Question

The area of a triangle is $5$. Two of its vertices are $A(2,1)$ and $B(3,-2)$. The third vertex lies on the line $y=x-2$. Find the third vertex.

Answer 1

Let the third vertex of $△ABC$ be $(x,y)$.

$\therefore$ Vertices of $△ABC$ are $A(2,1),B(3,-2),C(x,y)$ respectively.

As we know that,

$ar\left(△\right)=\frac{1}{2}\left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|$

$\therefore ar\left(△ABC\right)=\frac{1}{2}\left|\begin{array}{ccc}2& 1& 1\\ 3& -2& 1\\ x& y& 1\end{array}\right|$

$\Rightarrow ar\left(△ABC\right)=\frac{1}{2}[2((-2)-y)-1(3-x)+1(3y-(-2x))]$

$\Rightarrow ar\left(△ABC\right)=\frac{1}{2}[-4-2y-3+x+3y+2x]$

$\Rightarrow ar\left(△ABC\right)=\frac{1}{2}(3x+y-7)$

Since vertex C lies on the line

$y=x-2..........\left(1\right)$

$\therefore ar\left(△ABC\right)=\frac{1}{2}(3x+x-2-7)$

$\Rightarrow ar\left(△ABC\right)=\frac{1}{2}(4x-9)$

As given that, $ar\left(△ABC\right)=5$

$\therefore \frac{1}{2}(4x-9)=5$

$\Rightarrow 4x-9=10$

$\Rightarrow 4x=19$

$\Rightarrow x=\frac{19}{4}$

On sustituting the value of x in ${eq}^n\left(1\right)$, we have

$y=\frac{19}{4}-2$

$\Rightarrow y=\frac{19-8}{4}=\frac{11}{4}$

Hence, the third vertex will be $(\frac{19}{4},\frac{11}{4})$.