The vertices of ∆abc are a(-2, 2), b(6, 2), and c(0, 8). the perimeter of ∆abc is units is? what is the area?

Best is to draw a sketch of the three points.
Next step is to find the distances BC, CD, DB.
The perimeter is the sum of the three distances.

The distances are found using the distance formula:
D=sqrt((y2-y1)^2+(x2-x1)^2)
order of (x1,y1), (x2,y2) is not important.

Given A(2,8),B(16,2),C(6,2)
we calculate 
AB=sqrt((16-2)^2+(2-8)^2)=sqrt(14^2+6^2)=sqrt(232);
BC=sqrt((6-16)^2+(2-2)^2)=sqrt(10^2+0)=10
CA=sqrt((2-6)^2+(8-2)^2)=sqrt(4^2+6^2)=sqrt(16+36)=sqrt(52)

Perimeter=AB+BC+CA=32.443 units

For the area, we note that BC is horizontal (parallel to the x-axis), so 
area = (1/2)bast * height
=(1/2)10*(ya-yb)
=(1/2)10*(8-2)
=(1/2)10*6
=30 unit^2

Why are u not just putting the numbers this is to much