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: (C)

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.