Find the point on y axis which is equidistant from the points (3,2) and (4,6)

Find the point on y axis which is equidistant from the points (3,2) and (4,6)

2 thoughts on “Find the point on y axis which is equidistant from the points (3,2) and (4,6)”

  1. Answer:

    (0, 4.875)

    Step-by-step explanation:

    let the required coordinates be (x, y)

    (x1, y1) = (3, 2)

    (x2, y2) = (4, 6)

    distances are equal from both the points so

    d1 = d2

    [tex]\sqrt{(x1-x)^{2}+(y1-y)^2 } = \sqrt{(x2-x)^{2}+(y2-y)^2 }[/tex]

    [tex]\sqrt{(3-x)^{2}+(2-y)^2 } = \sqrt{(4-x)^{2}+(6-y)^2 }[/tex]

    as the point is in the y-axis then x=0, so

    [tex]\sqrt{(3)^{2}+(2-y)^2 } = \sqrt{(4)^{2}+(6-y)^2 }[/tex]

    [tex]\ (3)^{2}+(2-y)^2 = \ (4)^{2}+(6-y)^2[/tex]

    [tex]9 + 4 – 4y + y^{2} = 16 + 36 -12y + y^2[/tex]

    [tex]13 – 4y = 52 -12y[/tex]

    [tex]12y – 4y = 52 -13[/tex]

    [tex]8y = 39[/tex]

    [tex]y = 4.875[/tex]

    so the final coordinate is (0, 4.875).

  2. Given : point on y axis which is equidistant from the points (3,2) and (4,6)

    To Find :  Point

    Solution:

    Let say point on y axis  is P ( 0 , y)

    as x coordinate will be zero on y axis.

    A  (3,2)  ,  B  (4,6)

    PA = PB

    PA² = PB²

    Apply distance formula and squaring

    (0 – 3)² + (y – 2)²  = (0 – 4)² + (y – 6)²  

    => 9 + y² -4y + 4 = 16  + y² – 12y  + 36

    =>  8y = 39

    => y = 39/8

    => y  = 4.875

    point on y axis which is equidistant from the points (3,2) and (4,6)

    is ( 0 , 39/8)   or  ( 0 , 4.875)

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