Answer:
Part 1) [tex]m_G_H=-\frac{1}{2}[/tex]
Part 2) [tex]m_H_J=\frac{3}{4}[/tex]
Part 3) [tex]m_J_G=2[/tex]
Part 4) The triangle GHJ is a right triangle , because the slopes GH and JG are opposite reciprocal
Step-by-step explanation:
we know that
The formula to calculate the slope between two points is equal to
[tex]m=\frac{y2-y1}{x2-x1}[/tex]
Part 1)
Find the slope of segment GH
we have
G(-1, 3). H(1, 2)
substitute in the formula
[tex]m=\frac{2-3}{1+1}[/tex]
[tex]m=\frac{-1}{2}[/tex]
[tex]m_G_H=-\frac{1}{2}[/tex]
Part 2)
Find the slope of segment HJ
we have
H(1, 2),J(-3,-1)
substitute in the formula
[tex]m=\frac{-1-2}{-3-1}[/tex]
[tex]m=\frac{-3}{-4}[/tex]
[tex]m_H_J=\frac{3}{4}[/tex]
Part 3)
Find the slope of segment JG
we have
J(-3,-1),G(-1, 3)
substitute in the formula
[tex]m=\frac{3+1}{-1+3}[/tex]
[tex]m=\frac{4}{2}[/tex]
[tex]m_J_G=2[/tex]
Part 4) we know that
If two lines are perpendicular, then their slopes are opposite reciprocal (the product of their slopes is equal to -1)
In this problem we have
[tex]m_G_H=-\frac{1}{2}[/tex]
[tex]m_J_G=2[/tex]
Find the product
[tex]m_G_H*m_J_G=(-\frac{1}{2})(2)=-1[/tex]
so
Segments GH and JG are perpendicular
therefore
The triangle GHJ is a right triangle , because the slopes GH and JG are opposite reciprocal
