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The normal form of line sqrt(3)x=y+4=0 i...

The normal form of line `sqrt(3)x=y+4=0` is

A

only I is true

B

only II is true

C

both I and II are true

D

neither I nor II are true

Text Solution

Verified by Experts

The correct Answer is:
B
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Knowledge Check

  • Normal form of the line x+y+sqrt(2)=0 is

    A
    `x"cos"(pi)/4+y"sin"(pi)/4=1`
    B
    `"cos"(3pi)/4+y"sin"(3pi)/4=1`
    C
    `x"cos"(5pi)/4+y"sin"(5pi)/4=1`
    D
    `x"cos"(7pi)/4+y"sin"(7pi)/4=1`

  • Normal form of the line x-y+sqrt(2)=0 is

    A
    `x"cos"(pi)/4+y"sin"(pi)/4=1`
    B
    `x"cos"(3pi)/4+y"sin"(3pi)/4=1`
    C
    `x"cos"(5pi)/4+y"sin"(5pi)/4=1`
    D
    `x"cos"(7pi)/4+y"sin"(7pi)/4=1`

  • The normal form of the line x+y+1=0 is

    A
    `xcos(45^(@))+ysin(135^(@))=1/sqrt(2)`
    B
    `xcos(45^(@))+ysin(45^(@))=1/sqrt(2)`
    C
    `xcos(225^(@))+ysin(225^(@))=1/sqrt(2)`
    D
    `xcos(45^(@))+ysin(45^(@))=-1/sqrt(2)`

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    Quadrilateral formed by teh lines sqrt(3)x+y=0, sqrt(3)y+x=0, sqrt(3)x+y=1, sqrt(3)y+x=1 is

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    If x cos alpha +y sin alpha =p is the normal form of the equation of a strainght line x+ sqrt(3) y +4=0 and a,b are respectively x,y - intercepts of this line, then sqrt(3) pibp- 3a alpha=

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