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

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Question

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

Video Solution
Coordinate Geometry
Ex 7.1 | Question 7

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 $$(-7, 0)$$.

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