Question

Find the areas of the triangles the coordinates of whose angular points are$(-a,\frac{\pi }{6}),(a,\frac{\pi }{2})$, and $(-2a,-\frac{2\pi }{2})$

Answer 1

Given polar co-ordinates
$\left(i\right)$ $(-a,\frac{\pi }{6})$

Then, $r=-a$ and $\theta =\frac{\pi }{6}$

$x=r\cos \theta =-a\times \cos \left(\frac{\pi }{6}\right)=\frac{-\sqrt{3}a}{2}$

$y=r\sin \theta =-a\times \sin \left(\frac{\pi }{6}\right)=\frac{-a}{2}$

$A(\frac{-\sqrt{3}a}{2},\frac{-a}{2})$

$\left(ii\right)$ $(-2a,-\pi )$

Then, $r=-2a$ and $\theta =-\pi$

$x=r\cos \theta =-2a\times \cos (-\pi )=2a$

$y=r\sin \theta =-2a\times \sin (-\pi )=0$

$B(2a,0)$

$\left(iii\right)$ $(a,\frac{\pi }{2})$

Then, $r=a$ and $\theta =\frac{\pi }{2}$

$x=r\cos \theta =a\times \cos \left(\frac{\pi }{2}\right)=0$

$y=r\sin \theta =a\times \sin \left(\frac{\pi }{2}\right)=a$

$C(0,a)$

Now we have points $A(\frac{-\sqrt{3}a}{2},\frac{-a}{2})$, $B(2a,0)$ and $C(0,a)$

Area of triangle having angular points $(x_1,y_1)$ $(x_2,y_2)$ $(x_3,y_3)$ is given by the formula

$△=\frac{1}{2}|x_1{y}_2+x_2{y}_3+x_3{y}_1-x_2{y}_1-x_3{y}_2-x_1{y}_3|$

Then, $A\left(△ABC\right)=\frac{1}{2}|0+2a^2+0-(-a^2)-0+\frac{\sqrt{3}{a}^2}{2}|$

$\Rightarrow A\left(△ABC\right)=a^2\left(\frac{6+\sqrt{3}}{4}\right)$