QJEEM18Ma93fe: JEE Main 2018 Mathematics - Class 11 JEEM18Ma93fe Solution: In a triangle ABC, coordinates of A are (1, 2) and the…

Median through C is x = 4

So the coordinate of C is 4. Let C = (4, y), then the midpoint of A(1, 2) and

C(4, y) is D which lies on the median through B.



$$\therefore$$$$$$ D = $$\left(\frac{1+4}{2},\frac{2+y}{2}\right)$$

Now,    $$\frac{1+4+2+y}{2}$$ = 5  $$\Rightarrow$$ y = 3.

So, C $$\equiv$$ (4, 3).

The centroid of the triangle is the intersection of the mesians, Here the medians x = 4 and x + 4 and x + y = 5 intersect at G (4, 1).

The area of triangle $$\mathrm{\Delta }$$ABC = 3 $$\times$$ $$\mathrm{\Delta }$$AGC

= 3 $$\times$$ $$\frac{1}{2}$$ [1(1 $$-$$ 3) + 4(3 $$-$$ 2) + 4(2 $$-$$ 1)] = 9.