We know that ABC is a triangle and if its a triangle every vertice conects to the other 2 vertices
So we can do the distance between AB, AC and BC then sum them
dAB = [tex]\sqrt{(x_a-x_b)^2+(y_a-y_b)^2}[/tex]
dAB = [tex]\sqrt{(-5-(-2))^2+(-2-(-2))^2}[/tex]
dAB = [tex]\sqrt{(-3)^2+0^2}[/tex]
dAB = 3
dAC = [tex]\sqrt{(-5-(-5))^2+(-2-2)^2}[/tex]
dAC = [tex]\sqrt{0^2+(-4)^2}[/tex]
dAC = 4
dBC = [tex]\sqrt{(-2-(-5))^2+(-2-2)^2[/tex]
dBC = [tex]\sqrt{3^2+(-4)^2}[/tex]
dBC = [tex]\sqrt{9 + 16}[/tex]
dBC = [tex]\sqrt{25}[/tex]
dBC = 5
3 + 4 + 5 = 12
So alternative A.
