Verified Solution • 2025–26 Session moderate

QJEEM23Ma791c: JEE Main 2023 Mathematics - Class 11 JEEM23Ma791c Solution: The straight lines and pass through the origin and trisect the…

The straight lines l1\mathrm{l_{1}} and l2\mathrm{l_{2}} pass through the origin and trisect the line segment of the line L : 9x+5y=459 x+5 y=45 between the axes. If m1\mathrm{m}_{1} and m2\mathrm{m}_{2} are the slopes of the lines l1\mathrm{l_{1}} and l2\mathrm{l_{2}} , then the point of intersection of the line y=(m1+m2)x\mathrm{y=\left(m_{1}+m_{2}\right)}x with L lies on :

Multiple Choice Options

  • A 6x−y=156 x-y=15
  • B 6x+y=106 x+y=10
  • C y−x=5\mathrm{y}-x=5
  • D y−2x=5y-2 x=5

Step-by-Step Verified Solution

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

JEE Main 2023 (Online) 6th April Morning Shift Mathematics - Straight Lines and Pair of Straight Lines Question 34 English Explanation
Also given that m1m_1 and m2m_2 are the slopes of the lines l1l_1 and l2l_2 , respectively.

Let AA be the point of intersection of line l1l_1 and LL , then

co-ordinates of AA are (2×5+1×02+1,2×0+1×92+1)=(103,3)\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 BB be the point of intersection of line l2l_2 and LL , then

co-ordinates of BB are (1×5+2×01+2,1×0+2×91+2)=(53,6)\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 l1,(m1)=3−0103−0=910l_1,\left(m_1\right)=\frac{3-0}{\frac{10}{3}-0}=\frac{9}{10}

and slope of lines l2,(m2)=6−053−0=185l_2,\left(m_2\right)=\frac{6-0}{\frac{5}{3}-0}=\frac{18}{5}

∴ line yamp;=(m1+m2)xamp;=(910+185)x=4510x=92x.......(ii)\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)

amp;9x+5(92x)=45amp;⇒x+5x2=5amp;⇒7x2=5⇒x=107amp; and y=92×107=457amp;∴ Point of intersection =(107,457)amp; lies on line y−x=5\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}

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring