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 \(PR\).

 

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 ,

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