Question

Triangle ABC has vertices located at A( 0, 2), B (2,5), and C (-1, 7). Find the length of each side of the triangle Show your work. (4 points)

Answer 1

Answer:

• AB=BC=√13 units

,

• AC=√26 units

Explanation:

Given the coordinates of ABC: A(0, 2), B(2,5), and C(-1, 7)

To find the length of each side of the triangle, we use the distance formula:

$Distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Length of AB: A(0, 2), B(2,5)

$\begin{gathered} AB=\sqrt[]{(2-0)^2+(5-2)^2} \\ =\sqrt[]{(2)^2+(3)^2} \\ =\sqrt[]{4+9} \\ =\sqrt[]{13}\text{ units} \end{gathered}$

Length of BC: B(2,5), C(-1, 7)

$\begin{gathered} BC=\sqrt[]{(-1-2)^2+(7-5)^2} \\ =\sqrt[]{(-3)^2+(2)^2} \\ =\sqrt[]{9+4} \\ =\sqrt[]{13}\text{ units} \end{gathered}$

Length of AC: A(0, 2), C(-1, 7)

$\begin{gathered} AC=\sqrt[]{(-1-0)^2+(7-2)^2} \\ =\sqrt[]{(-1)^2+(5)^2} \\ =\sqrt[]{1+25} \\ =\sqrt[]{26}\text{ units} \end{gathered}$

The lengths of the sides of the triangle are:

• AB=√13 units

,

• BC=√13 units

,

• AC=√26 units