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