Question

The area of a triangle is 5 sq units. Two of its vertices are (2, 1) and (3, –2). If the third vertex is $\left(\frac{7}{2},y\right),$ find the value of y.

Answer 1

Let $A\left(x_1,y_1\right)=A\left(2,1\right),B\left(x_2,y_2\right)=B\left(3,-2\right)\mathrm{and}C\left(x_3,y_3\right)=C\left(\frac{7}{2},y\right).$
Now
$$\begin{gathered} \mathrm{Area}\left(∆ABC\right)=\frac{1}{2}\left|x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1\mathit{-}{y}_{\mathit{2}}\right)\right| \\ \Rightarrow 5=\frac{1}{2}\left|2\left(-2-y\right)+3\left(y-1\right)+\frac{7}{2}\left(1+2\right)\right| \\ \Rightarrow 10=\left|-4-2y+3y-3+\frac{21}{2}\right| \\ \Rightarrow 10=\left|y+\frac{7}{2}\right| \\ \Rightarrow 10=y+\frac{7}{2}\mathrm{or}-10=y+\frac{7}{2} \\ \Rightarrow y=\frac{13}{2}\mathrm{or}y=\frac{-27}{2} \end{gathered}$$
Hence, y = $\frac{13}{2}\mathrm{or}\frac{-27}{2}.$