Question

Equation of a straight line passing through the origin and making an acute angle with $x-$axis twice the size of the angle made by the line $y=\left(0.2\right)x$ with the $x-$axis, is:

• A. $y=\left(0.4\right)x$
• B. $y=\left(5/12\right)x$
• C. $6y-5x=0$
• D. $6y+5x=0$

Answer 1

The correct option is B $y=\left(5/12\right)x$

Given $e{q}^n$ of line: $y=\left(0.2\right)x$

Let this line makes $\angle \theta$ from X-axis.

This line has slope= $0.2\Rightarrow tan\theta =0.2\Rightarrow \theta ={\tan }^{-1}\left(0.2\right).$

Now, we have to find $e{q}^n$ of line making an angle $2\theta$ with X-axis $i.e$ $2{\tan }^{-1}\left(0.2\right)$[Acute]

$\Rightarrow 2ta{n}^{-1}\left(0.2\right)\to ta{n}^{-1}\left(0.2\right)+ta{n}^{-1}\left(0.2\right)$

$\Rightarrow ta{n}^{-1}\left(\frac{0.2+0.2}{1-\left(0.2\right)\left(0.2\right)}\right)$ [Using Identity: $ta{n}^{-1}a+ta{n}^{-1}b=ta{n}^{-1}\left(\frac{a+b}{1-ab}\right)$]

$\Rightarrow ta{n}^{-1}\left(\frac{5}{12}\right)$

Thus, slope of line making angle of $ta{n}^{-1}\left(\frac{5}{12}\right)$ from X-axis is:

$\Rightarrow$$tan\left(ta{n}^{-1}\right(\frac{5}{12}\left)\right)=\frac{5}{12}$ [As we know: $tan\left(ta{n}^{-1}a\right)=a]$

$\therefore$ $e{q}^n$ of line$\Rightarrow$ $y=\left(\frac{5}{12}\right)x$

Hence, Option (B) is correct.