Question

Prove that (4, – 1), (6, 0), (7, 2) and (5, 1) are the vertices of a rhombus. Is it a square? [3 MARKS]

Answer 1

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.