Question 1026787
First note that the vertices D and E are the endpoints of a segment that is parallel to the x-axis. (In fact the DE lies on the horizontal line y = 1.) The length of this segment is 7-2 = 5 units, which is also the base b of the triangle.

Now the area of a triangle is given by {{{A = (1/2)bh}}}.

==> {{{(1/2)*5h=  15}}} as per given information

==> 5h = 30, or h = 6 units.

This means that as long as a point (x,y) has a distance of 6 units from the line y = 1, it is a third vertex F to a triangle with an area of 15 square units.

Thus, all points (x,7) and (x,-5) for arbitrary x value are all possible points for F.