Question

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

Answer 1

For a point to be equidistant from $2$ given point it means that the distance between the point and the given point will be equal. And we can equate both the distances since both are equal.

Complete step-by-step answer:
Since the required point is on $x$-axis, the $y$-coordinate will be equal to $0$.
Therefore we can assume that the required point is$\left( {x,{\text{ }}0} \right)$.
Distance between $2$ points is given by the distance formula.
If the $2$ points are $\left( {{x_1},{\text{ }}{y_1}} \right)$ and $\left( {{x_2},{y_2}} \right)$ then the distance between them is given by $\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}}$.
Here, we can say that $\left( {x,{\text{ }}0} \right)$ is equidistant to $\left( {2, - 5} \right)$ and $\left( { - 2,9} \right)$.
We can equate the distances between them using the distance formula because both are equal. Solving this quadratic equation will be the value of $x$.
Therefore, we can find the required point by replacing $x$.
Using the distance formula,
$\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}}$
We can write that $\left( {{x_1},{y_1}} \right)$ will be $\left( {2, - 5} \right)$ and $\left( { - 2,9} \right)$.
Also we can write to equate them,
$\sqrt {{{\left( {x - 2} \right)}^2} + {{\left( {0 + 5} \right)}^2}} = \sqrt {{{\left( {x + 2} \right)}^2} + {{\left( {0 - 9} \right)}^2}}$
Squaring on both sides,
${\left( {\sqrt {{{\left( {x - 2} \right)}^2} + {{\left( {0 + 5} \right)}^2}} } \right)^2} = {\left( {\sqrt {{{\left( {x + 2} \right)}^2} + {{\left( {0 - 9} \right)}^2}} } \right)^2}$
We get,
${\left( {x - 2} \right)^2} + {\left( 5 \right)^2} = {\left( {x + 2} \right)^2} + {\left( { - 9} \right)^2}$
Expanding ${\left( {x - 2} \right)^2}$and ${\left( {x + 2} \right)^2}$ on both sides
${x^2} - 4x + 4 + 25 = {x^2} + 4x + 4 + 81$
We can cancelling the equal terms, we get
$- 8x = 81 - 25$
On subtracting, we get
$- 8x = 56$
Divided into it,
$x = - 7$
Therefore, the required point is $\left( { - 7,0} \right)$.

Note: A point is a basic relationship shown on a graph. Each point is defined by a pair of numbers containing two coordinates. A coordinate is one of a set of numbers used to identify the location of a point on a graph. Each point is identified by both an $x$ and $y$ coordinate.
And the coordinate grid has two perpendicular lines, or axes, labelled as number lines. The horizontal axis is known as the $x$-axis, while the vertical axis is called the $y$-axis. The point where the $x$-axis and $y$-axis intersect is called the origin. The numbers on a coordinate grid are used to locate points.