Question

In $△ABC$, the length of side $AB$ is $2$ units and $\angle ABC=\frac{\pi }{3}.$ If $B$ and the mid-point of $BC$ have the coordinates $(0,0)$ and $(2,0)$ respectively, then the orthocentre of $△ABC$ is

• A. $(1,3\sqrt{3})$
• B. $(\frac{5}{3},\frac{1}{\sqrt{3}})$
• C. $(1,\frac{1}{\sqrt{3}})$
• D. $(1,\sqrt{3})$

Answer 1

The correct option is D $(1,\sqrt{3})$

$B\equiv (0,0)$, mid-point of $BC$ is $(2,0)$
So, $C\equiv (4,0)$
Using parametric form of straight line, we get
$A\equiv (0+2\cos \frac{\pi }{3},0+2\sin \frac{\pi }{3})$
$\therefore$ Coordinates of $A$ are $(1,\sqrt{3})$

Now, $A{B}^2+A{C}^2=B{C}^2$
$\Rightarrow \angle BAC={90}^{\circ }$
$\therefore A$ is the orthocentre.