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

JEE Main · 2023 · 06 Apr 2023 · Shift 1 · Straight Lines

The straight lines

l_{1}

and

l_{2}

pass through the origin and trisect the line segment of the line L :

9 x + 5 y = 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

Answer: (C)

The line

9 x + 5 y = 45

meets the axes at

A (5, 0)

and

B (0, 9)

. Trisecting the segment

A B

gives points

P (\frac{10}{3}, 3)

and

Q (\frac{5}{3}, 6)

. Lines through the origin:

m_{1} = \frac{3}{10/3} = \frac{9}{10}

and

m_{2} = \frac{6}{5/3} = \frac{18}{5}

. Then

m_{1} + m_{2} = \frac{9}{10} + \frac{18}{5} = \frac{9 + 36}{10} = \frac{45}{10} = \frac{9}{2}

. The line

y = \frac{9}{2} x

intersects

9 x + 5 y = 45

9 x + 5 \cdot \frac{9 x}{2} = 45

, so

18 x + 45 x = 90

, giving

x = \frac{10}{7}

and

y = \frac{45}{7}

. Checking:

y - x = \frac{45}{7} - \frac{10}{7} = \frac{35}{7} = 5

, so the point lies on

y - x = 5

JEE Main · 2023 · 06 Apr 2023 · Shift 1 · Single correct option