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}`

∴ 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}`

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

Clearly, AC ≠ BD.

So, ABCD is not a square.

Type IV: On finding the unknown vertex from given points