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

Answer: (A)

$(-5/2,3/4)$

Answer 2

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.