# 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}

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