Question

Find the equation of the line which is passing through $(2,2\sqrt{3})$ and inclined with the $x$-axis at an angle of ${75}^{\circ }$.

Answer 1

We know that equation of line passing through point $(x_0,y_0)$ with slope $m$ is $(y-y_0)=m(x-x_0)$

Given :
$(x_0,y_0)=(2,2\sqrt{3})$ and $\theta ={75}^{\circ }$
slope $=m=\tan \theta$
$\therefore m=\tan \left({75}^{\circ }\right)=\frac{\sqrt{3}+1}{\sqrt{3}-1}$
Now, required equation is
$(y-2\sqrt{3})=\frac{\sqrt{3}+1}{\sqrt{3}-1}(x-2)$
$\Rightarrow (y-2\sqrt{3})(\sqrt{3}-1)=(\sqrt{3}+1)(x-2)$
$\Rightarrow y(\sqrt{3}-1)-2\sqrt{3}(\sqrt{3}-1)=x(\sqrt{3}+1)-2(\sqrt{3}+1)$
$\Rightarrow y(\sqrt{3}-1)-2\sqrt{3}\times \sqrt{3}+2\sqrt{3}=x(\sqrt{3}+1)-2\sqrt{3}-2$
$\Rightarrow y(\sqrt{3}-1)-6+2\sqrt{3}=x(\sqrt{3}+1)-2\sqrt{3}-2$
$\Rightarrow y(\sqrt{3}-1)=x(\sqrt{3}+1)-2\sqrt{3}-2+6-2\sqrt{3}$
$\Rightarrow y(\sqrt{3}-1)=x(\sqrt{3}+1)-4\sqrt{3}+4$
$\Rightarrow y(\sqrt{3}-1)-x(\sqrt{3}+1)=-4\sqrt{3}+4$
$\Rightarrow y(\sqrt{3}-1)-x(\sqrt{3}+1)=4(-\sqrt{3}+1)$
Multiply both sides with $(-1)$
$x(\sqrt{3}+1)-y(\sqrt{3}-1)=4(\sqrt{3}-1)$
Hence, the rquired equation is $x(\sqrt{3}+1)-y(\sqrt{3}-1)=4(\sqrt{3}-1)$