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Ex.7.4 Q8 Coordinate Geometry Solution - NCERT Maths Class 10

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Question

\(ABCD\) is a rectangle formed by the points \(A \;(- 1, - 1)\), \(B\; (- 1, 4)\), \(C \;(5, 4)\) and \(D\; (5, - 1)\). \(P\), \(Q\), \(R\) and \(S\) are the mid-points of \(AB\), \(BC\), \(CD\), and \(DA\) respectively. Is the quadrilateral \(PQRS\) is a square? A rectangle? Or a rhombus? Justify your answer.

 Video Solution
Coordinate Geometry
Ex 7.4 | Question 8

Text Solution

Reasoning:

The distance between the two points can be measured using the Distance Formula which is given by:

Distance Formula

\[\begin{align}= \sqrt {{{\left( {{x_1} - {x_2}} \right)}^2} + {{\left( {{y_1} - {y_2}} \right)}^2}} .\end{align}\]

What is known?

The \(x\) and \(y\) coordinates of the vertices of the rectangle.

What is unknown?

To find whether quadrilateral \(PQRS\) is a square? A rectangle? Or a rhombus?

Steps:

From the figure below,

\(P\) is the mid-point of side \(AB\). The co-ordinates of \(P\) can be calculated using the Mid-Point Formula as

\[\begin{align}M = \left( {\frac{{{x_1} + {x_2}}}{2},\frac{{{y_1} + {y_2}}}{2}} \right)\end{align}\]

Therefore, the coordinates of \(P\) are

\[\begin{align}\left[ {\frac{{ - 1 - 1}}{2},\,\frac{{ - 1 + 4}}{2}} \right) = \left( { - 1,\,\,\frac{3}{2}} \right] \end{align}\]

Similarly, the coordinates of \(Q\), \(R\) and \(S\) are calculated using the Mid-Point Formula as :

\((2,4),\left( {5,\frac{3}{2}} \right)\,{\text{and}}\,(2, - 1)\) respectively.

We know that the distance between the two points is given by the Distance Formula,

\[\begin{align} & \sqrt {{{\left( {{x_1} - {x_2}} \right)}^2} + {{\left( {{y_1} - {y_2}} \right)}^2}}  \ldots  (1) \end{align}\]

Distance between two points \(P\) and \(Q\),

Length of \(PQ\)

\[\begin{align} &= \sqrt {{{( - 1 - 2)}^2} + {{\left( {\frac{3}{2} - 4} \right)}^2}} \\ &= \sqrt {9 + \frac{{25}}{4}} \\& = \sqrt {\frac{{61}}{4}} \end{align}\]

Distance between two points \(Q\) and \(R\),

Length of \(QR\)

\[\begin{align} &= \sqrt {{{(2 - 5)}^2} + {{\left( {4 - \frac{3}{2}} \right)}^2}} \\ &= \sqrt {9 + \frac{{25}}{4}} \\ &= \sqrt {\frac{{61}}{4}} \end{align}\]

Distance between two points \(R\) and \(S\),

Length of \(RS\)

\[\begin{align} &= \sqrt {{{(5 - 2)}^2} + {{\left( {\frac{3}{2} + 1} \right)}^2}} \\ &= \sqrt {9 + \frac{{25}}{4}} \\ &= \sqrt {\frac{{61}}{4}} \end{align}\]

Distance between two points \(S\) and \(P\),

Length of \(SP\)

\[\begin{align} &= \sqrt {{{(2 + 1)}^2} + {{\left( { - 1 - \frac{3}{2}} \right)}^2}} \\& = \sqrt {9 + \frac{{25}}{4}} \\ &= \sqrt {\frac{{61}}{4}}\end{align}\]

Distance between two points \(P\) and \(R\) which form the diagonal are ,

Length of \(PR\)

\[\begin{align} &= \sqrt {{{( - 1 - 5)}^2} + \left( {\frac{3}{2}} \right) - \frac{{{3^2}}}{2}} \\ &= 6\end{align}\]

Distance between two points \(Q\) and \(S\) which form the diagonal is calculated using the Distance Formula as,

Length of \(QS\)

\[\begin{align}{\text{ }}&= \,\sqrt {{{(2 - 2)}^2} + {{(4 + 1)}^2}} \\ &= 5\end{align}\]

It can be observed that all sides of the given quadrilateral are of the same measure. However, the diagonals are of different lengths. Therefore, \(PQRS\) is a rhombus.

  
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