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

Question

The straight lines $$\mathrm{l_{1}}$$ and $$\mathrm{l_{2}}$$ pass through the origin and trisect the line segment of the line L : $9 x+5 y=45$ between the axes. If $$\mathrm{m}_{1}$$ and $$\mathrm{m}_{2}$$ are the slopes of the lines $$\mathrm{l_{1}}$$ and $$\mathrm{l_{2}}$$, then the point of intersection of the line $$\mathrm{y=\left(m_{1}+m_{2}\right)}x$$ with L lies on :

Accepted Answer

Given line $L: 9 x+5 y=45$ ..........(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 $\left(\frac{2 \times 5+1 \times 0}{2+1}, \frac{2 \times 0+1 \times 9}{2+1}\right)=\left(\frac{10}{3}, 3\right)$

And let $B$ be the point of intersection of line $l_2$ and $L$, then

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

Now, slope of line $l_1,\left(m_1\right)=\frac{3-0}{\frac{10}{3}-0}=\frac{9}{10}$

and slope of lines $l_2,\left(m_2\right)=\frac{6-0}{\frac{5}{3}-0}=\frac{18}{5}$

$$
\begin{aligned}
\therefore \text { line } y & =\left(m_1+m_2\right) x \\
& =\left(\frac{9}{10}+\frac{18}{5}\right) x=\frac{45}{10} x=\frac{9}{2} x .......(ii)
\end{aligned}
$$

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

$$
\begin{aligned}
& 9 x+5\left(\frac{9}{2} x\right)=45 \\
& \Rightarrow x+\frac{5 x}{2}=5 \\
& \Rightarrow \frac{7 x}{2}=5 \Rightarrow x=\frac{10}{7} \\
& \text { and } y=\frac{9}{2} \times \frac{10}{7}=\frac{45}{7} \\
& \therefore \text { Point of intersection }=\left(\frac{10}{7}, \frac{45}{7}\right) \\
& \text { lies on line } y-x=5
\end{aligned}
$$

Solutions · 1