Question

The area of a triangle is $5$. Two of its vertices are $(2,1)$ & $(3,-2)$. The third vertex lies on $y=x+3$. The third vertex can be

• A. $(\frac{7}{2},\frac{13}{2})$ or $(-\frac{3}{2},\frac{3}{2})$
• B. $(-\frac{7}{2},-\frac{13}{2})$ or $(-\frac{3}{2},\frac{3}{2})$
• C. $(\frac{7}{2},-\frac{13}{2})$ or $(\frac{3}{2},\frac{3}{2})$
• D. None of these

Answer 1

The correct option is A $(\frac{7}{2},\frac{13}{2})$ or $(-\frac{3}{2},\frac{3}{2})$

We know that the area of the triangle whose vertices are $(x_1,y_1),(x_2,y_2),$ and $(x_3,y_3)$ is $\frac{1}{2}\left|x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right|$

Let the third vertex be $(x,x+3)$.

Then by applying formula for area of triangle we get,

$\frac{1}{2}|(4x+5-9|=5$

Or

$|4x-4|=10$

$\Rightarrow |2x-2|=5$

$\Rightarrow 2x-2=5$
$\Rightarrow x=\frac{7}{2}$ and

$\Rightarrow 2x-2=-5$
$\Rightarrow x=\frac{-3}{2}$.

Hence the points are $(\frac{7}{2},\frac{13}{2})$ or $(\frac{-3}{2},\frac{3}{2})$