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 \((-2, 9)\).

 

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 \((-7, 0)\).