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.
