The straight lines l_1 and l_2 pass through the origin and trisect the line segment of the…

Question

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. If ${m}_{1}$ and ${m}_{2}$ are the slopes of the lines ${l}_{1}$ and ${l}_{2}$, then the point of intersection of the line $y=({m}_{1}+{m}_{2})x$ with L lies on :

Accepted Answer

Given line $L:9x+5y=45$ ..........(i)

Also given that ${m}_{1}$ and ${m}_{2}$ are the slopes of the lines ${l}_{1}$ and ${l}_{2}$, respectively.

Let $A$ be the point of intersection of line ${l}_{1}$ and $L$, then

co-ordinates of $A$ are $(\frac{2	imes 5+1	imes 0}{2+1},\frac{2	imes 0+1	imes 9}{2+1})=(\frac{10}{3},3)$

And let $B$ be the point of intersection of line ${l}_{2}$ and $L$, then

co-ordinates of $B$ are $(\frac{1	imes 5+2	imes 0}{1+2},\frac{1	imes 0+2	imes 9}{1+2})=(\frac{5}{3},6)$

Now, slope of line ${l}_{1},({m}_{1})=\frac{3-0}{\frac{10}{3}-0}=\frac{9}{10}$

and slope of lines ${l}_{2},({m}_{2})=\frac{6-0}{\frac{5}{3}-0}=\frac{18}{5}$

$\begin{gathered} 	herefore y =({m}_{1}+{m}_{2})x \ \ =(\frac{9}{10}+\frac{18}{5})x=\frac{45}{10}x=\frac{9}{2}x.......(ii) \end{gathered}$

Point of intersection of lines (i) and (ii)

$\begin{gathered} 9x+5(\frac{9}{2}x)=45 \ \Rightarrow x+\frac{5x}{2}=5 \ \Rightarrow \frac{7x}{2}=5\Rightarrow x=\frac{10}{7} \ y=\frac{9}{2}	imes \frac{10}{7}=\frac{45}{7} \ 	herefore =(\frac{10}{7},\frac{45}{7}) \ y-x=5 \end{gathered}$

Solutions · 1