The portion of the line 4x + 5y = 20 in the first quadrant is...

Using section formula

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

We know that, $tan\theta = \left|\dfrac{{m}_{1} - {m}_{2}}{1 + {m}_{1}{m}_{2}}\right|$

$⇒ tan\theta = \dfrac{\dfrac{6}{5}}{1 + \dfrac{16}{25}}$

$⇒ tan\theta = \dfrac{30}{41}$