Where is the foot of the perpendicular from a point to a line?

Where is the foot of the perpendicular from a point to a line?

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.

Interactive diagram

Use this applet to explore the behaviour as $A$, $B$ and $P$ are moved.

Download GeoGebra file

What can you say about the position of $P$ when $N$ is on $XY$?

You can show the circle that has $AB$ as a diameter.

Solution

Here’s a diagram of the situation.

You might also like to experiment with the situation in a more dynamic way using the GeoGebra applet here.

The gradient of the line $AB$ is $\frac{-b}{a}$, so the line has equation $y=-\frac{b}{a}(x-a)$, which rearranges to $bx+ay=ab$ or $\frac{x}{a}+\frac{y}{b}=1$, which has a nice symmetry to it.

As the line $AB$ has gradient $-\frac{b}{a}$, it follows that the line perpendicular line $NP$ has gradient $\frac{a}{b}$. So $NP$, passing through $P(s,t)$ has equation $y-t = \frac{a}{b}(x - s)$, which we can rearrange to get $ax-by=as-bt$.

Now we know that

$AB$

and

$NP$

intersect at

$N$

, so to find the coordinates of

$N$

we solve the simultaneous equations

$$\begin{align} &&bx+ay&=ab&&\quad\\ \text{and}\quad&&ax-by&=as-bt. \end{align}$$

Adding $b\times(1)$ to $a\times(2)$ gives $(a^2+b^2)x=ab^2+a^2s-abt$ so $x=\frac{ab^2+a^2s-abt}{a^2+b^2}=\frac{ab^2-a(bt-as)}{a^2+b^2}.$

Similarly, subtracting $b\times(2)$ from $a\times(1)$ gives $(a^2+b^2)y=a^2b-abs+b^2t$ so $y=\frac{a^2b-abs+b^2t}{a^2+b^2}=\frac{a^2b+b(bt-as)}{a^2+b^2}$ as required.

The points $X$ and $Y$ have coordinates $(s,0)$ and $(0,t)$ respectively, so by the same argument as for $AB$, we see that the equation of the line $XY$ is $tx+sy = st.$

Now

$N$

lies on this line exactly when

$$\begin{align*} &t\frac{ab^2-a(bt-as)}{a^2+b^2} + s\frac{a^2b+b(bt-as)}{a^2+b^2}=st\\ \iff\quad& ab^2t-at(bt-as) + a^2bs + bs(bt-as) = st(a^2+b^2)\\ \iff\quad& ab^2t - abt^2 +a^2st + a^2bs + b^2st - abs^2 = a^2st + b^2st\\ \iff\quad& ab^2t - abt^2 + a^2bs - abs^2 = 0\\ \iff\quad& bt - t^2 + as - s^2 = 0\quad\text{as }ab\text{ is non-zero}\\ \iff\quad& t(b-t) + s(a-s) = 0. \end{align*}$$

Then assuming that $s$ and $a-s$ are non-zero, we can divide by them to see that $N$ lies on this line exactly when $\frac{t(b-t)}{s(a-s)}+1=0,$ which is equivalent to the condition we were asked to show (and in fact stronger, as we have shown that this is also a necessary condition).

A geometrical interpretation of this result is that $\dfrac{t - b}{s}$ and $\dfrac{t}{s - a}$ are the gradients of $BP$ and $AP$ respectively. So if $BP$ and $AP$ are perpendicular, then $N$ lies on $XY$.

And what does this mean? We know that the angle in a semi-circle is a right angle.

So $N$ lies on $XY$ exactly when $P$ lies on the circle with $AB$ as diameter.

You can explore this further by showing the circle in the GeoGebra applet.