The coordinates of the foot of the perpendicu | vedclass

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(A) Let the coordinates of the foot of the perpendicular be $(h, k)$.Since $(h, k)$ lies on the line $ax + by + c = 0$,we have $ah + bk + c = 0$.The s

Vedclass · Accepted Answer

$\left( \frac{b^2x_1 - aby_1 - ac}{a^2 + b^2}, \frac{a^2y_1 - abx_1 - bc}{a^2 + b^2} \right)$

Vedclass · Answer

(A) Let the coordinates of the foot of the perpendicular be $(h, k)$.Since $(h, k)$ lies on the line $ax + by + c = 0$,we have $ah + bk + c = 0$.The slope of the line $ax + by + c = 0$ is $m_1 = -\frac{a}{b}$.The slope of the line joining $(x_1, y_1)$ and $(h, k)$ is $m_2 = \frac{k - y_1}{h - x_1}$.Since the lines are perpendicular,$m_1 	imes m_2 = -1$,which implies $\left(-\frac{a}{b}ight) 	imes \left(\frac{k - y_1}{h - x_1}ight) = -1$,so $\frac{k - y_1}{h - x_1} = \frac{b}{a}$.Using the property $\frac{h - x_1}{a} = \frac{k - y_1}{b} = -\frac{ax_1 + by_1 + c}{a^2 + b^2}$,we solve for $h$ and $k$:$h = x_1 - a \left( \frac{ax_1 + by_1 + c}{a^2 + b^2} ight) = \frac{x_1(a^2 + b^2) - a^2x_1 - aby_1 - ac}{a^2 + b^2} = \frac{b^2x_1 - aby_1 - ac}{a^2 + b^2}$.$k = y_1 - b \left( \frac{ax_1 + by_1 + c}{a^2 + b^2} ight) = \frac{y_1(a^2 + b^2) - abx_1 - b^2y_1 - bc}{a^2 + b^2} = \frac{a^2y_1 - abx_1 - bc}{a^2 + b^2}$.Thus,the correct option is $A$.

The coordinates of the foot of the perpendicular from $(x_1, y_1)$ to the line $ax + by + c = 0$ are

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