Question

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

Answer 1

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:

& \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{aligned}$$ Next, we will find the distance between the points (0, y) and (-4,3), as follows: $$\begin{aligned} & \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{aligned}$$ Now, since the two distances are equal, so we have: $$\begin{aligned} & \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{aligned}$$ **So we get the required coordinate on the y-axis as $$\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.