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

Let $A\left(x_1=-2,y_1=5\right),B\left(x_2=k,y_2=-4\right)\mathrm{and}C\left(x_3=2k+1,y_3=10\right)$ be the vertices of
the triangle. So
$$\begin{gathered} \mathrm{Area}\left(∆ABC\right)=\frac{1}{2}\left[x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right] \\ \Rightarrow 53=\frac{1}{2}\left[\left(-2\right)\left(-4-10\right)+k\left(10-5\right)+\left(2k+1\right)\left(5+4\right)\right] \\ \Rightarrow 53=\frac{1}{2}\left[28+5k+9\left(2k+1\right)\right] \\ \Rightarrow 28+5k+18k+9=106 \end{gathered}$$

$$\begin{gathered} \Rightarrow 37+23k=106 \\ \Rightarrow 23k=106-37=69 \\ \Rightarrow k=\frac{69}{23}=3 \end{gathered}$$
Hence, k = 3.