Question

given triangle abc has vertices located at a(-2,2), b(3,3), and c(1,-3). part a. find the length of ab. part b. find the length of bc. part c. find the length of ac. part d. use the lengths of the sides of the triangle to classify the type of triangle abc is.

Explanation:

Part A

Step1: Recall 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 points \(A(-2,2)\) and \(B(3,3)\), \(x_1=-2,y_1 = 2,x_2=3,y_2=3\).

Step2: Substitute into formula

Substitute the values into the formula: \(AB=\sqrt{(3 - (-2))^2+(3 - 2)^2}=\sqrt{(5)^2+(1)^2}=\sqrt{25 + 1}=\sqrt{26}\)

Step1: Recall 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 points \(B(3,3)\) and \(C(1,-3)\), \(x_1 = 3,y_1=3,x_2=1,y_2=-3\).

Step2: Substitute into formula

Substitute the values: \(BC=\sqrt{(1 - 3)^2+(-3 - 3)^2}=\sqrt{(-2)^2+(-6)^2}=\sqrt{4 + 36}=\sqrt{40}=2\sqrt{10}\)

Step1: Recall 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 points \(A(-2,2)\) and \(C(1,-3)\), \(x_1=-2,y_1 = 2,x_2=1,y_2=-3\).

Step2: Substitute into formula

Substitute the values: \(AC=\sqrt{(1 - (-2))^2+(-3 - 2)^2}=\sqrt{(3)^2+(-5)^2}=\sqrt{9 + 25}=\sqrt{34}\)

Answer:

\(\sqrt{26}\)

Part B