JEE Main 2023 — Straight Lines Question with Solution

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, $L:9x+5y=45$

$\Rightarrow \frac{x}{5}+\frac{y}{9}=1$

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

$\Rightarrow C\equiv \left(\frac{10}{3},3\right)$ and $D\equiv \left(\frac{5}{3},6\right)$

Now finding the slope $m_1 \& m_2$ we get,

$m_1=\frac{3-0}{{\displaystyle \frac{10}{3}}-0} =\frac{9}{10} \& m_2=\frac{6\times 3}{5}=\frac{18}{5}$

So, equation of line $y=\left(m_1+m_2\right)x$ will be,

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

So, intersection point with $L$ will be,

$7y=45\Rightarrow y=\frac{45}{7}, x=\frac{10}{7}$

Hence, $y-x=\frac{45-10}{7}=5$