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Which point on y-axis is equidistant from $\left( {2,3} \right)$ and $\left( { - 4,1} \right)$?

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Hint- Here, we will be using distance formula.

Given, we have two points A$\left( {2,3} \right)$ and B$\left( { - 4,1} \right)$.
Since, we know that any point on the y-axis will have its x-coordinate as zero.
Let P$\left( {0,y} \right)$be that point that will lie on the y-axis.
We know that according to distance formula, the distance between any two points ${\text{A}}\left( {a,b} \right)$ and $B\left( {c,d} \right)$is given by $d = \sqrt {{{\left( {c - a} \right)}^2} + {{\left( {d - b} \right)}^2}} $
Also given that the point P$\left( {0,y} \right)$is equidistant from the points A$\left( {2,3} \right)$ and B$\left( { - 4,1} \right)$ which means that the distances AP and BP are equal.
i.e. $
  {\text{AP}} = {\text{BP}} \Rightarrow \sqrt {{{\left( {0 - 2} \right)}^2} + {{\left( {y - 3} \right)}^2}} = \sqrt {{{\left[ {0 - \left( { - 4} \right)} \right]}^2} + {{\left( {y - 1} \right)}^2}} \\
   \Rightarrow \sqrt {4 + {y^2} + 9 - 6y} = \sqrt {16 + {y^2} + 1 - 2y} \\
 $
Now squaring both the sides of above equation, we get
$
   \Rightarrow 4 + {y^2} + 9 - 6y = 16 + {y^2} + 1 - 2y \Rightarrow 6y - 2y = 4 + 9 - 1 - 16 \\
   \Rightarrow 4y = - 4 \Rightarrow y = - 1 \\
 $
Therefore, the required point on the y-axis is P$\left( {0, - 1} \right)$.


Note- In this problem if the point which is equidistant from the two given points instead of lying on y-axis, lies on x-axis then the coordinates of the required point would have been assumed as P$\left( {x,0} \right)$ because any point lying on the x-axis have its y coordinate as zero.

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