Find a point on the y-axis which is equidistant from the points A (6,5) and B (- 4,3)

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P (0, 9)

P (0, 5)

P (0, 7)

P (0, 4)

answer is A.

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Detailed Solution

Given points are A (6,5) and B (- 4,3)
Let us suppose the point on the y-axis which is equidistant from the given points be P (0, y) because as it lies on the y-axis so its x - coordinate will be 0
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For point P to be equidistant from points A and B, distance between points A and P will be equal to the distance between points B and P
AP = BP ……………… (equation 1)
According to distance formula, the distance between any two points X (a, b) and Y (c, d) is given by
XY =

\sqrt{{(c - a)}^{2} + {(d - b)}^{2}}

Now substituting the values in the above equation, we get
AP =

\sqrt{{(0 - 6)}^{2} + {(y - 5)}^{2}}

\sqrt{36 + {(y - 5)}^{2}}

BP =

\sqrt{{(0 - (- 4))}^{2} + {(y - 3)}^{2}}

\sqrt{16 + {(y - 3)}^{2}}

Now substituting the values of AP and BP in equation 1 we get

\sqrt{36 + {(y - 5)}^{2}}

\sqrt{16 + {(y - 3)}^{2}}

36 + {(y - 5)}^{2}

16 + {(y - 3)}^{2}

36 + y2 + 25 – 10y = 16 + y2 + 9 – 6y [since (a ± b)2 = a2 + b2 ± 2ab]
⇒ 36 + 25 − 10y + 6y – 9 – 16 = y2 – y2 ⇒36 − 4y = 0
⇒4y = 36
⇒y = 9
Therefore, the point on the y - axis which is equidistant from the points A (6, 5) and B (- 4, 3) is P (0, 9).
So, option (1) is correct.

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