In a triangle ABC, coordinates of A are (1, 2) and the equations of the medians through B …

Question

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

Accepted Answer

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( {{{1 + 4} \over 2},{{2 + y} \over 2}} \right) (

Now,    \]{{1 + 4 + 2 + y} \over 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 $$\Delta $ABC = 3$ \times $$\Delta $AGC

= 3$ \times $${1 \over 2}$$ [1(1 $$-$3) + 4(3$-$2)  + 4(2$-$

Solutions · 1