Question 519385
<pre>
A(-1,-7),B(5,-5),C(9,-17)

{{{drawing(400,2000/3,-2,10,-18,2,

graph(400,2000/3,-2,10,-18,2),

triangle(-1,-7,5,-5,9,-17),
locate(-1.3,-6.5,A),locate(5,-4.5,B),locate(9,-17,C)
 
 )}}}

It looks like a right triangle, because angle B looks
like a 90° angle.  But it's not acceptable to just say
it looks like a right triangle, we have to show that B
is a right angle.  We need to show that AB is perpendicular
to BC.  To do that we find their slopes, and see if their
product is -1, or if one is the reciprocal of the other
with the opposite sign:

We find the slope of AB 

m = {{{(y[2]-y[1])/(x[2]-x[1])}}} = {{{((-5)-(-7))/((5)-(-1))}}} = {{{(-5+7)/(5+1)}}} = {{{2/6}}} = {{{1/3}}}

We find the slope of BC  A(-1,-7),B(5,-5),C(9,-17)

m = {{{(y[2]-y[1])/(x[2]-x[1])}}} = {{{((-17)-(-5))/((9)-(5))}}} = {{{(-17+5)/(9-5)}}} = {{{(-12)/4}}} = {{{-3}}}

So we see that -3 is the reciprocal of {{{1/3}}} with the opposite sign
or, you can say the slopes have product {{{1/3}}}·(-3) = -1.

So AB and BC are perpendicular and therefore B is a right angle, and
therefore ABC is a right triangle.

Edwin</pre>