The area of a triangle is 3/2 sq. units. Two of its vertices are the points A (2,-3) and B (3,-2), the centroid of the triangle lies on the line 3x – y – 8 = 0. Find the third vertex C.

Give the explanation of your answer!!

Given vertex of point A(2,-3) B(3,-2) the centroid of the triangle lies on the line 3x - y - 8
the area of the triangle is 3 by 2 square units
now, 
the area of triangle =
1/2[x1(y2-y3)+x2(y3-y2)+x3(y1-y2) ]=3/2 square units
x1=2 y1=-3
x2=3 y2=-2
x3=x y3=y
solving of this is in the pic given upward
now we get eqn 1 
now, 
in the question we have that centroid of this Triangle lies on the line 3x-y-8=0.....(2) 
next steps are in the photographs second which have uploaded upward 
the vertices of the C point is (5.5,4.5)