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

## 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 \(PR\).

## 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 ,

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