Question

The two vertices of triangle are $(2,-1),(3,2)$ and the third vertex lies on $x+y=5$. The area of the triangle is $4$ units, then the third vertex is

• A. $(0,5)$ or $(1,4)$
• B. $(5,0)$ or $(4,1)$
• C. $(5,0)$ or $(1,4)$
• D. $(0,5)$ or $(4,1)$

Answer 1

The correct option is C $(5,0)$ or $(1,4)$

Since, the third vertex $(x_1,y_1)$ lie on the line $x+y=5$.
$\therefore x_1+y_1=5$
$y_1=5-x_1$
$\therefore$ Coordinate of $C$ is $(x_1,5-x_1)$.
Given, area of $△ABC=4units$

$\therefore \frac{1}{2}\left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_2& 1\end{array}\right|=areaoftraingle$
$\therefore \frac{1}{2}\left|\begin{array}{ccc}2& -1& 1\\ 3& 2& 1\\ x_1& 5-x_1& 1\end{array}\right|=4$
Using $R_2\to R_2-R_1$ and $R_3\to R_3-R_1$,
$\left|\begin{array}{ccc}2& -1& 1\\ 1& 3& 0\\ x_1-2& 6-x_1& 0\end{array}\right|=8$
$\Rightarrow 6-x_1-3(x_1-2)=\pm 8$
$\Rightarrow 6-x_1-3x_1+6=\pm 8$
$\Rightarrow 12-8=4x_1$ or $4x_1=20$
$\Rightarrow x_1=1$ or $x_1=5$
$\therefore y_1=5-1=4$ or $y_1=0$
$\therefore C(x_1,y_1)=C(1,4)$ or $(5,0)$