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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.
Solution
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)`
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