Answer:
The calculate a slope we use the equation:
[tex]m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}[/tex]; where [tex]m[/tex] is the slope, [tex](x_{1};y_{1})[/tex] are the coordinates of the first point, and [tex](x_{2};y_{2})[/tex] are the coordinates of the second point. The first and second points are arbitrary, we just choose a specific order b our own.
So, the problem is asking for the slope of AB, BC, CD and AD, which are sides of a quadrilateral figures. Now, we use each coordinates, replace them is the slope equations and get the result.
The slope of AB:
Coordinates are [tex]A(-1, -1)[/tex] and [tex]B(-3, 3)[/tex].
[tex]m_{AB} =\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{3-(-1)}{-3-(-1)}=\frac{3+1}{-3+1} =\frac{4}{-2}=-2[/tex]
The slope of BC:
Coordinates are [tex]B(-3, 3)[/tex] and [tex]C(1, 5)[/tex]
[tex]m_{BC} =\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{5-3}{1-(-3)} =\frac{2}{1+3}=\frac{2}{4}=\frac{1}{2}[/tex]
The slope of CD:
Coordinates are [tex]C(1, 5)[/tex] and [tex]D(5, 2)[/tex]
[tex]m_{CD} =\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{2-5}{5-1}=\frac{-3}{4}[/tex]
The slope of AD:
Coordinates are [tex]A(-1, -1)[/tex] and [tex]D(5, 2)[/tex]
[tex]m_{AD} =\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{2-(-1)}{5-(-1)} =\frac{2+1}{5+1}=\frac{3}{6}=\frac{1}{2}[/tex]
According to these results, sides AB and BC are perpendicular because the slopes have opposite signs and inverse numbers. In addition, BC and AD are parallel, because they have the same slope. The graph is attached.
