# 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 $$(-3, 4)$$.

Video Solution
Coordinate Geometry
Ex 7.1 | Question 10

## 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 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( {{x_1}\! - {x_2}} \right)}^2}\! +\! {{\left( {{y_1} - {y_2}}\! \right)}^2}}\;\;\dots(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}{(x\! -\! 3)^2} \!+\! {(y \!-\! 6)^2} &\!=\! {(x \!+\! 3)^2}\! +\! {(y \!-\! 4)^2}\\ \begin{bmatrix}{x^2} \!+ \!9\! -\! 6x\! +\!\\ {y^2} \!+\! 36\! -\! 12y \end{bmatrix}&\!=\! \begin{bmatrix}{x^2}\! +\! 9 \!+\! 6x\! +\! \\{y^2}\! +\! 16\! -\! 8y\end{bmatrix}\\6x\! +\! 6x\! + \!12y\! -\! 8 y &\!=\! 36\! -\! 16\\12x \!+\! 4y &\!=\!20\\3x \!+\! y &\!=\! 5\\3x\! +\! y\! -\! 5 &\!=\! 0\end{align}

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

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