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 = (

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Solution

Q.
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 [2023]

1 $y - x = 5$

2 $6 x + y = 10$

3 $y - 2 x = 5$

4 $6 x - y = 15$

Ans.
(1)

Let $l_{1} : y = m_{1} x$ and $l_{2} : y = m_{2} x$ be the two lines which trisects L.

Now, $9 x + 5 y = 45 \Rightarrow \frac{x}{5} + \frac{y}{9} = 1$ ...(i)

Now, L intersects coordinate axes at $A (5, 0)$ and $B (0, 9)$ .

Since $l_{1}$ divides $A B$ in the ratio $2 : 1$

$∴$ $M \equiv (\frac{5}{3}, 6)$

Also, $l_{2}$ divides $A B$ in the ratio $1 : 2$ .

$∴$ $N \equiv (\frac{10}{3}, 3)$

So, $l_{1} : 6 = m_{1} \times \frac{5}{3} \Rightarrow m_{1} = \frac{18}{5}$
and $l_{2} : 3 = m_{2} \times \frac{10}{3} \Rightarrow m_{2} = \frac{9}{10}$

So, equation of line $y = (m_{1} + m_{2}) x$ becomes

$y = (\frac{18}{5} + \frac{9}{10}) x \Rightarrow y = \frac{9}{2} x$ ...(ii)

Now point of intersection of (i) and (ii) is,

$\frac{x}{5} + \frac{x}{2} = 1 \Rightarrow \frac{7 x}{10} = 1 \Rightarrow x = \frac{10}{7}$ $So, y = \frac{45}{7}$

i.e., $(\frac{10}{7}, \frac{45}{7})$ is the point of intersection of (i) & (ii), which satisfies option (1).

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