Question

1. the points a(1, -7), b(1, -1), and c(-6, 2) form a triangle. graph the triangle on the set of axes below and find the perimeter in simplest radical form.

Explanation:

Step1: Find the lengths of the sides using 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 side \(AB\) with \(A(1,-7)\) and \(B(1,-1)\):
\(x_1 = 1,y_1=-7,x_2 = 1,y_2=-1\)
\(AB=\sqrt{(1 - 1)^2+(-1+7)^2}=\sqrt{0 + 36}=6\)
For side \(BC\) with \(B(1,-1)\) and \(C(-6,2)\):
\(x_1 = 1,y_1=-1,x_2=-6,y_2 = 2\)
\(BC=\sqrt{(-6 - 1)^2+(2 + 1)^2}=\sqrt{(-7)^2+3^2}=\sqrt{49+9}=\sqrt{58}\)
For side \(AC\) with \(A(1,-7)\) and \(C(-6,2)\):
\(x_1 = 1,y_1=-7,x_2=-6,y_2 = 2\)
\(AC=\sqrt{(-6 - 1)^2+(2 + 7)^2}=\sqrt{(-7)^2+9^2}=\sqrt{49 + 81}=\sqrt{130}\)

Step2: Calculate the perimeter

The perimeter \(P\) of \(\triangle ABC\) is \(P=AB + BC+AC\)
\(P=6+\sqrt{58}+\sqrt{130}\)

Answer:

\(6+\sqrt{58}+\sqrt{130}\)