Question

If two vertices of a triangle of area $a^2$ are $(0,0)$ and $(a,0)$ then the coordinates of the third vertex can be

• A. $(\frac{a}{2},2a)$
• B. $(\frac{3a}{2},-2a)$
• C. $(\frac{a}{2},\frac{a}{2})$
• D. $(\frac{a}{2},-a)$

Answer 1

Answer: (A), (B)

The correct options are
A $(\frac{a}{2},2a)$
B $(\frac{3a}{2},-2a)$

Let coordinate of third vertex be $(x,y)$

Area of triangle $\frac{1}{2}\left|\begin{array}{ccc}0& 0& 1\\ a& 0& 1\\ x& y& 1\end{array}\right|=a^2$

$\Rightarrow \frac{1}{2}\left(ay\right)=a^2$

$\Rightarrow y=2a$

Hence, the y-coordinate of the vertex is $2a$.