we know that
The formula to calculate the slope between two points is equal to
[tex]m=\frac{y2-y1}{x2-x1}[/tex]
we have
[tex]G(-1,3)\ H(1,2)\ J(-3,-1)[/tex]
Step 1
Find the slope GH
[tex]G(-1,3)\ H(1,2)[/tex]
substitute in the formula
[tex]m=\frac{2-3}{1+1}[/tex]
[tex]m=\frac{-1}{2}[/tex]
[tex]mGH=-\frac{1}{2}[/tex]
Step 2
Find the slope HJ
[tex]H(1,2)\ J(-3,-1)[/tex]
substitute in the formula
[tex]m=\frac{-1-2}{-3-1}[/tex]
[tex]m=\frac{-3}{-4}[/tex]
[tex]mHJ=\frac{3}{4}[/tex]
Step 3
Find the slope JG
[tex]J(-3,-1)\ G(-1,3)[/tex]
substitute in the formula
[tex]m=\frac{3+1}{-1+3}[/tex]
[tex]m=\frac{4}{2}[/tex]
[tex]mJG=2[/tex]
Step 4
Verify if triangle GHJ is a right triangle
we know that
if two lines are perpendicular, then the product of their slopes is equal to minus one
so
[tex]m1*m2=-1[/tex]
compare slope GH with slope JG
we have
[tex]mGH=-\frac{1}{2}[/tex]
[tex]mJG=2[/tex]
[tex]-\frac{1}{2}*2=-1[/tex] -----> side GH and side JG are perpendicular
The triangle GHJ is a right triangle
The answer in the attached figure
