# Ex.7.1 Q7 Coordinate Geometry Solution - NCERT Maths Class 10

## Question

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

## Text Solution

**Reasoning:**

The distance between the two points can be measured using the Distance Formula which is given by:

Distance Formula

\[\begin{align} = \sqrt {{{\left( {{x_{\text{1}}} - {x_{\text{2}}}} \right)}^2} + {{\left( {{y_{\text{1}}} - {y_{\text{2}}}} \right)}^2}} .\end{align}\]

**What is Known?**

The \(x\) and \(y\) co-ordinates of the points from which the point on the \(x\)-axis is equidistant.

**What is Unknown?**

The point on the \(x\)-axis which is equidistant from \((2, -5)\) and \((-2, 9)\).

**Steps:**

Given,

Since the point is on \(x\)-axis the co-ordinates are \((x, 0)\).

We have to find a point on \(x\)-axis which is equidistant from \(A\; (2, -5)\) and \(B \;(-2, 9)\).

We know that the distance between the two points is given by the Distance Formula,

\[\begin{align}&\sqrt {{{\left( {{x_1} - {x_2}} \right)}^2} + {{\left( {{y_1} - {y_2}} \right)}^2}}\;\;\;(1)\end{align}\]

To find the distance between \(PA\), substitute the values of \(P\; (x, 0)\) and \(A\; (2, -5)\) in Equation (1),

\[\begin{align} &= \sqrt {{{(x - 2)}^2} + {{(0 - ( - 5))}^2}} \\ &= \sqrt {{{(x - 2)}^2} + {{(5)}^2}} \end{align}\]

To find the distance between \(PB\), substitute the values of \(P \;(x, 0)\) and \(B \;(-2, 9)\) in Equation (1),

Distance

\[\begin{align}&=\sqrt {{{(x - ( - 2))}^2} + {{(0 - ( - 9))}^2}} \\& = \sqrt {{{(x + 2)}^2} + {{(9)}^2}} \end{align}\]

By the given condition, these distances are equal in measure.

Hence \(PA = PB\)

\(\begin{align}\!\sqrt {{{(\!x \!- \!2)}^2} \!+\! {{(5)}^2}} &= \!\sqrt {{{(x\! + \!2)}^2} \!+ \!{{(9)}^2}} \end{align}\)

Squaring on both sides

\(\begin{align} {(x - 2)^2} + 25 &= {(x + 2)^2} \!+\! 81 \\ {x^2} + 4 - 4x + 25 &= \!{x^2} \!+ \!4 +\! 4x \!+ \!81 \\ 8x &= 25 - \!81 \\ 8x &= - 56 \\x &= - 7\end{align}\)

Therefore, the point equidistant from the given points on the axis is \((-7, 0)\).