Prove that the points (3, 0), (4, 5), (−1, 4) and (−2 −1),
Question:

Prove that the points (3, 0), (4, 5), (−1, 4) and (−2 −1), taken in order, form a rhombus. Also, find its area.

Solution:

The distance d between two points  and  is given by the formula

$d=\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$

In a rhombus all the sides are equal in length. And the area ‘A’ of a rhombus is given as

$A=\frac{1}{2}$ (Product of both diagonals)

Here the four points are A(3,0), B(4,5), C(1,4) and D(2,1).

First let us check if all the four sides are equal.

$A B=\sqrt{(3-4)^{2}+(0-5)^{2}}$

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

 

$=\sqrt{1+25}$

$A B=\sqrt{26}$

$B C=\sqrt{(4+1)^{2}+(5-4)^{2}}$

$=\sqrt{(1)^{2}+(5)^{2}}$

 

$=\sqrt{1+25}$

$C D=\sqrt{26}$

$A D=\sqrt{(3+2)^{2}+(0+1)^{2}}$

$=\sqrt{(5)^{2}+(1)^{2}}$

 

$=\sqrt{25+1}$

$A D=\sqrt{26}$

Here, we see that all the sides are equal, so it has to be a rhombus.

Hence we have proved that the quadrilateral formed by the given four vertices is a.

Now let us find out the lengths of the diagonals of the rhombus.

$A C=\sqrt{(3+1)^{2}+(0-4)^{2}}$

$=\sqrt{(4)^{2}+(-4)^{2}}$

 

$=\sqrt{16+16}$

$A C=4 \sqrt{2}$

$B D=\sqrt{(4+2)^{2}+(5+1)^{2}}$

$=\sqrt{(6)^{2}+(6)^{2}}$

 

$=\sqrt{36+36}$

$B D=6 \sqrt{2}$

Now using these values in the formula for the area of a rhombus we have,

$A=\frac{(6 \sqrt{2})(4 \sqrt{2})}{2}$

$=\frac{(6)(4)(2)}{2}$

$A=24$

Thus the area of the given rhombus is 24 square units

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