Find the point on the x-axis which is equidistant from $\left( {2, - 5} \right)$ and $\left( { - 2,9} \right)$.

Hint: The point is on x-axis . So, its y-coordinate will be zero. Then, the point will be $\left( {a,0} \right)$. Also, the distance between two given points $\left( {{x_1},{y_1}} \right)$ and $\left( {{x_2},{y_2}} \right)$ can be calculated by the formula, $\sqrt {{{\left( {{x_1} - {x_2}} \right)}^2} + {{\left( {{y_1} - {y_2}} \right)}^2}} $.

Complete step-by-step answer:
The given points are $\left( {2, - 5} \right)$ and $\left( { - 2,9} \right)$.
 We have to find a point on x-axis. Therefore, its y-coordinate will be $0$.
Let x-coordinate of the point $ = a$
So, the point $ = \left( {a,0} \right)$
As mentioned in the question, the two points$\left( {2, - 5} \right)$ and $\left( { - 2,9} \right)$ are equidistant from $\left( {a,0} \right)$.
So, distance between $\left( {a,0} \right)$ and $\left( {2, - 5} \right)$=$\sqrt {{{\left( {{x_1} - {x_2}} \right)}^2} + {{\left( {{y_1} - {y_2}} \right)}^2}} $
$ = \sqrt {{{\left( {a - 2} \right)}^2} + {{\left( {0 + 5} \right)}^2}} $
Similarly, distance between $\left( {a,0} \right)$ and $\left( { - 2,9} \right)$$ = \sqrt {{{\left( {a + 2} \right)}^2} + {{\left( {0 - 9} \right)}^2}} $
Now, the points are equidistant.
So, $\sqrt {{{\left( {a - 2} \right)}^2} + {{\left( {0 + 5} \right)}^2}} $= $\sqrt {{{\left( {a + 2} \right)}^2} + {{\left( {0 - 9} \right)}^2}} $
Squaring both sides,
$ \Rightarrow $${\left( {a - 2} \right)^2} + {\left( {0 + 5} \right)^2}$= ${\left( {a + 2} \right)^2} + {\left( {0 - 9} \right)^2}$
$ \Rightarrow $${a^2} + 4 - 4a + 25 = {a^2} + 4 + 4a + 81$
$ \Rightarrow $${a^2} - 4a - {a^2} - 4a = 4 + 81 - 4 - 25$
$ \Rightarrow $$ - 8a = 56$
$ \Rightarrow $$a = \dfrac{{56}}{{ - 8}}$
$ \Rightarrow $$a = - 7$

Hence, the point on x-axis is $\left( { - 7,0} \right)$.

Note: If the point lies on x-axis then the y coordinate will be equal to zero. If the point lies on the y-axis then the x coordinate will be equal to zero. Here the mid-point formula can’t be used because it is not necessary that the mid-point lies on the x-axis.