If two straight lines pass through the origin and makes an angle $\tan^{-1}(1/2)$ with $3y=2x$, then find its equations.
Let $m$ be the gradient of the line then, $$\frac{1}{2}=\frac{m-2/3}{1-2m/3}$$ I don't know whether my approach is correct or not.
On
There are two small errors:
Thus ultimately, you have to solve the equation $$\left|\frac{m-2/3}{1+2m/3}\right|=\frac 12\iff 2\cdot|3m-2|=|3+2m|$$ Can you solve it?
On
It may be better to compute the gradient $m$ directly.
The angle between the line to be solved $y=mx$ and the $x$-axis is $\tan^{-1}(1/2)+\tan^{-1}(2/3)$. So,
$$m=\tan \left[ \tan^{-1}(1/2)+\tan^{-1}(2/3) \right]=\frac{1/2+2/3}{1-1/3}=\frac{7}{4}$$
If you solve your equation $1/2=(m-2/3)/(1+2m/3)$, you will get the same $m$.
Thus, the line is, $$y= \frac{7}{4} x $$
The gradient of the second line can be computed similarly and the line is $y=x/8$.
On
$tan^{-1}(\frac{1}{2}) = \theta$
$tan(\theta) = \frac{1}{2}$
Line $3y = 2x$ have the slope of $\frac{2}{3}$, suppose this line is called a and the two lines in question is b and c (with $m_{b} > m_{a}, m_{c} < m_{a})$. So $m_{a} = \frac{2}{3}$ $=>$ $tan(A) = \frac{2}{3}$, $m_{b} = tan(B)$, and $m_{c} = tan{C}$.
Hint : $B - A = tan^{-1}(\frac{1}{2})$
$A-C = tan^{-1}(\frac{1}{2}) $
$tan(B-A) = \frac{1}{2}$
$tan(A-C) = \frac{1}{2}$
$tan(B)$ and $tan(C)$ will be the slope of the line you want to find, with $tan(A) = \frac{2}{3}$.
Rather complicated but i think the easiest one you can do with logical thinking rather than looking up formula
Note that if a line makes an angle $\arctan \alpha$ with the positive $x$-axis, then its slope is given by the tangent of this angle of inclination, in this case $$\tan (\arctan \alpha)=\alpha.$$