Question

Show that the Perpendicular drawn from the point (4, 1) on the line segment joining (6, 5) and (2, – 1) divide it internally in the ratio 8: 5.

Answer 1

Let the points be A(4, 1), B(6, 5) and C(2, – 1)

AP ⊥ BC is drawn

Slope of line BC, m = \(\frac{−1\, −\,5}{2\, −\, 6}\) = \(\frac{3}{2}\)

Since AP ⊥ BC

∴ Slope of AP, m′ = − \(\frac{1}{m}\) = − \(\frac{2}{3}\)

∴ Equation of line AP is

y − 1 = − \(\frac{2}{3}\)(x − 4)

⇒ 2x + 3y – 11 = 0 …(1)

Let P divide BC in the ratio m : n

∴ Co-ordinates of P are \(\big(\frac{2m + 6n}{m + n}, \frac{−m + 5n}{m +n}\big)\)

Since P lies on AP, so (1)

⇒ 2.\(\frac{2m + 6n}{m + n}\) + 3.\(\big(\frac{−m + 5n}{m +n}\big)\) − 11 = 0

⇒ 4m + 12n - 3m + 15n - 11m – 11n = 0

⇒ \(\frac{m}{n}\) = \(\frac{16}{10}\)

⇒ m : n = 8 : 5