Question

The axes being inclined at an angle of ${60}^o$, find the inclination to the axis of $x$ of the straight lines whose equations are
(1) $y=2x+5$,
and (2) $2y=(\sqrt{3}-1)x+7$.

Answer 1

Given that:

Angle between axes is ${60}^{\circ }.$

If $\theta$ is angle between the axes then the transformation of points of Cartesian to Oblique coordinate system are as follows:

$X=x-y\cot \theta$ and

$Y=y\csc \theta$

$\left(1\right)$ Equation of line in oblique coordinate system:

$X=x-y\cot {60}^{\circ }=x-\frac{y}{\sqrt{3}}$

$Y=y\csc {60}^{\circ }=\frac{2y}{\sqrt{3}}$

Now, $\frac{2y}{\sqrt{3}}=2(x-\frac{y}{\sqrt{3}})+5$

or, $4y=2\sqrt{3}x+5\sqrt{3}$

So, Slope of the line $m_1=\frac{\sqrt{3}}{2}$

Slope of line $y=0$ i.e. $m_2=0$

Let $\alpha$ be the angle between x-axis and the straight line then

$\tan \alpha =\frac{m_1-m_2}{1+m_1{m}_2}=\frac{\frac{\sqrt{3}}{2}-0}{1+\frac{\sqrt{3}}{2}\times 0}$

or, $\alpha ={\tan }^{-1}\frac{\sqrt{3}}{2}$

$\left(2\right)$ Equation of line in oblique coordinate system:

$X=x-y\cot {60}^{\circ }=x-\frac{y}{\sqrt{3}}$

$Y=y\csc {60}^{\circ }=\frac{2y}{\sqrt{3}}$

Now, $2\times \frac{2y}{\sqrt{3}}=(\sqrt{3}-1)(x-\frac{y}{\sqrt{3}})+7$

or, $(\sqrt{3}+1)y=(\sqrt{3}-1)x+7$

So, Slope of the line $m_1=\frac{\sqrt{3}-1}{\sqrt{3}+1}$

Slope of line $y=0$ i.e. $m_2=0$

Let $\alpha$ be the angle between x-axis and the straight line then

$\tan \alpha =\frac{m_1-m_2}{1+m_1{m}_2}=\frac{\frac{\sqrt{3}-1}{\sqrt{3}+1}-0}{1+\frac{\sqrt{3}-1}{\sqrt{3}+1}\times 0}$

or, $\tan \alpha =\frac{\sqrt{3}-1}{\sqrt{3}+1}$

or, $\alpha ={15}^{\circ }$