# Ex.7.1 Q5 Coordinate Geometry Solution - NCERT Maths Class 10

## Question

In a classroom, \(4\) friends are seated at the points \(A\), \(B\), \(C\) and \(D\) as shown in the following figure. Champa and Chameli walk into the class and after observing for a few minutes Champa asks Chameli, “Don’t you think \(ABCD\) is a square?” Chameli disagrees. Using distance formula, find which of them is correct.

## Text Solution

**Reasoning:**

To prove that the points \(A\),\(B\),\(C\) and \(D\) from a square, the length of the four sides should be equal and the length of the two diagonals should be the same.

**What is Known?**

The \(x\) and \(y\) co-ordinates of the points between which the distance is to be measured can be deduced from the diagram.

**What is Unknown?**

To verify whether the positions of the four friends form a square or not.

**Steps:**

Let \(A \;(3, 4)\), \(B\; (6, 7)\), \(C \;(9, 4)\), and \(D \;(6, 1)\) be the positions of \(4\) friends.

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

\[\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}} \;\;\dots(1)\]

To find \(AB\) i.e. Distance between Points \(A \;(3, 4)\) and \(B\; (6, 7)\)

- \(x_1 = 3\)
- \(y_1 = 4\)
- \(x_2 = 6\)
- \(y_2 =7\)

By substituting the values in the Equation (1), we get

\[\begin{align}AB &= \sqrt {{{(3 - 6)}^2} + {{(4 - 7)}^2}} \\ &= \sqrt {{{( - 3)}^2} + {{( - 3)}^2}} \\ &= \sqrt {9 + 9} \\ &= \sqrt {18} \\ &= 3\sqrt 2 \end{align}\]

To find \(BC\) i.e. Distance between Points \(B (6, 7)\) and \(C (9, 4)\)

- \(x_1 = 6\)
- \(y_1 = 7\)
- \(x_2 = 9\)
- \(y_2 = 4\)

By substituting the values in the Equation (1), we get

\[\begin{align}BC& = \sqrt {{{(6 - 9)}^2} + {{(7 - 4)}^2}} \\ &= \sqrt {{{( - 3)}^2} + {{(3)}^2}} \\ &= \sqrt {9 + 9} \\& = \sqrt {18} \\ &= 3\sqrt 2 \end{align}\]

To find \(CD\) i.e. Distance between Points \(C \; (9, 4)\) and \(D\; (6, 1)\)

- \(x_1 = 9\)
- \(y_1 = 4\)
- \(x_2 = 6\)
- \(y_2 = 1\)

By substituting the values in the Equation (1)

\[\begin{align}CB& = \sqrt {{{(9 - 6)}^2} + {{(4 - 1)}^2}} \\ &= \sqrt {{{(3)}^2} + {{(3)}^2}} \\ &= \sqrt {9 + 9} \\ &= \sqrt {18} \\ &= 3\sqrt 2 \end{align}\]

To find \(AD\) i.e. Distance between Points \(B\; (3, 4)\) and \(D \;(6, 1)\)

- \(x_1 = 3\)
- \(y_1 = 4\)
- \(x_2 = 6\)
- \(y_2 = 1\)

By substituting the values in the Equation (1)

\[\begin{align}AD &= \sqrt {{{(3 - 6)}^2} + {{(4 - 1)}^2}} \\ &= \sqrt {{{( - 3)}^2} + {{(3)}^2}} \\ &= \sqrt {9 + 9} \\&= \sqrt {18} \\ &= 3\sqrt 2 \end{align}\]

To find \(AC\) i.e. Distance between Points \(A \;(3, 4)\) and \(C\; (9, 4)\)

- \(x_1 = 3\)
- \(y_1 = 4\)
- \(x_2 = 9\)
- \(y_2 = 4\)

By substituting the values in the Equation (1), we get

\(\begin{align}{\text{Diagonal}}\,AC &\!=\! \sqrt {{{(3\! - \!9)}^2} \!+ \!{{(4\! - \!4)}^2}} \\ &= \sqrt {{{( - 6)}^2} + {0^2}} \\ &= 6\end{align}\)

To find \(BD\) Distance between Points \(B \;(6, 7)\) and \(D \;(6, 1)\)

- \(x_1 = 6\)
- \(y_1 = 7\)
- \(x_2 = 6\)
- \(y_2 = 1\)

By substituting the values in the Equation (1)

\(\begin{align}\text{Diagonal} \,BD &\!=\! \sqrt {{{\!(6 \!- \!6)}^2} \!+\! {{(7\! -\! 1)}^2}} \\ &= \sqrt {{0^2} + {{( - 6)}^2}} \\ &= 6\end{align}\)

The four sides \(AB\), \(BC\), \(CD\), and \(AD\) are of same length and diagonals \(AC\) and \(BD\) are of equal length. Therefore, \(ABCD\) is a square and hence, Champa was correct