Formula: 1 Mark
Application: 1 Mark
Answer: 1 Mark
Let the given points be A(4, – 1), B(6, 0), C(7, 2) and D(5, 1) respectively. Then,
Coordinates of the mid-point of AC are
$(\frac{4+7}{2},\frac{-1+2}{2})=(\frac{11}{2},\frac{1}{2})$ 
Coordinates of the mid-point of BD are
$(\frac{6+5}{2},\frac{0+1}{2})=(\frac{11}{2},\frac{1}{2})$ 
Thus, AC and BD have the same mid-point.

Hence, ABCD is a parallelogram.

Now
Distance between the points is given by
$\sqrt{{(x_1-x_2)}^2+{(y_1-y_2)}^2}$ 
So, 
AB = $\sqrt{(6-4{)}^2+(0+1{)}^2}=\sqrt{5}$ 
BC = $\sqrt{(7-6{)}^2+(2-0{)}^2}=\sqrt{5}$ 
$\therefore$ AB = BC 
So, ABCD is a parallelogram whose adjacent sides are equal.
$\Rightarrow$  ABCD is a rhombus.
We have, 

AC = $\sqrt{(7-4{)}^2+(2+1{)}^2}=3\sqrt{2}$   and , 

BD = $\sqrt{(6-5{)}^2+(0-1{)}^2}=\sqrt{2}$ 

 
Clearly, the diagonals AC ≠ BD.
So, ABCD is not a square.