Question

$△ABC$ is a right angled triangle with $\angle B={90}^{\circ }.$ The coordinates of $A$ and $C$ are $(-2,-1)$ and $(6,5)$ respectively. If the area of $△ABC$ is $30$ sq. units, then the number of such points $B$, is

• A. $0$
• B. $1$
• C. $2$
• D. $\text{infinite}$

Answer 1

The correct option is A $0$

Clearly, $B$ lies on the circle with $A$ and $C$ as the end points of the diameter.


$AC=\sqrt{(-2-6{)}^2+(-1-5{)}^2}=10\therefore \text{Radius}=\frac{10}{2}=5\text{Area of}△ABC=30\Rightarrow \frac{1}{2}\times BD\times AC=30\Rightarrow \frac{1}{2}\times BD\times 10=30\Rightarrow BD=6$

Since, the value of $BD$ can't be greater than the radius of the circle,
$\therefore$ No such triangle is possible.