Question

Let the two vertices of a triangle are $(2,-1)$ and $(3,2)$ and third vertex lies on the line $x+y=5$. If the area of triangle is $4\text{sq. units}$, then the coordinates of the third vertex is/are

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

Answer 1

Answer: (B), (D)

The correct options are
B $(5,0)$
D $(1,4)$

Let $A=(2,-1)$ and $B=(3,2)$ and the third vertex be $C(x,5-x)$ as $C$ lies on the line $x+y=5.$

Now, Area of $△ABC=4\text{sq. units}$
$\Rightarrow \frac{1}{2}\left|\begin{array}{cc}2& -1\\ 3& 2\\ x& 5-x\\ 2& -1\end{array}\right|=4$
$\Rightarrow \frac{1}{2}|7+15-3x-2x-x-10+2x|=4\Rightarrow 6-2x=\pm 4$
$\Rightarrow x=1,5$
$\Rightarrow y=4,0$

Thus, the possible coordinates of the third vertex are $(1,4)$ and $(5,0).$