I just did a whole page of latex and lost the whole lot so now I will do a condensed version. 
Heureka, Your answer looks great but I don't understand it so I will do it my way.
Maybe you can explain your way?
Anyway here goes- again
Find the value of x such that the area of a triangle whose vertices have coordinates
P(6, 5), Q(8, 2) and R( x, 11) is 15 square units.
distance PQ=sqrt(13)
Equation of line PQ is
3x+2y-28=0
Now I need the perp distance from R to PQ
a=3, b=2, c=-28, x1=x, y1=11
$$\\d=\frac{|3x+22-28|}{\sqrt{13}}\\\\
d=\frac{|3x-6|}{\sqrt{13}}\\\\
d=\frac{|3x-6|}{\sqrt{13}}\\\\$$
$$\\\mbox{perpendicular distance from R to PQ}=\frac{|3x-6|}{\sqrt{13}}\\\\
Area=\frac{1}{2}\times \sqrt{13} \times \frac{|3x-6|}{\sqrt{13}}\\\\
15=\frac{1}{2} \times |3x-6|\\\\
30=|3x-6|\\\\
30=3x-6\qquad or \qquad 30=6-3x\\\\
36=3x\qquad\qquad or \qquad 24=-3x\\\\
x=12\qquad\qquad \;\;or \qquad x=-8\\\\$$
Below are the 2 possible triangles. :)
