Question

With oblique coordinates find the tangent of the angle between the straight lines
$y=mx+c$ and $my+x=d$.

Answer 1

When the axis is oblique then inclination with axis is given by $\tan \theta =\frac{m\sin \omega }{1+m\cos \omega }$

For $y=mx+c$

$\tan {\theta }_1=\frac{m\sin \omega }{1+m\cos \omega }......\left(i\right)$

For $my+x=d$

$my=-x+d$

$y=-\frac{1}{m}x+\frac{d}{m}\tan {\theta }_2=\frac{-\frac{1}{m}\sin \omega }{1+(-\frac{1}{m})\cos \omega }=\frac{-\sin \omega }{m-\cos \omega }......\left(ii\right)$

As you know, angle between the lines is the difference between the angle that they make with their axis.

$\tan ({\theta }_1-{\theta }_2)=\frac{\tan {\theta }_1-\tan {\theta }_2}{1+\tan {\theta }_1\tan {\theta }_2}$

Substituting $\left(i\right)$ and $\left(ii\right)$

$\tan ({\theta }_1-{\theta }_2)=\frac{\frac{m\sin \omega }{1+m\cos \omega }-\left(\frac{-\sin \omega }{m-\cos \omega }\right)}{1+\left(\frac{m\sin \omega }{1+m\cos \omega }\right)\left(\frac{-\sin \omega }{m-\cos \omega }\right)}\tan ({\theta }_1-{\theta }_2)=\frac{m^2\sin \omega -m\sin \omega \cos \omega +\sin \omega +m\sin \omega \cos \omega }{m-\cos \omega {+m}^2\cos \omega -m{\cos }^2\omega -m{\sin }^2\omega }\tan ({\theta }_1-{\theta }_2)=\frac{(m^2+1)\sin \omega }{(m^2-1)\cos \omega +m-m({\cos }^2\omega +{\sin }^2\omega )}\tan ({\theta }_1-{\theta }_2)=\frac{(m^2+1)\sin \omega }{(m^2-1)\cos \omega }{\theta }_1-{\theta }_2={\tan }^{-1}\left(\frac{(m^2+1)\sin \omega }{(m^2-1)\cos \omega }\right)$