The orthocentre of the triangle with vertices $A(0,0), B\left(0, \frac{3}{2}\right), C(-5,0)$ is

• A

  $\left(\frac{5}{2}, \frac{3}{4}\right)$
• B

  $\left(\frac{-5}{2}, \frac{3}{4}\right)$
• C

  $\left(-5, \frac{3}{2}\right)$
• Correct answer
  D

  $(0,0)$

Solution

Let $\triangle A O B$ is the given triangle. Slope of $A B=\frac{\frac{3}{2}-0}{0+5}=\frac{3}{10}$ Slope of $B O=\frac{0-0}{0+5}=0$ The equation of line passing through $A$ and perpendicular to $B O$ is $y-0=-0\left(x-\frac{3}{2}\right)$

$\Rightarrow \quad y=0$ and equation of line passing through 0 and perpendicular to $A B$ is $y-0=-\frac{10}{3}(x-0)$ $\Rightarrow \quad y=-\frac{10}{3} x$ The intersection point of Eqs. (i) and (ii) is $(0,0)$, which is the required orthocentre.

Hence, the correct answer is option: 'D'.