Question

Find the perimeter of △ABC with vertices A(−5, −5), B(3, −5), and C(−5, 1).

Answer 1

Check the picture below.

so notice, the sides AB and AC you can pretty much count them off the grid.

now, to get the CB side.

$\bf \textit{distance between 2 points}\\ \quad \\ \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) C&({{ -5}}\quad ,&{{ 1}})\quad % (c,d) B&({{ 3}}\quad ,&{{ -5}}) \end{array}\qquad % distance value d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2} \\\\\\ CB=\sqrt{[3-(-5)]^2+[-5-1]^2}\implies CB=\sqrt{(3+5)^2+(-5-1)^2} \\\\\\ CB=\sqrt{8^2+(-6)^2}\implies CB=\sqrt{100}\implies CB=10$

sum all three sides up, and that's the perimeter of the triangle.