Coordinate Geometry | Exercise 7.1

Question 5

In a classroom, 4 friends are seated at the points $A$ , $B$ , $C$ and $D$ as shown in Fig. 7.8. 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.

Solution

Understand the Question

The seating coordinates of the four friends from the classroom grid are $A(3, 4)$ , $B(6, 7)$ , $C(9, 4)$ , and $D(6, 1)$ .
To determine if quadrilateral $ABCD$ is a square, we use the distance formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$
A quadrilateral is a square if:
1. All four sides are equal: $AB = BC = CD = DA$
2. Both diagonals are equal: $AC = BD$

Step 1 · Identify the Coordinates

From the given classroom grid, the coordinates of the four points are: $A(3, 4), \quad B(6, 7), \quad C(9, 4), \quad D(6, 1)$ Diagram 1

Step 2 · Calculate the Lengths of the Four Sides

Using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ :

\begin{aligned} AB &= \sqrt{(6 - 3)^2 + (7 - 4)^2} \\[0.6em] &= \sqrt{(3)^2 + (3)^2} \\[0.6em] &= \sqrt{9 + 9} \\[0.6em] &= \sqrt{18} \end{aligned}

\begin{aligned} BC &= \sqrt{(9 - 6)^2 + (4 - 7)^2} \\[0.6em] &= \sqrt{(3)^2 + (-3)^2} \\[0.6em] &= \sqrt{9 + 9} \\[0.6em] &= \sqrt{18} \end{aligned}

\begin{aligned} CD &= \sqrt{(6 - 9)^2 + (1 - 4)^2} \\[0.6em] &= \sqrt{(-3)^2 + (-3)^2} \\[0.6em] &= \sqrt{9 + 9} \\[0.6em] &= \sqrt{18} \end{aligned}

\begin{aligned} DA &= \sqrt{(3 - 6)^2 + (4 - 1)^2} \\[0.6em] &= \sqrt{(-3)^2 + (3)^2} \\[0.6em] &= \sqrt{9 + 9} \\[0.6em] &= \sqrt{18} \end{aligned}

Since $AB = BC = CD = DA = \sqrt{18}$ , all four sides are equal.

Step 3 · Calculate Diagonal Lengths and Conclude

Calculate the diagonals $AC$ and $BD$ :

\begin{aligned} AC &= \sqrt{(9 - 3)^2 + (4 - 4)^2} \\[0.6em] &= \sqrt{(6)^2 + (0)^2} \\[0.6em] &= \sqrt{36 + 0} \\[0.6em] &= \sqrt{36} \\[0.6em] &= 6 \end{aligned}

\begin{aligned} BD &= \sqrt{(6 - 6)^2 + (1 - 7)^2} \\[0.6em] &= \sqrt{(0)^2 + (-6)^2} \\[0.6em] &= \sqrt{0 + 36} \\[0.6em] &= \sqrt{36} \\[0.6em] &= 6 \end{aligned}

Since all four sides are equal ( $AB = BC = CD = DA = \sqrt{18}$ ) and both diagonals are equal ( $AC = BD = 6$ ), $ABCD$ is a square.

Answer

Champa is correct ( $ABCD$ is a square).

Common Mistakes

Checking Only Sides: Showing that all four sides are equal proves $ABCD$ is a rhombus, but it could still fail to be a square. You must also check that the diagonals are equal ( $AC = BD$ ).
Swapping $x$ and $y$ Coordinates: Reading the grid row first instead of column first (e.g., writing $A(4, 3)$ instead of $A(3, 4)$ ).
Sign Error When Squaring: Forgetting that negative numbers squared are positive, e.g. $(-3)^2 = 9$ , not $-9$ .

Question 5

More questions in Exercise 7.1