Find a point on the x-axis which is equidistant

Question:

Find a point on the x-axis which is equidistant from the points (7, 6) and (−3, 4).

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 x−axis which is equidistant from both the points A(7,6) and B(3,4).

Let this point be denoted as C(x, y).

Since the point lies on the x-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

$A C=\sqrt{(7-x)^{2}+(6-y)^{2}}$

$=\sqrt{(7-x)^{2}+(6-0)^{2}}$

$A C=\sqrt{(7-x)^{2}+(6)^{2}}$

$B C=\sqrt{(-3-x)^{2}+(4-y)^{2}}$

$=\sqrt{(-3-x)^{2}+(4-0)^{2}}$

$B C=\sqrt{(-3-x)^{2}+(4)^{2}}$

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

$A C=B C$

$\sqrt{(7-x)^{2}+(6)^{2}}=\sqrt{(-3-x)^{2}+(4)^{2}}$

Squaring on both sides we have,

$(7-x)^{2}+(6)^{2}=(-3-x)^{2}+(4)^{2}$

$49+x^{2}-14 x+36=9+x^{2}+6 x+16$

$20 x=60$

$x=3$

Hence the point on the $x$-axis which lies at equal distances from the mentioned points is $(3,0)$.

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