Question

Prove that (4,-1), (6,0), (7,2) and (5,1) are the verices of a rhombus. Is it a square ?

Answer 1

Let the given points be A, B, C and D 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,

$AB=\sqrt{(6-4{)}^2+(0+1{)}^2}=\sqrt{5}$

$BC=\sqrt{(7-6{)}^2+(2-0{)}^2}=\sqrt{5}$

Therefore,

$AB=BC$

So, ABCD is a parallelogram whose adjacent sides are equal.

Hence, ABCD is a rhombus.

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

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

$AC\ne BD$

So, ABCD is not a square.