Q : 19     If the lines  \small y=3x+1  and  \small 2y=x+3  are equally inclined to the line \small y=mx+4 , find the value of  m

Answers (1)

Given equation of lines are 
\small y=3x+1 \ \ \ \ \ \ \ \ \ \ -(i)
\small 2y=x+3 \ \ \ \ \ \ \ \ \ \ -(ii)
\small y=mx+4 \ \ \ \ \ \ \ \ \ \ -(iii)
Now, it is given that line (i) and (ii)  are equally inclined to the line (iii)
Slope of line \small y=3x+1  is  ,  \small m_1=3
Slope of line \small 2y=x+3 is , \small m_2= \frac{1}{2}
Slope of line \small y=mx+4 is , \small m_3=m
Now, we know that
\tan \theta = \left | \frac{m_1-m_2}{1+m_1m_2} \right |
Now,
\tan \theta_1 = \left | \frac{3-m}{1+3m} \right |              and                  \tan \theta_2 = \left | \frac{\frac{1}{2}-m}{1+\frac{m}{2}} \right |

It is given that \tan \theta_1=\tan \theta_2
Therefore,
\left | \frac{3-m}{1+3m} \right |= \left | \frac{1-2m}{2+m} \right |
\frac{3-m}{1+3m}= \pm\left ( \frac{1-2m}{2+m} \right )
Now, if     \frac{3-m}{1+3m}= \left ( \frac{1-2m}{2+m} \right )
Then,
(2+m)(3-m)=(1-2m)(1+3m)
6+m-m^2=1+m-6m^2
5m^2=-5
m= \sqrt{-1}
Which is not  possible
Now,  if \frac{3-m}{1+3m}= -\left ( \frac{1-2m}{2+m} \right )
Then,

(2+m)(3-m)=-(1-2m)(1+3m)
6+m-m^2=-1-m+6m^2
7m^2-2m-7=0
m = \frac{-(-2)\pm \sqrt{(-2)^2-4\times 7\times (-7)}}{2\times 7}= \frac{2\pm \sqrt{200}}{14}= \frac{1\pm5\sqrt2}{7}

Therefore, the value of  m is \frac{1\pm5\sqrt2}{7}

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