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DeltaABC is a triangle with the point P ...

`DeltaABC` is a triangle with the point P on side BC such that 3BP=2PC, the point Q is on the line CA such that 4CQ=QA. Find the ratio in which the line joining the common point R of AP and BQ and the point S divides AB.

Text Solution

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The correct Answer is:
`6:1`
let S be the point of intersection of AB and CR. Let A be the origin and the position vectors of the points, B,C,P,Q,R and S be b,c,p,q,r and s respectively.
`thereforep=(3b+2c)/(5)`
and `q=(4c)/(5)` . .. (i)
`implies (5q-3b)/(2)=(5q)/(4) implies 10p-6b=5q`
i.e., `10p=5q+6b implies (10p)/(11)=(5q+6b)/(11)=r`

`implies (11r)/(10)=r=(3b+2c)/(5)` [using Eq. (ii)]
`11r=6b+4c` ltbgt `11r-4c=6b`
`(11r-4c)/(7)=(6)/(7)b=s`, thus s divides AB in the ratio 6:1.
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