Question

What is the length side of a triangle that has vertices at (-5, -1), (-5, 5), and (3, -1)?

Answer 1

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
`P\left(x_(1), y_(1)\right) \text { and } Q\left(x_(2), y_(2)\right)` is given by

`P Q=\sqrt{\left(x_(2)-x_(1)\right)^(2)+\left(y_(2)-y_(1)\right)^(2)}`

Now,

`\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)}=√(64+36)=√(100)=10} \\\\ {=\sqrt{(3-1)^(2)+(-1-(-1))^(2)}} \\\\ {=\sqrt{(5+3)^(2)+(0)^(2)}=\sqrt{(8)^(2)}=8}\end{array}}\end{array}`