Ex.7.1 Q10 Coordinate Geometry Solution - NCERT Maths Class 10

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Question

Find a relation between \(x\) and \(y\) such that the point \((x, y)\) is equidistant from the point \((3, 6)\) and \((-3, 4)\).

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 the known?

The \(x\) and \(y\) co-ordinates of the points between which the distance is to be measured.

What is the unknown?

The relation between \(x\) and \(y\) such that the point \((x, y)\) is equidistant from the point \((3, 6)\) and \((-3, 4)\)

Steps:

Let Point \(P\;(x, y)\) be equidistant from points \(A \;(3, 6)\) and \(B\; (-3, 4)\).

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

\(\begin{align}\sqrt {{{\left( {{{\text{x}}_1} - {{\text{x}}_2}} \right)}^2} + {{\left( {{{\text{y}}_1} - {{\text{y}}_2}} \right)}^2}} \qquad \qquad ...\;\rm {Equation}\;(1)\end{align}\)

Since they are equidistant, \(PA = PB\)

Hence by applying the distance formula for \(PA = PB\), we get

\(\begin{align} \sqrt {{{(x - 3)}^2} + {{(y - 6)}^2}} &= \sqrt {x - {{( - 3)}^2} + {{(y - 4)}^2}} \\\;\;\,\sqrt {{{(x - 3)}^2} + {{(y - 6)}^2}} &= \sqrt {{{(x + 3)}^2} + {{(y - 4)}^2}} \end{align}\)

By Squaring, \(\begin{align}PA^2=PB^2\end{align}\)

\(\begin{align}{({\text{x}} - 3)^2} + {({\text{y}} - 6)^2} &= {({\text{x}} + 3)^2} + {({\text{y}} - 4)^2}\\{{\text{x}}^2} + 9 - 6{\text{x}} + {{\text{y}}^2} + 36 - 12{\text{y}} &= {{\text{x}}^2} + 9 + 6{\text{x}} + {{\text{y}}^2} + 16 - 8{\text{y}}\\6{\text{x}} + 6{\text{x}} + 12{\text{y}} - 8 \,\rm y &= \text{36 - 16}\\12{\text{x}} + 4\,\rm y &={\text{ 20}}\\3{\text{x}} + {\text{y}} &= 5\\3{\text{x}} + {\text{y}} - 5 &= 0\end{align}\)

Thus, the relation between \(x\) and \(y\) is given by \(\begin{align}3{\text{x}} + {\text{y}}-5 = 0\end{align}\)