Let the angle between the lines 
x-2y+3=0 and kx-y+2=0 be 45°. If $k_1,k_2(k_1>k_2)$ are two distinct real values of k, then k₁ - 2 =
 

Given lines $x-2y+3=0\to \left(1\right)$
$kx-y+2=0\to \left(2\right)$
Slope of (1) $m_1=\frac{1}{2}$
Slope of (2) m₂ = k
$\tan \theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|\Rightarrow \tan {45}^0=\left|\frac{{\displaystyle \frac{{\displaystyle 1}}{{\displaystyle 2}}-k}}{{\displaystyle 1+\frac{{\displaystyle 1}}{{\displaystyle 2}}k}}\right|$

$\begin{array}{l}\Rightarrow k=3\text{ or }\frac{-1}{3}\\ k_1=3 \mathrm{and} k_2=\frac{-1}{3}\\ \Rightarrow k_1-2=3-2=1=-3k_2\end{array}$