A triangle is given with its vertices A (-1; 6), B (-5; 3) and C (-2; -1). Find the perimeter of triangle ABC.

We have the coordinates of the vertices of the triangle:

A (-1; 6), B (-5; 3), C (-2; -1).

We find the lengths of the sides of the triangle:

| AB | = ((-5 + 1) ^ 2 + (3 – 6) ^ 2) ^ (1/2) = (16 + 9) ^ (1/2) = 5;

| BC | = ((-2 + 5) ^ 2 + (-1 – 3) ^ 2) ^ (1/2) = (9 + 16) ^ (1/2) = 5;

| AC | = ((-2 + 1) ^ 2 + (-1 – 6) ^ 2) ^ (1/2) = (1+ 49) ^ (1/2) = 50 ^ (1/2).

As you can see, the triangle also turned out to be isosceles.

Find the perimeter of the triangle:

P (ABC) = 10 + 50 ^ (1/2).