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The distance between the lines (x+7y)^2+...

The distance between the lines `(x+7y)^2+4sqrt(2)(x+7y)-42=0` is_____________

Text Solution

Verified by Experts

The correct Answer is:
2

`(x+7y)=4sqrt(2)(x+7y)-42=0`
or `(x+7y)^(2)+7sqrt(2)(x+7y)-3sqrt(2)(x+7y)-42=0`
or `(x+7y)[x+7y+7sqrt(2)]-3sqrt(2)(x+7y+7sqrt(2))=0`
or `(x+7y+7sqrt(2))(x+7y-3sqrt(2))=0`
or `x+7y+7sqrt(2)=0andx+7y-3sqrt(2)=0`
or `d=|(7sqrt(2)+2sqrt(2))/(sqrt(1+49))|=(10sqrt(2))/(sqrt(50))=2`
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Knowledge Check

  • Distance between the lines 5x+3y-7=0 and 15x+9y+14=0 is

    A
    `35/sqrt(34)`
    B
    `1/(3sqrt(34))`
    C
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    D
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  • The angle between the lines sqrt3 x – y–2 = 0 and x – sqrt3 y+1=0 is

    A
    `90^(@)`
    B
    `60^(@)`
    C
    `45^(@)`
    D
    `30^(@)`
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