Let the co-ordinates of B are (p, q).
Then, slope of line AB, m1 = \(\frac{q − 1}{p + 3}\).
And, slope of line 2x – 3y = 4 is \(\frac{2}{3}\) = mL.
Since the lines are perpendicular, so m1 m2 = – 1
⇒ \(\frac{q − 1}{p + 3}\) × \(\frac{2}{3}\) = −1
⇒ 2q – 2 = – 3p – 9
⇒ 3p + 2q + 7 = 0 …(1)
The midpoint of AB is \(\big(\frac{p − 3}{2} , \frac{q + 1}{2}\big)\), which lies on the line AB.
∴ 2\(\big(\frac{p − 3}{2}\big)\) − 3. \(\big(\frac{q + 1}{2}\big)\) = 4
⇒ 2p – 6 – 3q – 3 = 8
⇒ 2p – 3q – 17 = 0 …(2)
Solving (1) and (2), we get p = 1, q = – 5.