Rectangular Coordinate Systems

Question

Let $A(0,2)$ and $B(6,0).$ Find all points $P$ on the line $y = 4$ such that $AP\perp BP.$
(a) by slope
(b) by Pythagorean Theorem

I tried getting the midpoint which is $M=(3,1)$ and then I looked for the slope $m= -1/3$ and the inverse of slope $3$ to find the equation of the line perpendicular to the points $A$ and $B$ which is $y = 3x -8.$

Answer 1

Take P $(x,4)$ Find slope of AP and BP in terms of $x $. Now use the relation between perpendicular slopes and equate and solve for $x $

Answer 2

If $AP$ and $BP$ are perpendicular, that means they form a right angle and $AB$ is the hypotenuse of a right triangle. So you can use the distance formula to find the distance between $(0, 2)$ and $(p, 4)$ and the distance between $(6, 0)$ and $(p, 4)$. Plug these distances in to the Pythagorean Theorem to see if they make a right triangle.

So we have these distances:
$AB=\sqrt{\left(6-0\right)^2+\left(0-2\right)^2}=\sqrt{36+4}=\sqrt{40}$
$AP=\sqrt{\left(p-0\right)^2+\left(4-2\right)^2}=\sqrt{p^2+4}$
$BP=\sqrt{\left(6-p\right)^2+\left(0-4\right)^2}=\sqrt{36-12p+p^2+16}=\sqrt{p^2-12p+52}$

Now we use the converse of the Pythagorean Theorem to see which values of $p$ (if any) make a right triangle. (This is the fun part because we make the square root symbols go away.)

$\left(\sqrt{p^2-12p+52}\right)^2+\left(\sqrt{p^2+4}\right)^2=\left(\sqrt{40}\right)^2$
$p^2-12p+52+p^2+4=40$
$2p^2-12p+56=40$
$2p^2-12p+16=0$

This quadratic equation is easy to factor: $2\left(p^2-6p+8\right)=0$
$2\left(p-2\right)\left(p-4\right)=0$

Thus we have $p=2$ or $p=4$. In the context of the problem, this means the points that make a right triangle (where $AP$ and $BP$ are perpendicular) are $(2,4)$ and $(4,4)$.