**Question:**

Which point on *y*-axis is equidistant from (2, 3) and (−4, 1)?

**Solution:**

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 $(0,-1)$.

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