Question

Show that the points $(-4,-7)$, $(-1,2)$ $(8,5)$ and $(5,4)$ taken in order are the vertices of a rhombus. Also find its area.

Answer 1

Let the points are $A(-4,-7),B(-1,2),C(8,5),D(5,-4)$

AB$=\sqrt{(-1-(-4){)}^2+(2-(-7){)}^2}$

$=\sqrt{3^2+9^2}$

$=\sqrt{90}$

$=3\sqrt{10}$

BC$=\sqrt{(8-(-1){)}^2+(5-2{)}^2}$

$=\sqrt{9^2+3^2}$

$=\sqrt{90}$

$=3\sqrt{10}$

CD$=\sqrt{(5-8{)}^2+(-4-\left(5\right){)}^2}$

$=\sqrt{3^2+9^2}$

$=\sqrt{90}$

$=3\sqrt{10}$

DA$=\sqrt{(-4-(5){)}^2+(-7-(-4){)}^2}$

$=\sqrt{9^2+3^2}$

$=\sqrt{90}$

$=3\sqrt{10}$

now ,

AC$=\sqrt{(8-(-4){)}^2+(5-(-7){)}^2}$

$=\sqrt{(12{)}^2+(12{)}^2}$

$=\sqrt{144}$

$=12\sqrt{2}$

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

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

$=\sqrt{72}$

$=6\sqrt{2}$

Since , $AB=CD=BC=DA,AC\ne BD$

so, it is rhombus.

Area of rhombus $=\frac{1}{2}\times 12\sqrt{2}\times 6\sqrt{2}=72$