QUESTION

# Equation of a straight line passing through the origin and making with x-axis an angle twice the size of the angle made by the line y=0.2x with the x-axis isa) $y=4.0x$b) $y=\dfrac{5}{12}x$c) $6y-5x=0$d) None of these

Hint: As the slope of a line is given by the tangent of the angle it makes with the x-axis, we can use the information given in the question to find the tangent of angle made by the line with the x-axis and then use the equation of the line using the slope.

It is given that the angle made by the line with the x-axis is twice that made by the line y = 0.2x with the x-axis.
Let the angle made by the line y=0.2x with the x-axis be $\theta$ and its slope be ${{m}_{1}}$. Then the angle made by the line to be found with the x-axis would be $2\theta$ and let its slope be ${{m}_{2}}$. ………………….(1.1)
Comparing the general equation of a straight line that is:
$y=mx+c$
where m is the slope and c is the intercept, we find that slope of the line y=0.2x is ${{m}_{1}}=0.2$ and the intercept $c=0$. …………..(1.2)
The tangent of sum of angles formula is given by $\tan (a+b)=\dfrac{\tan (a)+\tan (b)}{1-\tan (a)\tan (b)}.........(1.3)$
However, as the slope is equal to the tangent of the angle made with x-axis, from equation (1.1) and (1.2), we get,
Slope of the line $y=0.2x$ is ${{m}_{1}}=\tan (\theta )=0.2\text{ (from equation (1}\text{.2))}$ and
Slope of the required line is ${{m}_{2}}$ given by
\begin{align} & {{m}_{2}}=\tan (2\theta )=\dfrac{2\tan (\theta )}{1-{{\tan }^{2}}(\theta )}\text{ (from equation (1}\text{.3))} \\ & \Rightarrow {{\text{m}}_{2}}=\dfrac{2{{m}_{1}}}{1-m_{1}^{2}}=\dfrac{2\times 0.2}{1-{{\left( 0.2 \right)}^{2}}}=\dfrac{0.4}{1-0.04}=\dfrac{0.4}{0.96}=\dfrac{5}{12} \\ \end{align}
Thus, we find that the slope of the line is $\dfrac{5}{12}$. Comparing with the general equation $y=mx+c$ for a straight line, we find that the equation of the line should be of the form (as here the slope $m=\dfrac{5}{12}$)
$y=\dfrac{5}{12}x+c..........(1.4)$
Again, it is given that the line passes through origin, thus the point $\left( 0,0 \right)$should satisfy the line equation (1.4) and thus putting x=0, y=0 in equation (1.4), we get
$0=\dfrac{5}{12}\times 0+c\Rightarrow c=0$
Thus, using the value of c in equation (1.4), we obtain the equation of the line as
$y=\dfrac{5}{12}x$
Which matches option (b) and thus option (b) is the correct answer.

Note: While finding the slope, we could also have found out the angle from ${{m}_{1}}$ and from that we could also have obtained the angle made by the required line and its slope by again taking the tangent of the angle. That method would have taken a lot of steps and also it would have been difficult to calculate the angle from slope as it involves inverse trigonometric functions.