Name the type of quadrilateral formed by the following points and give reasons for your answer :

(4, 5), (7, 6), (4, 3), (1, 2)

Let A(4, 5), B(7, 6), C(4, 3), D(1, 2) are the given points.

By distance formula,

$d = \sqrt{(x$2 - x1)^2 + (y2 - y1)^2} \\[1em] \therefore AB = \sqrt{(7 - 4)^2 + (6 - 5)^2} \\[1em] = \sqrt{3^2 + 1^2} \\[1em] = \sqrt{9 + 1} \\[1em] = \sqrt{10} \text{ units}. \\[1em] BC = \sqrt{(4 - 7)^2 + (3 - 6)^2} \\[1em] = \sqrt{(-3)^2 + (-3)^2} \\[1em] = \sqrt{9 + 9} \\[1em] = \sqrt{18} \\[1em] = 2\sqrt{2} \text{ units}. \\[1em] CD = \sqrt{(1 - 4)^2 + (2 - 3)^2} \\[1em] = \sqrt{(-3)^2 + (-1)^2} \\[1em] = \sqrt{9 + 1} \\[1em] = \sqrt{10} \text{ units}. \\[1em] AD = \sqrt{(1 - 4)^2 + (2 - 5)^2} \\[1em] = \sqrt{3^2 + (-3)^2} \\[1em] = \sqrt{9 + 9} \\[1em] = \sqrt{18} \\[1em] = 2\sqrt{2} \text{ units}.

Calculating diagonals :

$AC = \sqrt{(4 - 4)^2 + (3 - 5)^2} \\[1em] = \sqrt{0 + (-2)^2} \\[1em] = \sqrt{4} \\[1em] = 2 \text{ units}. \\[1em] BD = \sqrt{(1 - 7)^2 + (2 - 6)^2}\\[1em] = \sqrt{(-6)^2 + (-4)^2} \\[1em] = \sqrt{36 + 16} \\[1em] = \sqrt{52} \text{ units}.$

Since, opposite sides are equal and diagonals are not equal.

Hence, (4, 5), (7, 6), (4, 3), (1, 2) are the vertices of a parallelogram.