# How do you find the perimeter of a triangle on a coordinate plane? A (-3, 6) B (-3, 2) C (3, 2) Round the solution to 2 decimal points.

Mar 17, 2018

$\text{perimeter "~~17.21" units to 2 dec. places}$

#### Explanation:

$\text{to find the perimeter of the triangle we require to}$
$\text{calculate the lengths of the 3 sides}$

$\text{perimeter } = A B + A C + B C$

$\textcolor{b l u e}{\text{calculate AB}}$

$\text{Note that the points A and B have the same value of}$
$\text{x-coordinate which means that AB is a vertical line}$

$\text{Thus the length of AB is the difference in y-coordinates}$

$\Rightarrow A B = 6 - 2 = 4$

$\textcolor{b l u e}{\text{calculate BC}}$

$\text{Note that the points B and C have the same value of }$
$\text{y-coordinate which means that BC is a horizontal line}$

$\text{Thus the length of BC is the difference in x-coordinates}$

$\Rightarrow B C = 3 - \left(- 3\right) = 6$

$\textcolor{b l u e}{\text{calculate AC}}$

$\text{to calculate AC use the "color(blue)"distance formula}$

•color(white)(x)d=sqrt((x_2-x_1)^2+(y_2-y_1)^2)

$\text{let "(x_1,y_1)=(-3,6)" and } \left({x}_{2} , {y}_{2}\right) = \left(3 , 2\right)$

$A C = \sqrt{{\left(3 + 3\right)}^{2} + {\left(2 - 6\right)}^{2}}$

$\textcolor{w h i t e}{A C} = \sqrt{36 + 16} = \sqrt{52}$

$\Rightarrow \text{perimeter "=4+6+sqrt52~~17.21" to 2 dec. places}$