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

Go back to  'Ex.7.1'

## 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( {{{\text{x}}_{\text{1}}} - {{\text{x}}_{\text{2}}}} \right)}^2} + {{\left( {{{\text{y}}_{\text{1}}} - {{\text{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( {{{\text{x}}_1} - {{\text{x}}_2}} \right)}^2} + {{\left( {{{\text{y}}_1} - {{\text{y}}_2}} \right)}^2}} \qquad\qquad ...\;{\text{Equation}}\,(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 {{{({\text{x}} - 2)}^2} + {{(0 - ( - 5))}^2}} \\ &= \sqrt {{{({\text{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),

\begin{align}{\text{Distance }} \\ &=\sqrt {{{({\text{x}} - ( - 2))}^2} + {{(0 - ( - 9))}^2}} \\& = \sqrt {{{({\text{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 $$(-7, 0)$$.

Learn from the best math teachers and top your exams

• Live one on one classroom and doubt clearing
• Practice worksheets in and after class for conceptual clarity
• Personalized curriculum to keep up with school