If the lines y = 3x + 1 and 2y = x + 3 are eq | 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 lines are equally incl

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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 lines are equally inclined to the third line,the angle between the first and third line is equal to the angle between the second and third line.Using the formula for the angle between two lines,$	an 	heta = |\frac{m_a - m_b}{1 + m_a m_b}|$,we have:$|\frac{3 - m}{1 + 3m}| = |\frac{m - 1/2}{1 + m/2}|$$|\frac{3 - m}{1 + 3m}| = |\frac{2m - 1}{2 + m}|$Case $1$: $\frac{3 - m}{1 + 3m} = \frac{2m - 1}{2 + m}$$(3 - m)(2 + m) = (2m - 1)(1 + 3m)$$6 + 3m - 2m - m^2 = 2m + 6m^2 - 1 - 3m$$6 + m - m^2 = 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 $2y = x + 3$ are equally inclined to the line $y = mx + 4$,then $m =$ ?

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