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Question

Two lines are given by(x-2y)2+k(x-2y)=0. The value of k so that the distance between them is 3, is


A

25

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B

13

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C

±35

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D

None of these

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Solution

The correct option is C

±35


Find the value of k:

Given,

The distance between the lines 3.

(x-2y)2+k(x-2y)=0

x-2yx-2y+k=0

x-2y=0 and x-2y+k=0

These are two parallel lines.

Distance between parallel lines=c1-c2a2+b2=3

k-012+-22=3k=±35

Hence, the correct option is (C).


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