# 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\(\begin{align} = \sqrt {{{\left( {{{\text{x}}_{\text{1}}} - {{\text{x}}_{\text{2}}}} \right)}^2} + {{\left( {{{\text{y}}_{\text{1}}} - {{\text{y}}_{\text{2}}}} \right)}^2}} .\end{align}\)

**What is the known?**

The \(x\) and \(y\) co-ordinates of the points \(P\), \(Q\) 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,

\[\begin{align}\sqrt {{{\left( {{{\text{x}}_1} - {{\text{x}}_2}} \right)}^2} + {{\left( {{{\text{y}}_1} - {{\text{y}}_2}} \right)}^2}} \qquad \qquad...\;{\text{Equation}}\,(1)\end{align}\]

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,

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

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 ,

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