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

## 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.

## 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 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 \(\begin{align}M = \left( {\frac{{{{\text{x}}_{\text{1}}} + {{\text{x}}_{\text{2}}}}}{{\text{2}}}{\text{,}}\frac{{{{\text{y}}_{\text{1}}} + {{\text{y}}_{\text{2}}}}}{{\text{2}}}} \right)\end{align}\)

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

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

\(\begin{align}\sqrt {{{\left( {{{\text{x}}_1} - {{\text{x}}_2}} \right)}^2} + {{\left( {{{\text{y}}_1} - {{\text{y}}_2}} \right)}^2}} & & \ldots \ldots \; \end{align}\) Equation (1)

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.