Question

Find the coordinates of a point on the y-axis which is equidistant from points (6, 5) and (-4, 3).

Answer 1

Hint-Here, we will consider the point on the y –axis to be (0, k). Then, we will find its distance from (6, 5) and (-4, 3) using the distance formula which is given as \[d=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}\]. Since both the distances are equal, we can find the value of k after equating these distances.

Complete Step-by-Step solution:
The distance formula is an useful tool in math for finding the distance between two points which can be arbitrarily represented as point A \[\left( {{x}_{1}},{{y}_{1}} \right)\] and B \[\left( {{x}_{2}},{{y}_{2}} \right)\]. The distance formula itself is actually derived using Pythagoras theorem.
According to the distance formula, distance between two points A and B is given as:
\[d=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}...........\left( 1 \right)\]
Using equation (1), distance between the required point on the y-axis which we have taken as (0, k) and the point with coordinate (6, 5) can be given as:
$\sqrt{{{\left( 6-0 \right)}^{2}}+{{\left( 5-k \right)}^{2}}}=\sqrt{36+{{\left( 5-k \right)}^{2}}}$
Also, the distance between the point (-4, 3) and the required point (0, k) can be given as:
 $\sqrt{{{\left( -4-0 \right)}^{2}}+{{\left( 3-k \right)}^{2}}}=\sqrt{16+{{\left( 3-k \right)}^{2}}}$
Since, it is given that the distances of the required point from both the points are same, so we can write:
$\sqrt{36+{{\left( 5-k \right)}^{2}}}=\sqrt{16+{{\left( 3-k \right)}^{2}}}$
On squaring this equation both sides, we get:
$\begin{align}
  & 36+{{\left( 5-k \right)}^{2}}=16+{{\left( 3-k \right)}^{2}} \\
 & \Rightarrow 36-16={{\left( 3-k \right)}^{2}}-{{\left( 5-k \right)}^{2}} \\
 & \Rightarrow 20=9+{{k}^{2}}-6k-\left( 25+{{k}^{2}}-10k \right) \\
 & \Rightarrow 20={{k}^{2}}-6k+9-{{k}^{2}}+10k-25 \\
 & \Rightarrow 20=4k-16 \\
 & \Rightarrow 20+16=4k \\
 & \Rightarrow k=\dfrac{36}{4} \\
 & \Rightarrow k=9 \\
\end{align}$
So, the value of k comes out to be 9.
Hence, the required point on the y-axis is (0, 9).

Note: Students should note here that any point on the y-axis is always of the form (0, k). The value of x-coordinate on the y-axis is always zero.