find the perimeter of the triangle whos verticies are the following specified points in a plane

A(1,-9)

B(5,6)

C(3,0)

Solution:

AB = sqrt(4^2+15^2) = sqrt(241)

BC = sqrt(2^2+6^2) = 2sqrt(10)

AC = sqrt(2^2+9^2) = sqrt(85)

So the perimeter is (sqrt(241)+2sqrt(10)+sqrt(85))