Question

The point on the y-axis equidistant from (-5,2) and (9,-2) is (0,-m), then m is:

Answer 1

Hint: In this question, we need to calculate the distance between the points using the distance formula. Then equate those distances which gives an equation in m this on further simplification gives the value of m.

Complete step-by-step answer:
\[\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}\]

DISTANCE FORMULA: Distance between two points (x1 , y1) and (x2 , y2) , is

\[\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}\]

Distance of the point (x1 , y1) from the origin is :

\[\sqrt{{{x}_{1}}^{2}+{{y}_{1}}^{2}}\]

Now, let us find the distance between the points (-5,2) and (0,-m).

On comparing the points considered with the points in the distance formula we get,

\[\begin{align}
  & {{x}_{1}}=-5,{{y}_{1}}=2 \\
 & {{x}_{2}}=0,{{y}_{2}}=-m \\
\end{align}\]

Let us assume that the distance between these two points as D.

\[\Rightarrow D=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}\]

On substituting the respective values we get,

\[\begin{align}
  & \Rightarrow D=\sqrt{{{\left( 0-\left( -5 \right) \right)}^{2}}+{{\left( -m-2 \right)}^{2}}} \\
 & \Rightarrow D=\sqrt{{{5}^{2}}+{{\left( m+2 \right)}^{2}}} \\
\end{align}\]

Let us now consider the points (9,-2) and (0,-m) then on comparing these with the points in the distance formula we get,

\[\begin{align}
  & {{x}_{1}}=9,{{y}_{1}}=-2 \\
 & {{x}_{2}}=0,{{y}_{2}}=-m \\
\end{align}\]

Let us consider the distance between these two points as E.

\[\begin{align}
  & \Rightarrow E=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}} \\
 & \Rightarrow E=\sqrt{{{\left( 0-\left( 9 \right) \right)}^{2}}+{{\left( -m-\left( -2 \right) \right)}^{2}}} \\
 & \Rightarrow E=\sqrt{{{9}^{2}}+{{\left( m-2 \right)}^{2}}} \\
\end{align}\]

Let us now equate E and D

\[\begin{align}
  & \Rightarrow E=D \\
 & \Rightarrow \sqrt{{{9}^{2}}+{{\left( m-2 \right)}^{2}}}=\sqrt{{{5}^{2}}+{{\left( m+2 \right)}^{2}}} \\
\end{align}\]

Now, by squaring on both sides we get,

\[\Rightarrow {{9}^{2}}+{{\left( m-2 \right)}^{2}}={{5}^{2}}+{{\left( m+2 \right)}^{2}}\]

Now, by expanding the terms on the both sides we get,

\[\Rightarrow 81+{{m}^{2}}-4m+4=25+{{m}^{2}}+4m+4\]

Let us now rearrange the terms and then cancel out the common terms.

\[\begin{align}
  & \Rightarrow 81-4m=25+4m \\
 & \Rightarrow 81-25=4m+4m \\
\end{align}\]

\[\begin{align}
  & \Rightarrow 56=8m \\
 & \Rightarrow m=\dfrac{56}{8} \\
 & \therefore m=7 \\
\end{align}\]


Note: It is important to note that to find the value of m we need to equate the distance between the points (-5,2) and (0,-m) with the distance between the points (9,-2) and (0,-m) because given that the both points are equidistant from (0,-m).

Instead of assigning different variables to the respective distances between the two points we can directly consider the distance formula and substitute the respective coordinates and then equate them there itself. Both the methods give the same result but in this case we can solve quickly.

While squaring and rearranging the terms from one side to the other we need to be careful about the respective terms and their corresponding signs because neglecting any one of them changes the equation in m which in further changes the value of m.