Question

Find a point on the y-axis which is equidistant from the points (5,2) and (-4,3).

Answer 1

Hint: Use the concept that any point on the y-axis is (0, y). We also need to apply the distance formula in this question. The distance formula is \[d=\sqrt{{{({{x}_{2}}-{{x}_{1}})}^{2}}+{{({{y}_{2}}-{{y}_{1}})}^{2}}}\]

Complete step by step answer:
In the question, we have to find a point on the y-axis which is equidistant from the points (5,2) and (-4,3). So, the required point has the coordinate of (0, y). Now, the abscissa of this point is zero because this is on the y-axis.
Now, distance (d) between points \[({{x}_{1}},{{y}_{1}})\]and \[({{x}_{2}},{{y}_{2}})\] will be:
\[d=\sqrt{{{({{x}_{2}}-{{x}_{1}})}^{2}}+{{({{y}_{2}}-{{y}_{1}})}^{2}}}\]
Now, the coordinate (0, y) on the y-axis is equidistant from points (5,2) and (-4,3).
So the distance between points (0, y) and (5,2) will be the same as the distance between the points (0, y) and (-4,3). The distance between the points (0, y) and (5,2) is as follows:
\[\begin{align}
  & \Rightarrow {{d}_{1}}=\sqrt{{{({{x}_{2}}-{{x}_{1}})}^{2}}+{{({{y}_{2}}-{{y}_{1}})}^{2}}} \\
 & \Rightarrow {{d}_{1}}=\sqrt{{{(5-0)}^{2}}+{{(2-y)}^{2}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\because {{x}_{2}}=5,\,\,{{x}_{1}}=0,\,\,{{y}_{2}}=2,\,{{y}_{1}}=y \\
 & \, \\
\end{align}\]
Next, we will find the distance between the points (0, y) and (-4,3), as follows:
\[\begin{align}
  & \Rightarrow d=\sqrt{{{({{x}_{2}}-{{x}_{1}})}^{2}}+{{({{y}_{2}}-{{y}_{1}})}^{2}}} \\
 & \Rightarrow {{d}_{2}}=\sqrt{{{(-4-0)}^{2}}+{{(3-y)}^{2}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\because {{x}_{2}}=-4,\,\,{{x}_{1}}=0,\,\,{{y}_{2}}=3,\,{{y}_{1}}=y \\
 & \, \\
\end{align}\]
Now, since the two distances are equal, so we have:
\[\begin{align}
  & \Rightarrow {{d}_{1}}={{d}_{2}} \\
 & \Rightarrow \sqrt{{{(5-0)}^{2}}+{{(2-y)}^{2}}}=\sqrt{{{(-4-0)}^{2}}+{{(3-y)}^{2}}} \\
 & \Rightarrow {{(5-0)}^{2}}+{{(2-y)}^{2}}={{(-4-0)}^{2}}+{{(3-y)}^{2}} \\
 & \Rightarrow {{y}^{2}}-4y+29={{y}^{2}}-6y+25\,\,\, \\
 & \Rightarrow -4y+29=-6y+25 \\
 & \Rightarrow y=-2 \\
\end{align}\]

So we get the required coordinate on the y-axis as \[\left( 0,\text{ }y \right)\equiv (0,-2)\].

Note: We should know that any coordinate on the y-axis is (0, y) and not (y,0). When we are finding the distance between two points (0, y) and (-4,3), then we can actually interchange \[{{x}_{1}}\] and \[\,{{y}_{1}}\]with \[{{x}_{2}}\]and \[{{y}_{2}}\] respectively, as there will be no change in the final result.