Find a point on y-axis which is equidistant

Question:

Find a point on y-axis which is equidistant form the points (5, −2) and (−3, 2).

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(5,2) and B(3,2).

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

$A C=\sqrt{(5-x)^{2}+(-2-y)^{2}}$

$=\sqrt{(5-0)^{2}+(-2-y)^{2}}$

$A C=\sqrt{(5)^{2}+(-2-y)^{2}}$

$B C=\sqrt{(-3-x)^{2}+(2-y)^{2}}$

$=\sqrt{(-3-0)^{2}+(2-y)^{2}}$

$B C=\sqrt{(-3)^{2}+(2-y)^{2}}$

We know that both these distances are the same. So equating both these we get,

$A C=B C$

$\sqrt{(5)^{2}+(-2-y)^{2}}=\sqrt{(-3)^{2}+(2-y)^{2}}$

Squaring on both sides we have,

$(5)^{2}+(-2-y)^{2}=(-3)^{2}+(2-y)^{2}$

$25+4+y^{2}+4 y=9+4+y^{2}-4 y$

$8 y=-16$

$y=-2$

Hence the point on the $y$-axis which lies at equal distances from the mentioned points is $(0,-2)$.

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