If the lines y = 3x + 1 and 2y = x + 3 make e | vedclass

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(D) The slopes of the given lines are $m_1 = 3$ and $m_2 = \frac{1}{2}$. Let the slope of the third line be $m_3 = m$. Since the line $y = mx + 4$ mak

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

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(D) The slopes of the given lines are $m_1 = 3$ and $m_2 = \frac{1}{2}$. Let the slope of the third line be $m_3 = m$. Since the line $y = mx + 4$ makes equal angles with the other two lines,we have: $|\frac{m - 3}{1 + 3m}| = |\frac{m - 1/2}{1 + m/2}|$ $|\frac{m - 3}{1 + 3m}| = |\frac{2m - 1}{2 + m}|$ Case $1$: $\frac{m - 3}{1 + 3m} = \frac{2m - 1}{2 + m}$ $(m - 3)(m + 2) = (2m - 1)(3m + 1)$ $m^2 - m - 6 = 6m^2 - m - 1$ $5m^2 = -5 \Rightarrow m^2 = -1$ (No real solution). Case $2$: $\frac{m - 3}{1 + 3m} = -(\frac{2m - 1}{2 + m})$ $(m - 3)(m + 2) = -(2m - 1)(3m + 1)$ $m^2 - m - 6 = -(6m^2 - m - 1)$ $m^2 - m - 6 = -6m^2 + m + 1$ $7m^2 - 2m - 7 = 0$ 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 $2y = x + 3$ make equal angles with the line $y = mx + 4$,then $m = $ . . .

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