Deriving the Formula for the Coordinates of the Foot of the Perpendicular on a Plane
Formula
Let the foot of the perpendicular from a point $P \left( x_{0} , x_{0} \right)$ to a line $l :y = ax + b$ be $Q$. The coordinates of $Q$ are as follows. $$ \left( {\frac{ 1 }{ 1 + a^{2} }} \left[ a^{2} y_{0} + a x_{0} + b \right] , {\frac{ 1 }{ 1 + a^{2} }} \left[ a y_{0} + x_{0} - ab \right] \right) $$
Explanation
수선의 발 refers to the intersection point between the line $l$ and the line passing through the point $P$ that is perpendicular to $l$.
Derivation
Slopes of two perpendicular lines: The product of the slopes of two perpendicular lines is always $-1$.
Since segment $\overline{PQ}$ is perpendicular to $l$, its slope must be $-1/a$. If we denote the coordinates of the foot of the perpendicular by $Q \left( x_{1} , y_{1} \right)$, then point $Q$ is the solution of the following system of equations. $$ \begin{align} y_{1} &= a x_{1} + b \\ y_{0} - y_{1} &= -\frac{1}{a} \left( x_{0} - x_{1} \right) \end{align} $$ To eliminate $y_{1}$, multiply the upper and lower equations by $a$ respectively and add them to obtain $$ \begin{align*} a y_{0} &= a^{2} x_{1} + ab - x_{0} + x_{1} \\ \implies a y_{0} + x_{0} - ab &= \left( 1 + a^{2} \right) x_{1} \end{align*} $$ Multiplying both sides of the lower equation by $- a^{2}$ yields $$ a^{2} y_{1} = a^{2} y_{0} + a x_{0} - a x_{1} $$ Adding this to the upper equation gives $$ \left( 1 + a^{2} \right) y_{1} = a^{2} y_{0} + a x_{0} + b $$
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