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.

Answer 1

From the figure co-ordinates of points can be expressed as shown below,

$$A(3 , 4) , B (6 , 7) , C (9 , 4) , D(6 , 1) $$
$$ AB = \sqrt{(6 - 3)^{2} + (7 - 4)^{2}} \\= \sqrt{3^{2} + 3^{2}} $$
$$ = \sqrt{9 + 9} \\= \sqrt{18} \\= 3\sqrt{2} $$
$$BC = \sqrt{(9 - 6)^{2} + (6 - 7)^{2}} \\= \sqrt{3^{2} + (-3)^{2}} $$
$$ = \sqrt{9 + 9} \\= \sqrt{18} \\= 3 \sqrt{2} $$
$$CD = \sqrt{(6 - 9)^{2} + (1 - 4)^{2}} \\= \sqrt{(-3)^{2} + (-3)^{2}} $$
$$ = \sqrt{9 + 9}\\ = \sqrt{18} \\= 3 \sqrt{2} $$
$$DA = \sqrt{(6 - 3)^{2} + (1 - 4)^{2}} \\= \sqrt{3^{2} + (-3)^{2}} $$
$$ = \sqrt{ 9 + 9} \\= \sqrt{18} \\= 3 \sqrt{2} $$

Four sides of the quadrilateral $$ABCD$$ are equal

$$AB = BC = CD = DA = 3 \sqrt{2} $$

$$Diagonal \space AC=\sqrt{(9-3)^2+(4-4)^2}=6$$

$$Diagonal \space BD=\sqrt{(6-6)^2+(1-7)^2}=6$$

$$Diagonal \space AC = Diagonal \space BD $$

Therefore $$ABCD$$ is a square

Therefore, Champa is correct between two