Question

Prove that $(4,-1),(6,0),(7,2)$ and $(5,1)$ are the vertices 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.

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, $AC\ne BD$.So, ABCD is not a square.