Question

The orthocentre of the $∆OAB$, where $O$ is the origin, $A\left(6,0\right)$ and $B\left(3,3\sqrt{3}\right)$ is

• A. $\left(\frac{9}{2},\frac{\sqrt{3}}{2}\right)$
• B. $\left(3,\sqrt{3}\right)$
• C. $\left(\sqrt{3},3\right)$
• D. $\left(3,-\sqrt{3}\right)$

Answer 1

The correct option is B

$\left(3,\sqrt{3}\right)$

Explanation for the correct option

Given that in $∆OAB$, $O$ is the origin, $A\left(6,0\right)$ and $B\left(3,3\sqrt{3}\right)$

Let $QB$ is the perpendicular drawn from the line $OA$ to $B$

Let $PO$ is the perpendicular drawn from the line $AB$ to $O$

The intersection of the line $QB$ and $PO$ is the orthocenter of the triangle .

Let $H$ be the point of orthocenter of the triangle $∆OAB$

The straight line $QB$ is perpendicular to the $x$ axis and meet $B$ at $\left(3,3\sqrt{3}\right)$

So the co-ordinate of $H$ is $\left(3,y\right)$

We know that two lines are perpendicular to each other is the product of their slope is equal to $-1$

Therefore,

$$\begin{gathered} \left(\mathrm{Slope}\mathrm{of}\mathrm{OP}\mathrm{i}.\mathrm{e}\mathrm{OH}\right)\times \left(\mathrm{Slope}\mathrm{of}\mathrm{BA}\right)=-1 \\ \Rightarrow \frac{y-0}{3-0}\times \frac{0-3\sqrt{3}}{6-3}=-1 \\ \Rightarrow \frac{y\sqrt{3}}{3}=1 \\ \Rightarrow y=\frac{3}{\sqrt{3}} \\ \Rightarrow y=\sqrt{3} \end{gathered}$$

So the required orthocentre $=\left(3,\sqrt{3}\right)$

Hence, option $\left(\mathrm{B}\right)$ is the correct answer.