The coordinates of the foot of the perpendicu | vedclass

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(A) Let the foot of the perpendicular be $(x, y)$.Using the formula for the foot of the perpendicular from $(x_1, y_1)$ to $ax + by + c = 0$:$\frac{x

Vedclass · Accepted Answer

$\left(\frac{4}{5}, \frac{13}{5}\right)$

Vedclass · Answer

(A) Let the foot of the perpendicular be $(x, y)$.Using the formula for the foot of the perpendicular from $(x_1, y_1)$ to $ax + by + c = 0$:$\frac{x - x_1}{a} = \frac{y - y_1}{b} = -\frac{ax_1 + by_1 + c}{a^2 + b^2}$Substituting the values $(x_1, y_1) = (1, 2)$ and $x - 3y + 7 = 0$:$\frac{x - 1}{1} = \frac{y - 2}{-3} = -\frac{1 - 3(2) + 7}{1^2 + (-3)^2}$$\frac{x - 1}{1} = \frac{y - 2}{-3} = -\frac{1 - 6 + 7}{1 + 9} = -\frac{2}{10} = -\frac{1}{5}$Now,solve for $x$ and $y$:$x - 1 = -\frac{1}{5} \Rightarrow x = 1 - \frac{1}{5} = \frac{4}{5}$$y - 2 = -3 	imes \left(-\frac{1}{5}ight) = \frac{3}{5} \Rightarrow y = 2 + \frac{3}{5} = \frac{13}{5}$Thus,the foot of the perpendicular is $\left(\frac{4}{5}, \frac{13}{5}ight)$.

The coordinates of the foot of the perpendicular from the point $(1, 2)$ on the line $x - 3y + 7 = 0$ are

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