The straight lines l _ 1 and l _ 2 pass through the origin and...

Given,

The straight lines ${l}_{1}$ and ${l}_{2}$ pass through the origin and trisect the line segment of the line $L: 9x + 5y = 45$ between the axes,

And ${m}_{1}$ and ${m}_{2}$ are the slopes of the lines ${l}_{1}$ and ${l}_{2}$,

Now on plotting the diagram we get,

Given equation of line, $L: 9x + 5y = 45$

$⇒ \dfrac{x}{5} + \dfrac{y}{9} = 1$

Now using the section formula between point $A\left(5, 0\right) \text{ and } B\left(0, 9\right)$ we get the value of point $C \text{ and } D$

$⇒ C ≡ \left(\dfrac{10}{3}, 3\right)$ and $D ≡ \left(\dfrac{5}{3}, 6\right)$

Now finding the slope ${m}_{1} \text{ and } {m}_{2}$ we get,

${m}_{1} = \dfrac{3 - 0}{\dfrac{10}{3} - 0} = \dfrac{9}{10} \text{ and } {m}_{2} = \dfrac{6 \times 3}{5} = \dfrac{18}{5}$

So, equation of line $y = \left({m}_{1} + {m}_{2}\right)x$ will be,

$y = \left(\dfrac{9}{10} + \dfrac{36}{10}\right)x = \dfrac{9}{2}x$

So, intersection point with $L$ will be,

$7y = 45 ⇒ y = \dfrac{45}{7}, x = \dfrac{10}{7}$

Hence, $y - x = \dfrac{45 - 10}{7} = 5$