Question

The orthocentre of the triangle with vertices $O(0, 0), A(0,3/2)$ and $B(-5, 0)$ is

• A. $(-5/2,3/4)$
• B. $(5/2,3/4)$
• C. $(0, 0)$
• D. $(-5, 3/2)$

Answer 1

The Correct Option is A

Let, $\Delta \, AOB $ is the given triangle 
Slope of $AB = \frac{\frac{3}{2}-0}{0+5} = \frac{3}{10}$ 
Slope of $BO = \frac{0-0}{0+54} =0$ 
The equation of line passing through A and perpendicular to BO is $ y-0 =- 0 \left(x - \frac{3}{2}\right) $ 


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