JEE MAIN - Mathematics (2023 - 6th April Morning Shift - No. 3)
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 :
$$6 x-y=15$$
$$6 x+y=10$$
$$\mathrm{y}-x=5$$
$$y-2 x=5$$
Explanation
Given line $L: 9 x+5 y=45$ ..........(i)
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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)
$$ \begin{aligned} & 9 x+5\left(\frac{9}{2} x\right)=45 \\\\ & \Rightarrow x+\frac{5 x}{2}=5 \\\\ & \Rightarrow \frac{7 x}{2}=5 \Rightarrow x=\frac{10}{7} \\\\ & \text { and } y=\frac{9}{2} \times \frac{10}{7}=\frac{45}{7} \\\\ & \therefore \text { Point of intersection }=\left(\frac{10}{7}, \frac{45}{7}\right) \\\\ & \text { lies on line } y-x=5 \end{aligned} $$
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)
$$ \begin{aligned} & 9 x+5\left(\frac{9}{2} x\right)=45 \\\\ & \Rightarrow x+\frac{5 x}{2}=5 \\\\ & \Rightarrow \frac{7 x}{2}=5 \Rightarrow x=\frac{10}{7} \\\\ & \text { and } y=\frac{9}{2} \times \frac{10}{7}=\frac{45}{7} \\\\ & \therefore \text { Point of intersection }=\left(\frac{10}{7}, \frac{45}{7}\right) \\\\ & \text { lies on line } y-x=5 \end{aligned} $$
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