Find the value of k, if the angle between the straight lines 4x–y+7=0 and kx–5y–9=0 is 45°.

If$\theta is$ angle between the st.lines
${\mathrm{a}}_1\mathrm{x}+{\mathrm{b}}_1\mathrm{y}+{\mathrm{c}}_1=0\text{ and }{\mathrm{a}}_2\mathrm{x}+{\mathrm{b}}_2\mathrm{y}+{\mathrm{c}}_2=0\text{ then}$
$\cos \mathrm{\theta }=\frac{{\mathrm{a}}_1{\mathrm{a}}_2+{\mathrm{b}}_1{\mathrm{b}}_2}{\sqrt{{\mathrm{a}}_1^2+{\mathrm{b}}_1^2}\sqrt{{\mathrm{a}}_2^2+{\mathrm{b}}_2^2}}$
Given st.lines 4x – y + 7 = 0; Kx – 5y – 9 = 0
$\begin{array}{l}\mathrm{Where} \mathrm{\theta }={45}^{\circ }{\text{ ,a}}_1{\text{=4,b}}_1{\text{=-1,a}}_2{\text{=k,b}}_2\text{=-5}\\ Now \cos {45}^{\circ }=\frac{4\mathrm{k}+5}{\sqrt{16+1}\sqrt{{\mathrm{k}}^2+25}}\\ \text{ S.O.B.S. }\\ \Rightarrow 15{\mathrm{K}}^2+80\mathrm{K}-375=0\\ \Rightarrow 3{\mathrm{K}}^2+16\mathrm{K}-75=0\Rightarrow (\mathrm{K}-3)(3\mathrm{K}+25)=0\\ \Rightarrow \mathrm{K}=3\text{ (or })\mathrm{k}=-25/3\\ \therefore \mathrm{K}=3\text{ (or) }\mathrm{k}=\frac{-25}{3}\end{array}$