Question 1030650
<pre><b>
We draw the triangle to see what it looks like:

 (3, -2), (7, 3), and (-7, 6) 

{{{drawing(400,800/3,-9,9,-4,8,
circle(3,-2,0.15),circle(3,-2,0.13),circle(3,-2,0.11),circle(3,-2,0.09),circle(3,-2,0.07),circle(3,-2,0.05),circle(3,-2,0.03),circle(3,-2,0.01),
green(triangle(3, -2, 7, 3, -7, 6)),
circle(7,3,0.15),circle(7,3,0.13),circle(7,3,0.11),circle(7,3,0.09),circle(7,3,0.07),circle(7,3,0.05),circle(7,3,0.03),circle(7,3,0.01),
locate(-9,6,"(-7,6)"), locate(3,-2,"(3,-2)"), locate(7,3,"(7,3)"),
circle(-7,6,0.15),circle(-7,6,0.13),circle(-7,6,0.11),circle(-7,6,0.09),circle(-7,6,0.07),circle(-7,6,0.05),circle(-7,6,0.03),circle(-7,6,0.01),
graph(400,800/3,-9,9,-4,8) )}}}

Looks suspiciously like the angle down at (3,-2) is a right
angle.   So let's see if we can prove that it is.

We find the slopes of its two sides and if that's a right angle 
then the slope of one with be the reciprocal of the other with
the opposite sign

We use the slope formula:

{{{m}}}{{{""=""}}}{{{(y[2]-y[1])/(x[2]-x[1])}}}
where (x<sub>1</sub>,y<sub>1</sub>) = (-7,6)
and where (x<sub>2</sub>,y<sub>2</sub>) = (3,-2)

{{{m}}}{{{""=""}}}{{{((-2)-(6))/((3)-(-7))}}}
{{{m}}}{{{""=""}}}{{{(-8)/(10)}}}{{{""=""}}}{{{-4/5}}}

We use the slope formula again:

{{{m}}}{{{""=""}}}{{{(y[2]-y[1])/(x[2]-x[1])}}}
this time with (x<sub>1</sub>,y<sub>1</sub>) = (7,3)
and where (x<sub>2</sub>,y<sub>2</sub>) = (3,-2)

{{{m}}}{{{""=""}}}{{{((-2)-(3))/((3)-(7))}}}
{{{m}}}{{{""=""}}}{{{(-5)/(-4)}}}{{{""=""}}}{{{5/4}}}

Indeed {{{-4/5}}} and {{{5/4}}} are reciprocals with
opposite signs, so the sides are perpendicular and
the angle with vertex (3,-2) is a right angle.

So the triangle is a right triangle.

Edwin</pre></b>