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

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

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