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).

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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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