The portion of the line 4 x + 5 y = 20 in the first quadrant is trisected by the lines L 1 and L 2 passing…

Using section formula

Co-ordinates of $\mathrm{A}=\left(\frac{1\times 5+2\times 0}{1+2},\frac{1\times 0+2\times 4}{1+2}\right)=\left(\frac{5}{3},\frac{8}{3}\right)$
Co-ordinates of $\mathrm{B}=\left(\frac{2\times 5+1\times 0}{1+2},\frac{2\times 0+1\times 4}{1+2}\right)=\left(\frac{10}{3},\frac{4}{3}\right)$
Slope of $\mathrm{OA}={\mathrm{m}}_1=\frac{{\displaystyle \frac{8}{3}}-0}{{\displaystyle \frac{5}{3}}-0}=\frac{8}{5}$
Slope of $\mathrm{OB}={\mathrm{m}}_2=\frac{{\displaystyle \frac{4}{3}-0}}{{\displaystyle \frac{10}{3}-0}}=\frac{2}{5}$

We know that, $\tan \theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|$

$\Rightarrow \tan \theta =\frac{\frac{6}{5}}{1+\frac{16}{25}}$

$\Rightarrow \tan \theta =\frac{30}{41}$