Q6(i):

Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer: (i) (– 1, – 2), (1, 0), (– 1, 2), (– 3, 0)

Given: The coordinates of the four points forming a quadrilateral are $A(-1, -2)$, $B(1, 0)$, $C(-1, 2)$, and $D(-3, 0)$.

To Find: The type of quadrilateral formed by these points, with supporting reasons.

Step 1: Distance Formula Application
To determine the type of quadrilateral, we calculate the lengths of the four sides ($AB, BC, CD, DA$) and the two diagonals ($AC, BD$) using the distance formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.

Step 2: Calculating Side Lengths
For $A(-1, -2)$ and $B(1, 0)$:
$AB = \sqrt{(1 - (-1))^2 + (0 - (-2))^2} = \sqrt{(2)^2 + (2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}$ units.

For $B(1, 0)$ and $C(-1, 2)$:
$BC = \sqrt{(-1 - 1)^2 + (2 - 0)^2} = \sqrt{(-2)^2 + (2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}$ units.

For $C(-1, 2)$ and $D(-3, 0)$:
$CD = \sqrt{(-3 - (-1))^2 + (0 - 2)^2} = \sqrt{(-2)^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}$ units.

For $D(-3, 0)$ and $A(-1, -2)$:
$DA = \sqrt{(-1 - (-3))^2 + (-2 - 0)^2} = \sqrt{(2)^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}$ units.

Step 3: Calculating Diagonal Lengths
For $A(-1, -2)$ and $C(-1, 2)$:
$AC = \sqrt{(-1 - (-1))^2 + (2 - (-2))^2} = \sqrt{0^2 + 4^2} = \sqrt{16} = 4$ units.

For $B(1, 0)$ and $D(-3, 0)$:
$BD = \sqrt{(-3 - 1)^2 + (0 - 0)^2} = \sqrt{(-4)^2 + 0^2} = \sqrt{16} = 4$ units.

Step 4: Conclusion based on Properties
Since all four sides are equal ($AB = BC = CD = DA = 2\sqrt{2}$) and both diagonals are equal ($AC = BD = 4$), the quadrilateral satisfies the necessary and sufficient conditions for a square.

Final Answer: The quadrilateral formed is a square.