Let the angle between the lines x-2y+3=0 and | vedclass

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(C) The slopes of the lines $x-2y+3=0$ and $kx-y+2=0$ are $m_1 = \frac{1}{2}$ and $m_2 = k$ respectively.Given the angle $	heta = 45^{\circ}$,we use

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$-3k_2$

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(C) The slopes of the lines $x-2y+3=0$ and $kx-y+2=0$ are $m_1 = \frac{1}{2}$ and $m_2 = k$ respectively.Given the angle $	heta = 45^{\circ}$,we use the formula $	an 	heta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} ight|$.$	an 45^{\circ} = \left| \frac{\frac{1}{2} - k}{1 + \frac{k}{2}} ight| = 1$.$|1 - 2k| = |2 + k|$.Case $1$: $1 - 2k = 2 + k$ $\Rightarrow 3k = -1$ $\Rightarrow k = -\frac{1}{3}$.Case $2$: $1 - 2k = -(2 + k)$ $\Rightarrow 1 - 2k = -2 - k$ $\Rightarrow k = 3$.Since $k_1 > k_2$,we have $k_1 = 3$ and $k_2 = -\frac{1}{3}$.Then $k_1 - 2 = 3 - 2 = 1$.Checking the options: $-3k_2 = -3(-\frac{1}{3}) = 1$.Thus,$k_1 - 2 = -3k_2$.

Let the angle between the lines $x-2y+3=0$ and $kx-y+2=0$ be $45^{\circ}$. If $k_1, k_2$ $(k_1 > k_2)$ are two distinct real values of $k$,then $k_1-2=$

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