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Question

Find the point on y-axis which is at a distance of 4 units from a point P(4,2).


A

(0,1)

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B

(2,0)

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C

(0,2)

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D

(0,0)

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Solution

The correct option is C

(0,2)


Any point on the y-axis is of the form (0,a).

Now, distance from P(4,2) to (0,a) is given by (40)2+(2a)2

According to the question,

42+(2a)2=4

16+4+a24a=4

204a+a2=4

204a+a2=16

a24a+4=0

(a2)2=0

a2=0

a=2

Therefore the required point is (0,2).


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