In a triangle ABC, coordinates of A are (1,2) and the equations of the medians through B and C are x+y=5 and x=4 respectively. Then area of ΔABC (in sq. units) is?

Median through C is x=4. So clearly the x coordinate of C is 4. So let C=(4,y), then the midpoint of A(1,2) and C(4,y) which is D lies on the median through B by definition. Clearly, D=((1+4)/2, (2+y)/2). Now, we have, (3+y)/2 = 5 ⇒ y=7. So, C=(4,7).
The centroid of the triangle is the intersection of the medians. It is easy to see that the medians x=4 and x+y=5 intersect at G=(4,1).
The area of triangle ΔABC = 3 × ΔAGC = 3 × (1/2) × 3 × 2 = 9 (In this case, it is easy as the points G and C lie on the same vertical line x=4. So the base GC=6 and the altitude from A is 3 units)
So the answer is option A (Corrected based on the calculation. Previous solution had an error in calculating y coordinate of C and area calculation).