Two lines passing through the point (2, 3) intersects each other at an angle of 60°.
Question:

Two lines passing through the point $(2,3)$ intersects each other at an angle of $60^{\circ}$. If slope of one line is 2 , find equation of the other line.

Solution:

It is given that the slope of the first line, $m_{1}=2$.

Let the slope of the other line be $m_{2}$.

The angle between the two lines is $60^{\circ}$.

$\therefore \tan 60^{\circ}=\left|\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\right|$

$\Rightarrow \sqrt{3}=\left|\frac{2-m_{2}}{1+2 m_{2}}\right|$

$\Rightarrow \sqrt{3}=\pm\left(\frac{2-m_{2}}{1+2 m_{2}}\right)$

$\Rightarrow \sqrt{3}=\frac{2-m_{2}}{1+2 m_{2}}$ or $\sqrt{3}=-\left(\frac{2-m_{2}}{1+2 m_{2}}\right)$

$\Rightarrow \sqrt{3}\left(1+2 m_{2}\right)=2-m_{2}$ or $\sqrt{3}\left(1+2 m_{2}\right)=-\left(2-m_{2}\right)$

$\Rightarrow \sqrt{3}+2 \sqrt{3} m_{2}+m_{2}=2$ or $\sqrt{3}+2 \sqrt{3} m_{2}-m_{2}=-2$

$\Rightarrow \sqrt{3}+(2 \sqrt{3}+1) m_{2}=2$ or $\sqrt{3}+(2 \sqrt{3}-1) m_{2}=-2$

$\Rightarrow m_{2}=\frac{2-\sqrt{3}}{(2 \sqrt{3}+1)}$ or $m_{2}=\frac{-(2+\sqrt{3})}{(2 \sqrt{3}-1)}$

Case I:

$m_{2}=\left(\frac{2-\sqrt{3}}{2 \sqrt{3}+1}\right)$

The equation of the line passing through point $(2,3)$ and having a slope of $\frac{(2-\sqrt{3})}{(2 \sqrt{3}+1)}$ is

$(y-3)=\frac{2-\sqrt{3}}{2 \sqrt{3}+1}(x-2)$

$(2 \sqrt{3}+1) y-3(2 \sqrt{3}+1)=(2-\sqrt{3}) x-2(2-\sqrt{3})$

$(\sqrt{3}-2) x+(2 \sqrt{3}+1) y=-4+2 \sqrt{3}+6 \sqrt{3}+3$

$(\sqrt{3}-2) x+(2 \sqrt{3}+1) y=-1+8 \sqrt{3}$

In this case, the equation of the other line is $(\sqrt{3}-2) x+(2 \sqrt{3}+1) y=-1+8 \sqrt{3}$.

Case II :

$m_{2}=\frac{-(2+\sqrt{3})}{(2 \sqrt{3}-1)}$

The equation of the line passing through point $(2,3)$ and having a slope of $\frac{-(2+\sqrt{3})}{(2 \sqrt{3}-1)}$ is

$(y-3)=\frac{-(2+\sqrt{3})}{(2 \sqrt{3}-1)}(x-2)$

$(2 \sqrt{3}-1) y-3(2 \sqrt{3}-1)=-(2+\sqrt{3}) x+2(2+\sqrt{3})$

$(2 \sqrt{3}-1) y+(2+\sqrt{3}) x=4+2 \sqrt{3}+6 \sqrt{3}-3$

$(2+\sqrt{3}) x+(2 \sqrt{3}-1) y=1+8 \sqrt{3}$

In this case, the equation of the other line is $(2+\sqrt{3}) x+(2 \sqrt{3}-1) y=1+8 \sqrt{3}$

Thus, the required equation of the other line is $(\sqrt{3}-2) x+(2 \sqrt{3}+1) y=-1+8 \sqrt{3}$ or $(2+\sqrt{3}) x+(2 \sqrt{3}-1) y=1+8 \sqrt{3}$.

Administrator

Leave a comment

Please enter comment.
Please enter your name.