Question

Find the value of k so that the area of the triangle with vertices A(k + 1, 1), B(4, −3) and C(7, −k) is 6 square units.

Answer 1

Let $A\left(x_1,y_1\right)=A\left(k+1,1\right),B\left(x_2,y_2\right)=B\left(4,-3\right)\mathrm{and}C\left(x_3,y_3\right)=C\left(7,-k\right)$. Now
$$\begin{gathered} \mathrm{Area}\left(∆ABC\right)=\frac{1}{2}\left[x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1\mathit{-}{y}_2\right)\right] \\ \Rightarrow 6=\frac{1}{2}\left[\left(k+1\right)\left(-3+k\right)+4\left(-k-1\right)+7\left(1+3\right)\right] \\ \Rightarrow 6=\frac{1}{2}\left[k^2-2k-3-4k-4+28\right] \\ \Rightarrow k^2-6k+9=0 \end{gathered}$$
$\Rightarrow {\left(k-3\right)}^2=0\Rightarrow k=3$
Hence, k = 3.