Which point on y-axis is equidistant from (2, 3) and (−4, 1)?
The distance $d$ between two points $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ is given by the formula
$d=\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$
Here we are to find out a point on the y-axis which is equidistant from both the points A (2, 3) and B (−4, 1).
Let this point be denoted as C(x, y).
Since the point lies on the y-axis the value of its ordinate will be 0. Or in other words we have.
Now let us find out the distances from ‘A’ and ‘B’ to ‘C’
$A C=\sqrt{(2-x)^{2}+\left(3-y^{2}\right)}$
$=\sqrt{(2-0)^{2}+(3-y)^{2}}$
$A C=\sqrt{(2)^{2}+(3-y)^{2}}$
$B C=\sqrt{(-4-x)^{2}+(1-y)^{2}}$
$=\sqrt{(-4-0)^{2}+(1-y)^{2}}$
$B C=\sqrt{(-4)^{2}+(1-y)^{2}}$
We know that both these distances are the same. So equating both these we get,
$A C=B C$
$\sqrt{(2)^{2}+(3-y)^{2}}=\sqrt{(-4)^{2}+(1-y)^{2}}$
Squaring on both sides we have,
$(2)^{2}+(3-y)^{2}=(-4)^{2}+(1-y)^{2}$
$4+9+y^{2}-6 y=16+1+y^{2}-2 y$
$4 y=-4$
$y=-1$
Hence the point on the $y$-axis which lies at equal distances from the mentioned points is $(0,-1)$.
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.