Question

Find a relation between x and y such that the point (x,y) is equidistant from the point (3,6) and (-3,4).

Answer 1

Hint: Use the distance formula, \[d=\sqrt{{{({{x}_{2}}-{{x}_{1}})}^{2}}+{{({{y}_{2}}-{{y}_{1}})}^{2}}}\]. Here \[({{x}_{1}},{{y}_{1}})\]and \[({{x}_{2}},{{y}_{2}})\] are two points between which the distance (d) is to be found.

Complete step by step solution:
In the question, we have to find the relation between x and y such that the point (x,y) is equidistant from the point (3,6) and (-3,4).
Now, distance (d) between points \[({{x}_{1}},{{y}_{1}})\]and \[({{x}_{2}},{{y}_{2}})\]is found using the formula:
\[d=\sqrt{{{({{x}_{2}}-{{x}_{1}})}^{2}}+{{({{y}_{2}}-{{y}_{1}})}^{2}}}\]
Now, it is given that the point (x,y) is equidistant from the point (3,6) and (-3,4).
So the distance between points (x,y) and (3,6) will be same as the distance between the points (x,y) and (-3,4).So, the distance between the points (x,y) and (3,6) is as follows:
\[\begin{align}
  & \Rightarrow {{d}_{1}}=\sqrt{{{({{x}_{2}}-{{x}_{1}})}^{2}}+{{({{y}_{2}}-{{y}_{1}})}^{2}}} \\
 & \Rightarrow {{d}_{1}}=\sqrt{{{(3-x)}^{2}}+{{(6-y)}^{2}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\because {{x}_{2}}=3,\,\,{{x}_{1}}=x,\,\,{{y}_{2}}=6,\,{{y}_{1}}=y \\
 & \, \\
\end{align}\]
Next, we will find the distance between the points (x,y) and (-3,4), as follows:
\[\begin{align}
  & \Rightarrow d=\sqrt{{{({{x}_{2}}-{{x}_{1}})}^{2}}+{{({{y}_{2}}-{{y}_{1}})}^{2}}} \\
 & \Rightarrow {{d}_{2}}=\sqrt{{{(-3-x)}^{2}}+{{(4-y)}^{2}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\because {{x}_{2}}=-3,\,\,{{x}_{1}}=x,\,\,{{y}_{2}}=4,\,{{y}_{1}}=y \\
 & \, \\
\end{align}\]
Now, since the two distances are equal, so we have:
\[\begin{align}
  & \Rightarrow {{d}_{1}}={{d}_{2}} \\
 & \Rightarrow \sqrt{{{(3-x)}^{2}}+{{(6-y)}^{2}}}=\sqrt{{{(-3-x)}^{2}}+{{(4-y)}^{2}}} \\
\end{align}\]
Squaring both sides of the equation, we get:
\[\begin{align}
  & \Rightarrow {{(3-x)}^{2}}+{{(6-y)}^{2}}={{(-3-x)}^{2}}+{{(4-y)}^{2}} \\
 & \Rightarrow {{y}^{2}}-12y+{{x}^{2}}+45-6x={{y}^{2}}-8y+{{x}^{2}}+6x+25\, \\
 & \Rightarrow -12x=4y-20 \\
 & \Rightarrow 3x+y=5 \\
\end{align}\]
So the required relation is\[3x+y=5\].

Note: When finding the distance between coordinates of two points (0, y) and (-4,3), then we can actually we can interchange \[{{x}_{1}}\] and \[\,{{y}_{1}}\]with \[{{x}_{2}}\]and \[{{y}_{2}}\] respectively, as there will be no change in the final result.