Question

The axes being inclined at an angle of ${120}^o$, find the tangent of the angle between the two straight lines
$8x+7y=1$ and $28x-73y=101$.

Answer 1

Given that:

Angle between axes is ${120}^{\circ }.$

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

$Y=y\csc \theta$

Equation of line in oblique coordinate system:

$X=x-y\cot {120}^{\circ }=x+\frac{y}{\sqrt{3}}$

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

First Equation:

$8(x+\frac{y}{\sqrt{3}})+7\times \frac{2y}{\sqrt{3}}=1$

or, $8\sqrt{3}x+8y+14y=\sqrt{3}$

or, $8\sqrt{3}x+22y=\sqrt{3}$

Slope of the line i.e $m_1=-\frac{4\sqrt{3}}{11}$

Second Equation:

$28(x+\frac{y}{\sqrt{3}})-73\times \frac{2y}{\sqrt{3}}=101$

or, $28\sqrt{3}x+28y-146y=101\sqrt{3}$

or, $28\sqrt{3}x-118y=101\sqrt{3}$

Slope of the line i.e $m_2=\frac{14\sqrt{3}}{59}$

Let $\alpha$ be the angle between the straight lines then

$\tan \alpha =\frac{m_1-m_2}{1+m_1{m}_2}=\left|\frac{-\frac{4\sqrt{3}}{11}-\frac{14\sqrt{3}}{59}}{1+\frac{-4\sqrt{3}}{11}\times \frac{14\sqrt{3}}{59}}\right|$

or, $\tan \alpha =\left|\frac{-236\sqrt{3}-154\sqrt{3}}{649-168}\right|=\left|\frac{-390\sqrt{3}}{481}\right|=\frac{30\sqrt{3}}{37}$

or, $\alpha ={\tan }^{-1}\frac{30\sqrt{3}}{37}$