Review question

# Where is the foot of the perpendicular from a point to a line? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R5202

## Question

In the $x$$y$ plane, the point $A$ has coordinates $(a,0)$ and the point $B$ has coordinates $(0,b)$, where $a$ and $b$ are positive. The point $P$, which is distinct from $A$ and $B$, has coordinates $(s,t)$. $X$ and $Y$ are the feet of the perpendiculars from $P$ to the $x$-axis and $y$-axis respectively, and $N$ is the foot of the perpendicular from $P$ to the line $AB$. Show that the coordinates $(x,y)$ of $N$ are given by $x = \frac{ab^2 - a(bt - as)}{a^2 + b^2}, \quad y = \frac{a^2b + b(bt - as)}{a^2 + b^2}.$

Show that, if $\left(\frac{t - b}{s}\right)\left(\frac{t}{s - a}\right) = -1,$ then $N$ lies on the line $XY$.

Give a geometrical interpretation of this result.