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 $\tan \theta =\frac{m-\frac{1}{2}}{1+\frac{m}{2}}$
$\Rightarrow$ $\frac{3-m}{1+3m}=\frac{2m-1}{2+m}$
$\Rightarrow$ $7m^2-2m-7=0$
$\therefore m=\frac{2\pm \sqrt{4+196}}{2\times 7}=\frac{1}{7}(1\pm 5\sqrt{2})$