Question 647850
<pre>
There are two ways to do this:

1. Find the lengths of all the sides by the distance formula and showing that
they satisfy the Pythagorean theorem c²=a²+b²  

2. Find the slopes of the sides and observe that one of them is the 
"negative reciprocal" of another one.

The second way is easier.

If you make a sketch you can tell which two sides are the perpendicular
ones:
(-1,3), (3,5), (5,1)

{{{drawing(400,400,-3,7,-3,7, graph(400,400,-3,7,-3,7),
circle(-1,3,.1), circle(3,5,.1), circle(5,1,.1),
locate(-2.2,3,"(-1,3)"), locate(3.1,5,"(3,5)"), locate(5.1,1,"(5,1)"),
red(line(-1,3,3,5),line(3,5,5,1)), green(line(-1,3,5,1))
  )}}}

It has to be the red ones.  So we don't need to bother finding the
slope of the green line, (the hypotenuse).

Slope formula
m = {{{(y[2]-y[1])/(x[2]-x[1])}}} with (x<sub>1</sub>, y<sub>1</sub>) = (-1,3) and (x<sub>2</sub>, y<sub>2</sub>) = (3,5)

Substituting:
For the red line on the left:
m = {{{((5)-(3))/((3)-(-1))}}} = {{{(5-3)/(3+1)}}} = {{{2/4}}} = {{{1/2}}}

m = {{{(y[2]-y[1])/(x[2]-x[1])}}} with (x<sub>1</sub>, y<sub>1</sub>) = (-1,3) and (x<sub>2</sub>, y<sub>2</sub>) = (3,5)

Substituting:
For the red line on the right:
m = {{{((1)-(5))/((5)-(3))}}} = {{{(1-5)/(5-3)}}} = {{{(-4)/2}}} = {{{-2}}}

And since those slopes {{{1/2}}} and {{{-2}}} are "negative reciprocals",
those lines with those slopes are perpendicular.  [You can tell that they are
negative reciprocals either because you observe that one is the reciprocal of
the other one with the sign changed or because you observe that if you multiply
them you will get -1, which is true of all fractions multiplied by their
 negative reciprocals.

Edwin</pre>