Question

The equation to a straight line referred to axes inclined at ${30}^o$ to one another is $y=2x+1$. Find its equation referred to axes inclined at ${45}^o$, the origin and axis of $x$ being unchanged.

Answer 1

To change one set of coordinate axes inclined at an angle $\omega$ to another system at ${\omega }^{\prime }$ and origin remain unchanged we substitute

$x\sin \omega =x^{\prime }\sin (\omega -\theta )+y^{\prime }\sin (\omega -{\omega }^{\prime }-\theta )y\sin \omega =x^{\prime }\sin \theta +y^{\prime }\sin ({\omega }^{\prime }+\theta )$

Here $\omega ={30}^{\circ },{\omega }^{\prime }={45}^{\circ }$and $\theta =0^{\circ }$

$\Rightarrow x\sin {30}^{\circ }=x^{\prime }\sin ({30}^{\circ }-0^{\circ })+y^{\prime }\sin ({30}^{\circ }-{45}^{\circ }-0^{\circ })\Rightarrow \frac{1}{2}x=\frac{1}{2}{x}^{\prime }-\frac{\sqrt{3}-1}{2\sqrt{2}}{y}^{\prime }\Rightarrow x=x^{\prime }-\frac{\sqrt{3}-1}{\sqrt{2}}{y}^{\prime }.......\left(i\right)\Rightarrow y\sin {30}^{\circ }=x^{\prime }\sin 0^{\circ }+y^{\prime }\sin ({45}^{\circ }+0^{\circ })\Rightarrow \frac{1}{2}y=\frac{1}{\sqrt{2}}{y}^{\prime }y=\sqrt{2}{y}^{\prime }........\left(ii\right)$

Now equation of given line is $y=2x+1$

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

$\sqrt{2}{y}^{\prime }=2(x^{\prime }-\frac{\sqrt{3}-1}{\sqrt{2}}{y}^{\prime })+1\sqrt{2}{y}^{\prime }+2\left(\frac{\sqrt{3}-1}{\sqrt{2}}{y}^{\prime }\right)=2x^{\prime }+1\sqrt{6}{y}^{\prime }=2x^{\prime }+12x^{\prime }-\sqrt{6}{y}^{\prime }+1=0$