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Q : 12     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. 

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Let the slope of two lines are m_1 \ and \ m_2   respectively
It is given the lines intersects each other at an angle of  60^{\circ}  and slope of the line is 2
Now,
m_1 = m\ and \ m_2= 2 \ and \ \theta = 60\degree
\tan \theta = \left | \frac{m_2-m_1}{1+m_1m_2} \right |
\tan 60\degree = \left | \frac{2-m}{1+2m} \right |
\sqrt3 = \left | \frac{2-m}{1+2m} \right |
\sqrt3 = \frac{2-m}{1+2m} \ \ \ \ \ \ or \ \ \ \ \ \ \ \ \ \sqrt 3 = -\left ( \frac{2-m}{1+2m} \right )
m = \frac{2-\sqrt3}{2\sqrt3+1} \ \ \ \ \ \ or \ \ \ \ \ \ \ \ \ \ m = \frac{-(2+\sqrt3)}{2\sqrt3-1}
Now, the equation of line passing through point (2 ,3) and with slope  \frac{2-\sqrt3}{2\sqrt3+1}  is 
(y-3)= \frac{2-\sqrt3}{2\sqrt3+1}(x-2)
(2\sqrt3+1)(y-3)=(2-\sqrt3)(x-2)
x(\sqrt3-2)+y(2\sqrt3+1)=-1+8\sqrt3                        -(i)

Similarly,
Now , equation of line passing through point (2 ,3) and with slope  \frac{-(2+\sqrt3)}{2\sqrt3-1}  is 
(y-3)=\frac{-(2+\sqrt3)}{2\sqrt3-1}(x-2)
(2\sqrt3-1)(y-3)= -(2+\sqrt3)(x-2)
x(2+\sqrt3)+y(2\sqrt3-1)=1+8\sqrt3                             -(ii)

Therefore, equation of line is    x(\sqrt3-2)+y(2\sqrt3+1)=-1+8\sqrt3     or      x(2+\sqrt3)+y(2\sqrt3-1)=1+8\sqrt3

Posted by

Gautam harsolia

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