Show that the Perpendicular drawn from the point (4, 1) on the line se

1.	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» 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

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»

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