Question

If the lines $ y=3x+1 $ and $ 2y=x+3 $ are equally inclined to the line $ y=mx+4,\left( \frac{1}{2} < m < 3 \right), $ then the values of $m$ are

• A. $ \frac{1}{7}(1\pm 5\sqrt{3}) $
• B. $ \frac{1}{7}(1\pm 5\sqrt{5}) $
• C. $ \frac{1}{7}(1\pm 5\sqrt{2}) $
• D. $ \frac{1}{7}(1\pm 2\sqrt{5}) $
• E. $ \frac{1}{7}(1\pm 3\sqrt{2}) $

Answer 1

Answer: (C)

As $ m\in \left( \frac{1}{2},3 \right) $
$ \therefore $ Line $ y=mx+4 $ lies between
$ y=3x+1 $ and $ 2y=x+3 $
Slopes of given lines are
$ {{m}_{2}}=3,m=m $ and $ {{m}_{1}}=\frac{1}{2} $
$ \therefore $ $ \tan \theta =\frac{3-m}{1+3m} $ and $ \tan \theta =\frac{m-\frac{1}{2}}{1+\frac{m}{2}} $
$ \Rightarrow $ $ \frac{3-m}{1+3m}=\frac{2m-1}{2+m} $
$ \Rightarrow $ $ 7{{m}^{2}}-2m-7=0 $
$ \therefore \,\,\,m=\frac{2\pm \,\sqrt{4+196}}{2\times 7}\,=\frac{1}{7}\,(1\pm \,5\sqrt{2}) $