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

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(A) The equations of the given lines are:$y=3x+1$ .... $(1)$$2y=x+3$ or $y=\frac{1}{2}x+\frac{3}{2}$ .... $(2)$$y=mx+4$ .... $(3)$The slopes of these

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

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(A) The equations of the given lines are:$y=3x+1$ .... $(1)$$2y=x+3$ or $y=\frac{1}{2}x+\frac{3}{2}$ .... $(2)$$y=mx+4$ .... $(3)$The slopes of these lines are $m_1=3$,$m_2=\frac{1}{2}$,and $m_3=m$.Since lines $(1)$ and $(2)$ are equally inclined to line $(3)$,the angle between $(1)$ and $(3)$ is equal to the angle between $(2)$ and $(3)$.Using the formula for the angle between two lines,$	an 	heta = \left| \frac{m_a - m_b}{1 + m_a m_b} ight|$,we have:$\left| \frac{3-m}{1+3m} ight| = \left| \frac{\frac{1}{2}-m}{1+\frac{1}{2}m} ight|$$\left| \frac{3-m}{1+3m} ight| = \left| \frac{1-2m}{2+m} ight|$Case $1$: $\frac{3-m}{1+3m} = \frac{1-2m}{2+m}$$(3-m)(2+m) = (1-2m)(1+3m)$$6 + 3m - 2m - m^2 = 1 + 3m - 2m - 6m^2$$5m^2 + 5 = 0 \Rightarrow m^2 = -1$ (No real solution).Case $2$: $\frac{3-m}{1+3m} = -\left( \frac{1-2m}{2+m} ight)$$(3-m)(2+m) = -(1-2m)(1+3m)$$6 + m - m^2 = -(1 + m - 6m^2)$$6 + m - m^2 = -1 - m + 6m^2$$7m^2 - 2m - 7 = 0$Using the quadratic formula $m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$m = \frac{2 \pm \sqrt{(-2)^2 - 4(7)(-7)}}{2(7)} = \frac{2 \pm \sqrt{4 + 196}}{14} = \frac{2 \pm \sqrt{200}}{14} = \frac{2 \pm 10\sqrt{2}}{14} = \frac{1 \pm 5\sqrt{2}}{7}$.

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

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