Question

The orthocentre of a triangle with vertices $\left(0,0\right),\left(3,4\right)$ and $\left(4,0\right)$ is

• A. $\left(3,12\right)$
• B. $\left(3,\frac{3}{4}\right)$
• C. $\left(3,9\right)$
• D. None of these

Answer 1

The correct option is B

$\left(3,\frac{3}{4}\right)$

Explanation for the correct option

Let $O\left(0,0\right),A\left(4,0\right)$ and $B\left(3,4\right)$ be the vertices of the triangle.

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 $\left(3,4\right)$

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,

$$\begin{gathered} \left(\mathrm{Slope}\mathrm{of}\mathrm{OP}\mathrm{i}.\mathrm{e}\mathrm{OH}\right)\times \left(\mathrm{Slope}\mathrm{of}\mathrm{BA}\right)=-1 \\ \Rightarrow \frac{y-0}{3-0}\times \frac{4-0}{3-4}=-1 \\ \Rightarrow \frac{4y}{3}=1 \\ \Rightarrow y=\frac{3}{4} \end{gathered}$$

So, the required orthocentre $=\left(3,\frac{3}{4}\right)$

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