 1(4) - 1}{1^2 + 1^2}$.$\frac{h - 2}{1} = \frac{k - 4}{1} = -\frac{5}{2}$.$h - 2 = -2.5 \implies h = -0.5 = -\frac{1}{2}$.$k - 4 = -2.5 \implies k = 1.5 = \frac{3}{2}$.Thus,the foot of the perpendicular is $\left( -\frac{1}{2}, \frac{3}{2} \right)$.

The foot of the perpendicular drawn from $(2, 4)$ to the line $x + y = 1$ is

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