The vertices of ∆ABC are A (-2, 2), B (6, 2), and C (0, 8). What is the perimeter of ∆ABC in units?

Ok here what you need to use is the distance formula. So the process should look like this:

L (AB) = √[ (6 - - 2) ² + (2 - 2) ²] = 8

L (BC) = √[ (0 - 6) ² + (8 - 2) ²] = 8.4853

L (AC) = √[ (0 - - 2) ² + (8 - 2) ²] = 6.3246

8 + 8.4853 + 6.3246 = 14.8099

I hope this is what you were looking for.