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.

Answer 1

Let the point A(3,2), B(-5,-5), C(2,-3) and D(4,4) be the vertices of a rhombus ABCD.

We know that all the sides of a rhombus ABCD are equal.

So by distance formula,

Distance between two points = $\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$

$AB=\sqrt{(-5+3{)}^2+(-5-2{)}^2}=\sqrt{4+49}=\sqrt{53}$.

$BC=\sqrt{(2+5{)}^2+(-3+5{)}^2}=\sqrt{49+4}=\sqrt{53}$.

$CD=\sqrt{(4-2{)}^2+(4+3{)}^2}=\sqrt{4+49}=\sqrt{53}$.

$DA=\sqrt{(4+3{)}^2+(4-2{)}^2}=\sqrt{49+4}=\sqrt{53}$.

$\therefore AB=BC=CD=AD$

Hence, ABCD is a rhombus (Proved)

Area of the rhombus ABCD

$=\frac{1}{2}\mid (x_1{y}_2+x_2{y}_3+x_3{y}_4+x_4{y}_1)-(x_2{y}_1+x_3{y}_2+x_4{y}_3+x_1{y}_4)\mid$

$=\frac{1}{2}\mid (15+15+8+8)-(-10-10-12-12)$

$=\frac{1}{2}\mid 46+44\mid =\frac{1}{2}\mid 90\mid =45$ sq. units