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The equation of the locus of points equi...

The equation of the locus of points equidistant from `(-1-1)` and `(4,2)` is

A

`3x-5y-7=0`

B

`5x+3y-9=0`

C

`4x+3y+2=0`

D

`x-3y+5=0`

Text Solution

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The correct Answer is:
B
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Knowledge Check

  • Locus of the points equidistant from the point (-1, -1) and (4, 2) is

    A
    `5x-3y-9=0`
    B
    `5x+3y+9=0`
    C
    `5x+3y-9=0`
    D
    `5x-3y+9=0`

  • If the equation of the locus of a point equidistant from the points (a_1,b_1) and (a_2,b_2) is (a_1-a_2)x+(b_2+b_2)y+c=0 , then the value of C is

    A
    `a_1^2-a_2^2+b_1^2-b_2^2`
    B
    `sqrt(a_1^2+b_1^2-a_2^2-b_2^2)`
    C
    `(1)/(2)(a_1^2+a_2^2+b_1^2+b_2^2)`
    D
    `(1)/(2)(a_1^2+b_2^2+a_1^2+b_2^2)`

  • The equation of the locus of a poinnt which is equidistant from the axes is

    A
    y = 2x
    B
    x = 2y
    C
    `y = pm x`
    D
    `2y + x =0 `

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