Question

A(20, 0) and B(10, –20) are two fixed points. Find the co-ordinates of the point P in AB such that : 3PB = AB. Also, find the co-ordinates of some other point Q in AB such that : AB = 6 AQ.

Answer 1

Given, 3PB =AB

`=>(AB)/(PB) = 3/1`

`=> (AB - PB)/(PB) = (3 - 1)/1`

`=>(AP)/(PB) = 2/1`

Using section formula,

Coordinates of P are

`P(x, y) = P((2 xx 10 + 1 xx 20)/(2 + 1),(2 xx (-20) + 1 xx 0)/(2 + 1))`

= `P(40/3, -40/3)`

Given, AB = 6AQ

`=> (AQ)/(AB) = 1/6`

`=>(AQ)/(AB - AQ) = 1/(6 - 1)`

`=>(AQ)/(QB) = 1/5`

Using section formula,

Coordinates of Q are

`Q(x, y) = Q((1 xx 10 + 5 xx 20)/(1 + 5),(1 xx (-20) + 5 xx 0)/(1 + 5))`

= `Q(110/6, -20/6)`

= `Q(55/3, -10/3)`