Two vertices of a triangle are F (1, - 4) and G (6, - 4). List two possible coordinates of the third vertex so that the triangle has an area of 20 square units.

The two vertices F, G of the triangle both lie on the line y=-4.

The distance between F and G is b = (6-1) = 5 units.

In order for the triangle to have an area of 20 square units, it must have a height given by

20 = (b*h) / 2

solve to get

h=2*20/5=8 units.

Thus the vertices of the triangle (area = 20 sq. units) must lie on the line

y=-4+8=4

or

y=-4-8=-12

Thus two possible position of the third vertex that will give an area of 20 sq. units would be

(6,4) or (6,-12), both of which would give a right triangle.