Show that the points (−3, 2), (−5,−5), (2, −3) and

Question:

Show that the points (−3, 2), (−5,−5), (2, −3) and (4, 4) are the vertices of a rhombus. Find the area of this rhombus.

Solution:

The distance $d$ between two points $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ 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(3,2), B(5,5), C(2,3) and D(4,4)

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

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

$=\sqrt{(2)^{2}+(7)^{2}}$

$=\sqrt{4+49}$

$A B=\sqrt{53}$

$B C=\sqrt{(-5-2)^{2}+(-5+3)^{2}}$

$=\sqrt{(-7)^{2}+(-2)^{2}}$

$=\sqrt{49+4}$

$B C=\sqrt{53}$

$C D=\sqrt{(2-4)^{2}+(-3-4)^{2}}$

$=\sqrt{(-2)^{2}+(-7)^{2}}$

$=\sqrt{4+49}$

$C D=\sqrt{53}$

$A D=\sqrt{(-3-4)^{2}+(2-4)^{2}}$

$=\sqrt{(-7)^{2}+(-2)^{2}}$

$=\sqrt{49+4}$

$A D=\sqrt{53}$

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.

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

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

$=\sqrt{25+25}$

 

$=\sqrt{50}$

$A C=5 \sqrt{2}$

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

$=\sqrt{(-9)^{2}+(-9)^{2}}$

$=\sqrt{81+81}$

 

$=\sqrt{162}$

$B D=9 \sqrt{2}$

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

$A=\frac{(5 \sqrt{2})(9 \sqrt{2})}{2}$

$=\frac{(5)(9)(2)}{2}$

$A=45$

Thus the area of the given rhombus is 45 square units.

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