# Ex.7.1 Q9 Coordinate Geometry Solution - NCERT Maths Class 10

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## Question

If $$Q \;(0, 1)$$ is equidistant from $$P \;(5, -3)$$ and $$R\; (x, 6)$$, find the values of $$x.$$ Also find the distances  $$QR$$ and $$PR$$.

Video Solution
Coordinate Geometry
Ex 7.1 | Question 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( {{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 $$P$$, $$Q$$ and $$R$$ between which the distance is to be measured.

What is the unknown?

The value of $$x$$ and the distance $$QR$$ and $$PR$$.

Steps:

Given,

Since $$Q\; (0, 1)$$ is equidistant from $$P\; (5, -3)$$and $$R \;(x, 6)$$,\begin{align}PQ = QR\end{align}

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}

Hence by applying the distance formula for the \begin{align}PQ = QR,\end{align}we get

\begin{align}\!\sqrt {{{(5\! -\! 0)}^2}\! +\! {{( - 3\! - \!1)}^2}} \!&=\! \sqrt {{{(0 - x)}^2} + {{(1 - 6)}^2}} \\\sqrt {{{(5)}^2} + {{( - 4)}^2}} &=\! \sqrt {{{( - x)}^2} + {{( - 5)}^2}} \end{align}

By squaring both the sides,

\begin{align}25 + 16 &= {x^2} + 25\\\;\;16 &= {x^2}\\\;\;x &= \pm 4\end{align}

Therefore, point $$R$$ is $$(4, 6)$$ or $$(-4, 6)$$.

Case (1),

When point R is $$(4, 6)$$,

Distance between $$P \;(5, -3)$$ and $$R\; (4, 6)$$ can be calculated using the Distance Formula as ,

\begin{align}PR &= \sqrt {{{(5 - 4)}^2} + {{( - 3 - 6)}^2}} \\ &= \sqrt {{1^2} + {{( - 9)}^2}} \\ &= \sqrt {1 + 81} \\ &= \sqrt {82}\end{align}

Distance between $$Q\; (0, 1)$$ and $$R\; (4, 6)$$ can be calculated using the Distance Formula as ,

\begin{align}QR &= \sqrt {{{(0 - 4)}^2} + {{(1 - 6)}^2}} \\ &= \sqrt {{{( - 4)}^2} + {{( - 5)}^2}} \\ &= \sqrt {16 + 25} \\& = \sqrt {41}\end{align}

Case (2),

When point R is $$(-4, 6)$$,

Distance between $$P\; (5, -3)$$ and $$R\; (-4, 6)$$ can be calculated using the Distance Formula as,

\begin{align}PR& = \sqrt {{{(5 - ( - 4))}^2} + {{( - 3 - 6)}^2}} \\& = \sqrt {{{(9)}^2} + {{( - 9)}^2}} \\ &= \sqrt {81 + 81} \\& = 9\sqrt 2 \end{align}

Distance between $$Q \;(0, 1)$$ and $$R\; (-4, 6)$$ can be calculated using the Distance Formula as ,

\begin{align}QR& = \sqrt {{{(0 - ( - 4))}^2} + {{(1 - 6)}^2}} \\& = \sqrt {{{(4)}^2} + {{( - 5)}^2}} \\ &= \sqrt {16 + 25} \\&= \sqrt {41} \end{align}

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