Question

If the area of the $△ABC$ is $68$ sq.units and the vertices are $A(6,7),B(-4,1)$ and $C(a,-9)$ taken in order, then find the value of $a$.

Answer 1

Area of $ABC=5$ sq. Units

Area of a triangle with vertices $(x_1,y_1)$ ; $(x_2,y_2)$ and $(x_3,y_3)$ is given as $\left|\frac{x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)}{2}\right|$

According to the given condition

$\Rightarrow \frac{1}{2}\left(a(7-1)+6(1+9)-4(-9-7)\right)=68$

$\Rightarrow 136=60+64+6a$

$\Rightarrow 136-124=6a$

$\Rightarrow a=2$