Question

find the perimeter of the triangle with vertices at A(0,0), B(-4,-3), and C(-8,0)

Answer 1

Answer:

18units

Explanation:

Given the vertices of a triangle A(0,0), B(-4,-3), and C(-8,0)​

Perimeter of the triangle = AB + BC + AC

Using the distance formula;

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

Get AB;

$\begin{gathered} AB=\sqrt[]{(-3_{}-0_{})^2+(-4_{}-0_{})^2} \\ AB=\sqrt[]{9_{}+16} \\ AB=\sqrt[]{25} \\ AB=5\text{units} \end{gathered}$

Get the distance BC;

$\begin{gathered} BC=\sqrt[]{(0-(-3))^2_{}+(-8-(-4))^2} \\ BC=\sqrt[]{3^2+(-4)^2} \\ BC=\sqrt[]{9+16} \\ BC=\sqrt[]{25} \\ BC=5\text{units} \end{gathered}$

Get the distance AC:

$\begin{gathered} AC=\sqrt[]{(0-0)^2+(-8-0)^2} \\ AC=\sqrt[]{(-8)^2} \\ AC=\sqrt[]{64} \\ AC=8\text{units} \end{gathered}$

Get the perimeter;

Perimeter of the triangle = 5 + 5 + 8

Perimeter of the triangle = 18units