Question

The equation of lines $L_1$ and $L_2$ are $y=mx$ and $y=nx$, respectively. Suppose $L_1$ makes twice as large of an angle with the horizontal (measured counterclockwise from the positive x-axis) as does $L_2$ and that $L_1$ has $4$ times the slope of $L_2$. If $L_1$ is not horizontal, then the value of the product $\left(mn\right)$ equals:

• A. $\frac{\sqrt{2}}{2}$
• B. $-\frac{\sqrt{2}}{2}$
• C. $2$
• D. $-2$

Answer 1

The correct option is C $2$

Given the equation of $L_1$ is $y=mx$ ....... $\left(1\right)$ and that of $L_2$ is $y=nx$ ........ $\left(2\right)$

Now, the angle made by $\left(1\right)$ with the positive direction of $x$-axis is ${\tan }^{-1}m$ and that made by line $\left(2\right)$ is ${\tan }^{-1}n$.

Now, according to the problem,

${\tan }^{-1}m=2{\tan }^{-1}n$

or, ${\tan }^{-1}m={\tan }^{-1}\frac{2n}{1-n^2}$

or, $m=\frac{2n}{1-n^2}$........(1).

Again, by the problem $m=4n$.

Using this in $\left(3\right)$ we get,

$4n=\frac{2n}{1-n^2}$

or, $n^2=\frac{1}{2}$.

Now, $mn=4n^2=4\times \frac{1}{2}=2$.