The line 2x – 3y = 4 is the perpendicular bisector of the line segme

1.	The line 2x – 3y = 4 is the perpendicular bisector of the line segment AB. If co-ordinates of A are (– 3, 1), find the coordinates of B.
Answer» 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.

The line 2x – 3y = 4 is the perpendicular bisector of the line segment AB. If co-ordinates of A are (– 3, 1), find the coordinates of B.

Answer»

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.