If the lines y=3x+1 and 2y=x+3 are equally in | vedclass

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(D) The given lines are $y=3x+1$ (slope $m_1=3$) and $2y=x+3$ or $y=\frac{1}{2}x+\frac{3}{2}$ (slope $m_2=\frac{1}{2}$). Let the third line be $y=mx+4

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$\frac{1 \pm 5 \sqrt{2}}{7}$

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(D) The given lines are $y=3x+1$ (slope $m_1=3$) and $2y=x+3$ or $y=\frac{1}{2}x+\frac{3}{2}$ (slope $m_2=\frac{1}{2}$). Let the third line be $y=mx+4$ with slope $m$. Since the lines are equally inclined to the third line,the angle between the first and third line equals the angle between the second and third line: $\left|\frac{m_1-m}{1+m_1m}\right| = \left|\frac{m_2-m}{1+m_2m}\right|$ Substituting the slopes: $\left|\frac{3-m}{1+3m}\right| = \left|\frac{\frac{1}{2}-m}{1+\frac{1}{2}m}\right| = \left|\frac{1-2m}{2+m}\right|$ Case $1$: $\frac{3-m}{1+3m} = \frac{1-2m}{2+m}$ $(3-m)(2+m) = (1-2m)(1+3m)$ $6+m-m^2 = 1+m-6m^2$ $5m^2+5 = 0 \Rightarrow m^2 = -1$ (No real solution). Case $2$: $\frac{3-m}{1+3m} = -\left(\frac{1-2m}{2+m}\right)$ $(3-m)(2+m) = -(1-2m)(1+3m)$ $6+m-m^2 = -(1+m-6m^2) = -1-m+6m^2$ $7m^2-2m-7 = 0$ Using the quadratic formula $m = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$: $m = \frac{2 \pm \sqrt{4-4(7)(-7)}}{14} = \frac{2 \pm \sqrt{4+196}}{14} = \frac{2 \pm \sqrt{200}}{14} = \frac{2 \pm 10\sqrt{2}}{14} = \frac{1 \pm 5\sqrt{2}}{7}$.

If the lines $y=3x+1$ and $2y=x+3$ are equally inclined to the line $y=mx+4$,then the value of $m$ is equal to

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