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

(A) Let the points $A(4,5), B(7,6), C(4,3),$ and $D(1,2)$ represent the vertices of the quadrilateral.
Using the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$:
$AB = \sqrt{(7-4)^2 + (6-5)^2} = \sqrt{3^2 + 1^2} = \sqrt{9+1} = \sqrt{10}$
$BC = \sqrt{(4-7)^2 + (3-6)^2} = \sqrt{(-3)^2 + (-3)^2} = \sqrt{9+9} = \sqrt{18} = 3\sqrt{2}$
$CD = \sqrt{(1-4)^2 + (2-3)^2} = \sqrt{(-3)^2 + (-1)^2} = \sqrt{9+1} = \sqrt{10}$
$DA = \sqrt{(4-1)^2 + (5-2)^2} = \sqrt{3^2 + 3^2} = \sqrt{9+9} = \sqrt{18} = 3\sqrt{2}$
Diagonal $AC = \sqrt{(4-4)^2 + (3-5)^2} = \sqrt{0^2 + (-2)^2} = \sqrt{4} = 2$
Diagonal $BD = \sqrt{(1-7)^2 + (2-6)^2} = \sqrt{(-6)^2 + (-4)^2} = \sqrt{36+16} = \sqrt{52} = 2\sqrt{13}$
Since opposite sides are equal ($AB=CD$ and $BC=DA$) and diagonals are unequal $(AC \neq BD)$,the quadrilateral is a parallelogram.