Two vertices of a triangle are (0, 2) and (4, 3) . If its...

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

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

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

$⇒$ Equation of line $AC$ is

$4x + 3y = 6 . . . \left(1\right)<br /><br />Similarly, since$AH$is parallel to$y -$axis hence$BC$will be parallel to$x -$axis and<br /><br />Equation of line$BC$is<br /><br />$y = 3 . . . \left(2\right)$<br /><br />On solving \left(1\right)$ and \left(2\right)$<br /><br />$h = - \dfrac{3}{4}, k = 3$<br /><br />$∴The point $C\left(h, k\right)$ ≡ $\left(- \dfrac{3}{4}, 3\right)$ lies in the ${2}^{nd}$ quadrant.