The side lengths of triangle are 6 units, 8 units and 10 units.
SOLUTION:
Given that, we have to find what is the length side of a triangle that has vertices at (-5, -1), (-5, 5), and (3, -1)
We know that, distance between two points [tex]P\left(x_{1}, y_{1}\right) \text { and } Q\left(x_{2}, y_{2}\right)[/tex] is given by
[tex]P Q=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}[/tex]
Now,
[tex]\begin{array}{l}{\text { Distance between }(-5,-1) \text { and }(-5,5)=\sqrt{(-5-(-5))^{2}+(5-(-1))^{2}}} \\\\ {\qquad \begin{array}{l}{=\sqrt{(-5-(-5))^{2}+(5-(-1))^{2}}} \\\\ {=\sqrt{(0)^{2}+(5+1)^{2}}=\sqrt{(6)^{2}}=6} \\\\ {=\sqrt{(-5)^{2}+(5+1)^{2}+(5-(-1))^{2}}} \\\\ {=\sqrt{(-8)^{2}+(5+1)^{2}}=\sqrt{64+36}=\sqrt{100}=10} \\\\ {=\sqrt{(3-1)^{2}+(-1-(-1))^{2}}} \\\\ {=\sqrt{(5+3)^{2}+(0)^{2}}=\sqrt{(8)^{2}}=8}\end{array}}\end{array}[/tex]
