Let the third vertex be (h,k).

Since H is orthocenter of ΔABC
Since the slope of BH is 43 hence the slope of AC will be −34
⇒ Equation of line AC is
4x+3y=6...(1)
Similarly, since AH is parallel to y−axis hence BC will be parallel to x−axis and
Equation of line BC is
y=3...(2)
On solving (1) and (2)
h=−43,k=3
∴The point C(h,k)≡(−43,3) lies in the 2nd quadrant.