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

## 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 \(\begin{align}3{\text{x}} + {\text{y}}-5 = 0\end{align}\)