1. Let the triangle have vertices $A(0,2),\quad B(4,3),\quad C(x,y)$ and orthocenter at $H(0,0).$

2. Since the orthocenter is the intersection point of altitudes, the altitude from each vertex passes through the origin.

   • Altitude from $A(0,2)$ is the line through $(0,2)$ and $(0,0)$, i.e. $x=0.$ Therefore side $BC$ must be perpendicular to $x=0$, so $BC$ is horizontal. Hence, $y=3$ because $B=(4,3)$ lies on side $BC$.

   So the third vertex must be of the form $C=(x,3).$

3. Now consider the altitude from $B(4,3)$. Since it passes through the origin, its slope is $m_{BH}=\frac{0-3}{0-4}=\frac{-3}{-4}=\frac{3}{4}.$ Therefore side $AC$ must have slope $m_{AC}=-\frac{4}{3}$ because it is perpendicular to the altitude.

4. But $A=(0,2),\quad C=(x,3),$ so slope of $AC$ is $m_{AC}=\frac{3-2}{x-0}=\frac{1}{x}.$ Thus, $\frac{1}{x}=-\frac{4}{3}$ which gives $x=-\frac{3}{4}.$

5. Hence the third vertex is $C\left(-\frac{3}{4},3\right).$

6. Since its coordinates satisfy $x<0,\quad y>0,$ the point lies in the second quadrant.

7. Checking options:

   • A: third ❌
   • B: fourth ❌
   • C: second ✅
   • D: first ❌

Therefore, the correct option is C.