Question

For what value of $k(k>0)$ is the area of the triangle with vertices $(-2,5),(k,-4)$ and $(2k+1,10)$ equal to $53$ square units?

Answer 1

Area of triangle $=53$ square units

Area of $\Delta ABC=\frac{1}{2}\left[x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right]$

$53=\frac{1}{2}[-2(-4-10)+k(10-5)+(2k+1\left)\right(5+4\left)\right]$

$53=\frac{1}{2}[-2\times (-14)+k\times 5+(2k+1)\times 9]$

$106=23k+37$

$k=\frac{69}{23}=3$

The value of k is $3$.