Question

on a coordinate plane,△fgh is formed with the vertices at point f(0, -2), point g(1, 0), and point h(2, -3). which expression can be used to determine the perimeter, in units, of △fgh?

Explanation:

Step1: Use the distance formula

The distance formula between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\).
For \(FG\) with \(F(0,-2)\) and \(G(1,0)\):
\(d_{FG}=\sqrt{(1 - 0)^2+(0-(-2))^2}=\sqrt{(1 - 0)^2+(0 + 2)^2}\)
For \(GH\) with \(G(1,0)\) and \(H(2,-3)\):
\(d_{GH}=\sqrt{(2 - 1)^2+(-3-0)^2}=\sqrt{(2 - 1)^2+(-3-0)^2}\)
For \(FH\) with \(F(0,-2)\) and \(H(2,-3)\):
\(d_{FH}=\sqrt{(2 - 0)^2+(-3-(-2))^2}=\sqrt{(2 - 0)^2+(-3 + 2)^2}\)
The perimeter of \(\triangle FGH\) is \(P=d_{FG}+d_{GH}+d_{FH}=\sqrt{(0 - 1)^2+(-2-0)^2}+\sqrt{(1 - 2)^2+(0+3)^2}+\sqrt{(0 - 2)^2+(-2 + 3)^2}\)

Answer:

\(\sqrt{(0 - 1)^2+(-2-0)^2}+\sqrt{(1 - 2)^2+(0+3)^2}+\sqrt{(0 - 2)^2+(-2 + 3)^2}\)