Question:

The point on the \( y \)-axis which is equidistant from the points \( (5, -2) \) and \( (-3,2) \) is:

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A point \( (x,y) \) is equidistant from two points \( (x_1, y_1) \) and \( (x_2, y_2) \) if:
\[ \sqrt{(x - x_1)^2 + (y - y_1)^2} = \sqrt{(x - x_2)^2 + (y - y_2)^2}. \]
Updated On: Oct 27, 2025
  • \( (0,3) \)
  • \( (-2,0) \)
  • \( (0,-2) \)
  • \( (2,2) \)
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The Correct Option is C

Solution and Explanation

A point \( (0,y) \) on the \( y \)-axis is equidistant from two given points \( (x_1, y_1) \) and \( (x_2, y_2) \) if:
\[ \sqrt{(x_1 - 0)^2 + (y_1 - y)^2} = \sqrt{(x_2 - 0)^2 + (y_2 - y)^2} \] Substituting \( (5,-2) \) and \( (-3,2) \):
\[ \sqrt{(5 - 0)^2 + (-2 - y)^2} = \sqrt{(-3 - 0)^2 + (2 - y)^2} \] \[ \sqrt{25 + (y+2)^2} = \sqrt{9 + (y-2)^2} \] Squaring both sides:
\[ 25 + (y+2)^2 = 9 + (y-2)^2. \] Expanding:
\[ 25 + y^2 + 4y + 4 = 9 + y^2 - 4y + 4. \] Cancel \( y^2 \):
\[ 25 + 4y + 4 = 9 - 4y + 4. \] \[ 29 + 4y = 13 - 4y. \] \[ 8y = -16 \Rightarrow y = -2. \] Thus, the required point is \( (0, -2) \).
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