Question 1012777
There are two ways to show that the triangle is a right triangle.
You could find that two of the sides are perpendicular, or
you could find that the square of the length of one side is the sum of the squares of the lengths of the other two sides.
 
USING THE LENGTHS OF THE SIDES:
{{{PQ^2=(2-(-4))^2+(5-(-3))^2=(2+4)^2+(5+3)^2=6^2+8^2=36+64=100}}}
{{{QR^2=(4-2)^2+(5-1)^2=2^2+4^2=4+16=20}}}
{{{PR^2=(4-(-4))^2+(1-(-3))^2=(4+4)^2+(1+3)^2=8^2+4^2=64+16=80}}}
{{{QR^2+PR^2=20+80=100=PQ^2}}}
So, by the converse of the Pythagorean theorem,
{{{QR}}} and {{{PR}}} are the legs of a right triangle with hypotenuse {{{PQ}}} .
 
PROVING THAT TWO SIDES ARE PERPENDICULAR:
IF you have learned about slope of a line,
you may also have learned that if two lines have slopes whose product is {{{-1}}} , those lines are perpendicular.
Slope of {{{PR}}}={{{(1-(-3))/(4-(-4))=(1+3)/(4+4)=4/8=1/2}}}
Slope of {{{QR}}}={{{(5-1)/(2-4)=4/(-2)=-2}}}
The product of the slopes is {{{(1/2)(-2)=-1}}} ,
so {{{PR}}} and {{{QR}}} are perpendicular,
which means that triangle {{{PQR}}} is a right triangle,
with a right angle at {{{R}}} and hypotenuse {{{PQ}}} .
 
MIDPOINT OF THE HYPOTENUSE:
The coordinates of {{{M(x[M],y[M])}}} , the midpoint of hypotenuse {{{PQ}}} , are found by averaging the coordinates of {{{P}}} and {{{Q}}} :
{{{x[M]=(x[P]+x[Q])/2=(-4+2)/2_-2/2=-1}}}
{{{y[M]=(y[P]+y[Q])/2=(-3+5)/2_2/2=1}}}
 
DISTANCES FROM {{{M(-1,1)}}} TO {{{P}}} , {{{Q}}} , and {{{R}}} :
To show that the distances are the same, we can just show that their squares are the same.
{{{PM^2=(-4-(-1))^2+(-3-1)^2=(-4+1)^2+(-4)^2=(-3)^2+16=9+16=25}}}
{{{QM^2=(2-(-1))^2+(5-1)^2=(2+1)^2+4^2=3^2+16=9+16=25}}}
{{{RM^2=(4-(-1))^2+(1-1)^2=(4+1)^2+0^2=5^2+0=25+0=25}}} .
So, the distances from {{{M}}} to {{{P}}} , {{{Q}}} , and {{{R}}} are all {{{sqrt(25)=5}}} .