Find the intercepts of $L: 9x+5y=45$.

• On the $x$-axis, $y=0$: $9x=45 \Rightarrow x=5$ So one endpoint is $(5,0)$.

• On the $y$-axis, $x=0$: $5y=45 \Rightarrow y=9$ So the other endpoint is $(0,9)$.

Thus, the line segment between the axes is the segment joining $(5,0)$ and $(0,9)$.

Find the trisection points of this segment.

A point dividing the segment from $(5,0)$ to $(0,9)$ in the ratio $1:2$ is $\left(\frac{2\cdot 5+1\cdot 0}{3},\frac{2\cdot 0+1\cdot 9}{3}\right)=\left(\frac{10}{3},3\right).$

The other trisection point dividing in the ratio $2:1$ is $\left(\frac{1\cdot 5+2\cdot 0}{3},\frac{1\cdot 0+2\cdot 9}{3}\right)=\left(\frac{5}{3},6\right).$

Since $l_1$ and $l_2$ pass through the origin and these trisection points, their slopes are:

$m_1=\frac{3}{10/3}=\frac{9}{10}, \qquad m_2=\frac{6}{5/3}=\frac{18}{5}.$

Compute $m_1+m_2$.

$m_1+m_2=\frac{9}{10}+\frac{18}{5}=\frac{9}{10}+\frac{36}{10}=\frac{45}{10}=\frac{9}{2}.$

So the required line is $y=\left(m_1+m_2\right)x=\frac{9}{2}x.$

Find its intersection with $L: 9x+5y=45$.

Substitute $y=\frac{9}{2}x$ into $9x+5y=45$: $9x+5\left(\frac{9}{2}x\right)=45$ $9x+\frac{45}{2}x=45$ $\frac{18+45}{2}x=45$ $\frac{63}{2}x=45$ $x=\frac{90}{63}=\frac{10}{7}.$

Then $y=\frac{9}{2}\cdot \frac{10}{7}=\frac{45}{7}.$

So the intersection point is $\left(\frac{10}{7},\frac{45}{7}\right).$

Check which option passes through $\left(\frac{10}{7},\frac{45}{7}\right)$.

• A: $6x-y=15$ $6\cdot \frac{10}{7}-\frac{45}{7}=\frac{60-45}{7}=\frac{15}{7}\neq 15$ Not correct.

• B: $6x+y=10$ $6\cdot \frac{10}{7}+\frac{45}{7}=\frac{105}{7}=15\neq 10$ Not correct.

• C: $y-x=5$ $\frac{45}{7}-\frac{10}{7}=\frac{35}{7}=5$ Correct.

• D: $y-2x=5$ $\frac{45}{7}-2\cdot \frac{10}{7}=\frac{25}{7}\neq 5$ Not correct.

Therefore, the correct option is $\boxed{\text{C: } y-x=5}.$