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Q6(iii):
Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer: (iii) (4, 5), (7, 6), (4, 3), (1, 2)

Solution :

Given: The coordinates of the four points are $A(4, 5)$, $B(7, 6)$, $C(4, 3)$, and $D(1, 2)$.

To Find: The type of quadrilateral formed by these points, if any, and the reasons for the classification.

A(4,5) B(7,6) C(4,3) D(1,2)

Step 1: Formula Definition
To determine the nature of the quadrilateral, we use the Distance Formula to find the lengths of the four sides ($AB, BC, CD, DA$) and the two diagonals ($AC, BD$).
The Distance Formula between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by:
$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Step 2: Calculating Side Lengths
- Length of AB: $\sqrt{(7 - 4)^2 + (6 - 5)^2} = \sqrt{3^2 + 1^2} = \sqrt{9 + 1} = \sqrt{10}$ units.
- Length of BC: $\sqrt{(4 - 7)^2 + (3 - 6)^2} = \sqrt{(-3)^2 + (-3)^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}$ units.
- Length of CD: $\sqrt{(1 - 4)^2 + (2 - 3)^2} = \sqrt{(-3)^2 + (-1)^2} = \sqrt{9 + 1} = \sqrt{10}$ units.
- Length of DA: $\sqrt{(4 - 1)^2 + (5 - 2)^2} = \sqrt{3^2 + 3^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}$ units.

Step 3: Calculating Diagonal Lengths
- Length of AC: $\sqrt{(4 - 4)^2 + (3 - 5)^2} = \sqrt{0^2 + (-2)^2} = \sqrt{4} = 2$ units.
- Length of BD: $\sqrt{(1 - 7)^2 + (2 - 6)^2} = \sqrt{(-6)^2 + (-4)^2} = \sqrt{36 + 16} = \sqrt{52} = 2\sqrt{13}$ units.

Step 4: Analysis and Conclusion
From the calculations:
1. Opposite sides are equal: $AB = CD = \sqrt{10}$ and $BC = DA = 3\sqrt{2}$.
2. Diagonals are unequal: $AC = 2$ and $BD = \sqrt{52}$.
Since the opposite sides are equal and the diagonals are not equal, the quadrilateral satisfies the properties of a parallelogram.

Final Answer: The points form a parallelogram.


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