Question

Name the type of quadrilateral formed by the points $(4,5),(7,6),(4,3),(1,2)$.

Answer 1

Say, Coordinates of the points $A=(4,5),B=(7,6),C=(4,3),D=(1,2)$
To determine the type of quadrilateral let's evaluate the length of sides of the quadrilateral whose coordinates are given using the distance formula.
We know that,
$d=\sqrt{(x_1-x_2{)}^2-(y_1-y_2{)}^2}$
$AB=\sqrt{(4-7{)}^2+(5-6{)}^2}=\sqrt{(-3{)}^2+(-1{)}^2}AB=\sqrt{9+1}=\sqrt{10}\text{unit}$
$BC=\sqrt{(7-4{)}^2+(6-3{)}^2}=\sqrt{3^2+3^2}BC=\sqrt{9+9}=\sqrt{18}=3\sqrt{2}\text{unit}$
$CD=\sqrt{(4-1{)}^2+(3-2{)}^2}=\sqrt{3^2+1^2}CD=\sqrt{9+1}=\sqrt{10}\text{unit}$
$AD=\sqrt{(4-1{)}^2+(5-2{)}^2}=\sqrt{3^2+3^2}AD=\sqrt{9+9}=\sqrt{18}=3\sqrt{2}\text{unit}$

Here, we observe
$AB=CD=\sqrt{10}\text{unit}AD=BC=3\sqrt{2}\text{unit}$
This means the quadrilateral may be rectangle or parallelogram. Lets evaluate the length of diagonal.
$AC=\sqrt{(4-4{)}^2+(5-3{)}^2}=\sqrt{0+2^2}AD=\sqrt{4}=3\sqrt{2}\text{unit}$
$BD=\sqrt{(7-1{)}^2+(6-2{)}^2}=\sqrt{6^2+4^2}BD=\sqrt{36+16}=\sqrt{52}=2\sqrt{13}\text{unit}$

Since diagonals are unequal.

Therefore, the quadrilateral would be a parallelogram only.