Question

the coordinates of the vertices of triangle ABC are a (-2,2), b (5,-3), c (-4,-1). Identify the perimeter of triangle ABC.

Answer 1

The formula of a distance between two points:

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

$A(-2,\ 2);\ B(5,\ -3)\\\\AB=\sqrt{(5-(-2))^2+(-3-2)^2}=\sqrt{49+25}=\sqrt{74}\\\\A(-2,\ 2);\ C(-4,\ -1)\\\\AC=\sqrt{(-4-(-2))^2+(-1-2)^2}=\sqrt{4+9}=\sqrt{13}\\\\B(5,\ -3);\ C(-4,\ -1)\\BC=\sqrt{(-4-5)^2+(-1-(-3))^2}=\sqrt{81+4}=\sqrt{85}\\\\P_{\triangle ABC}=\sqrt{74}+\sqrt{13}+\sqrt{85}$