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ABCD is a rectangle with `A(-1,2),B(3,7)` and `AB:BC=4:3`. If P is the centre of the rectangle, then the distance of P from each corner is equal to

A

`sqrt(14)/(2)`

B

`3sqrt(41)/(4)`

C

`2sqrt(41)/(3)`

D

`5sqrt(41)/(8)`

Text Solution

Verified by Experts

The correct Answer is:
D

Since ABCD is a rectangle, `(AC)^2=(AB)^2+(BC)^2`
`=(AB)^2[1+(9)/(16)]=(25)/(16)(AB)^2`
and `PA=(AC)/(2)=(5AB)/(8)`
`(5)/(8)sqrt((-1-3)^2+(2-7)^2)=5(sqrt41)/(8)`
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