Question

find the area of triangle ABC whose vertices are a (4,4 )B (0,0) and c(6,2 )

Answer 1

First, move every point left by its y-coordinate. This means (4,4) goes to (0,4); (6,2) goes to (4,2); and (0,0) stays put.

Next, move every point down by half its x-coordinate. This means (0,4) and (0,0) stay put, while (4,2) goes to (4,0).

Now we have half of a 4x4 square, so the area is 8.

Answer 2

Area=0.5 [4 (0-2)+0 (2-4)+6 (4-0)]

=1/2 [4 (-2)+0+6 ×4]

=1/2 (-8+24)

=1/2×16

=8