Two vertices of a triangle are 0 , 2 and 4 , 3 . If its orthocenter is at the origin, then its third vertex…

Let the third vertex be $\left(h,k\right)$.

Since $H$ is orthocenter of $\Delta ABC$

Since the slope of $BH$ is $\frac{3}{4}$ hence the slope of $AC$ will be $-\frac{4}{3}$

$\Rightarrow$ Equation of line $AC$ is

$4x+3y=6 ...\left(1\right)$

Similarly, since $AH$ is parallel to $y-$axis hence $BC$ will be parallel to $x-$axis and

Equation of line $BC$ is

$y=3 ...\left(2\right)$

On solving $\left(1\right)$ and $\left(2\right)$

$h=-\frac{3}{4}, k=3$

$\therefore$The point $C\left(h,k\right)\equiv \left(-\frac{3}{4},3\right)$ lies in the $2^{\mathrm{nd}}$ quadrant.