Question

Which point on the x-axis is equidistant from (5, 9) and (−4, 6)?

Answer 1

The distance d between two points ${\displaystyle (x_1,y_1)}$ and ${\displaystyle (x_2,y_2)}$  is given by the formula

${\displaystyle d=\sqrt{{(x_1-x_2)}^2+{(y_1-y_2)}^2}}$

Here we are to find out a point on the x−axis which is equidistant from both the points A (5, 9) and B (−4, 6)

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 y = 0

Now let us find out the distances from ‘A’ and ‘B’ to ‘C’

${\displaystyle AC=\sqrt{{(5-x)}^2+{(9-y)}^2}}$

${\displaystyle =\sqrt{{(5-x)}^2+{(9-0)}^2}}$

${\displaystyle AC=\sqrt{{(5-x)}^2+{\left(9\right)}^2}}$

${\displaystyle BC=\sqrt{{(-4-x)}^2+{(6-y)}^2}}$

${\displaystyle =\sqrt{{(4+x)}^2+{(6-0)}^2}}$

${\displaystyle BC=\sqrt{{(4+x)}^2+{\left(6\right)}^2}}$

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

AC = BC

${\displaystyle \sqrt{{(5-x)}^2+{\left(9\right)}^2}=\sqrt{{(4+x)}^2+{\left(6\right)}^2}}$

Squaring on both sides we have,

${\displaystyle {(5-x)}^2+{\left(9\right)}^2={(4+x)}^2+{\left(6\right)}^2}$

${\displaystyle 25+x^2-10x+81=16+x^2+8x+36}$

18x = 54

x = 3

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