Find the separate equations of a pair of lines represented by the equation x^2 2xy cosec + y^2 = 0 . Also find the angle between them.

Solution

The given equation of the pair of lines is: $$x^2-2xy\mathrm{cosec}\theta +y^2=0$$

Step 1: Rewrite the equation in standard form

The general equation of a pair of lines is: $$a{x}^2+2hxy+b{y}^2+2gx+2fy+c=0$$

Comparing with the given equation: $$x^2-2xy\mathrm{cosec}\theta +y^2=0$$ We have:

• $a=1$
• $2h=-2\mathrm{cosec}\theta ⟹h=-\mathrm{cosec}\theta$
• $b=1$
• $g=0$, $f=0$, $c=0$

Step 2: Find the slope of the lines

The slopes $m_1$ and $m_2$ of the lines are given by the quadratic equation: $$b{m}^2+2hm+a=0$$ Substituting the values: $$1\cdot m^2+2(-\mathrm{cosec}\theta )m+1=0$$ $$m^2-2m\mathrm{cosec}\theta +1=0$$

Step 3: Solve for $m$

Using the quadratic formula: $$m=\frac{2\mathrm{cosec}\theta \pm \sqrt{(2\mathrm{cosec}\theta {)}^2-4\cdot 1\cdot 1}}{2}$$ $$m=\frac{2\mathrm{cosec}\theta \pm \sqrt{4{\mathrm{cosec}}^2\theta -4}}{2}$$ $$m=\frac{2\mathrm{cosec}\theta \pm 2\sqrt{{\mathrm{cosec}}^2\theta -1}}{2}$$ $$m=\mathrm{cosec}\theta \pm \sqrt{{\mathrm{cosec}}^2\theta -1}$$

Using the identity ${\mathrm{cosec}}^2\theta -1={\cot }^2\theta$: $$m=\mathrm{cosec}\theta \pm \cot \theta$$

Step 4: Write the separate equations of the lines

The equations of the lines can be written in the form: $$y=mx$$ Thus, the two lines are: $$y=(\mathrm{cosec}\theta +\cot \theta )x$$ $$y=(\mathrm{cosec}\theta -\cot \theta )x$$

Rewriting in standard form: $$y=\left(\frac{1}{\sin \theta }+\frac{\cos \theta }{\sin \theta }\right)x$$ $$y=\left(\frac{1}{\sin \theta }-\frac{\cos \theta }{\sin \theta }\right)x$$

Simplifying: $$y=\frac{1+\cos \theta }{\sin \theta }x$$ $$y=\frac{1-\cos \theta }{\sin \theta }x$$

Multiplying both sides by $\sin \theta$: $$y\sin \theta =(1+\cos \theta )x$$ $$y\sin \theta =(1-\cos \theta )x$$

Rewriting: $$(1+\cos \theta )x-y\sin \theta =0$$ $$(1-\cos \theta )x-y\sin \theta =0$$

Step 5: Find the angle between the lines

The angle $\alpha$ between two lines with slopes $m_1$ and $m_2$ is given by: $$\tan \alpha =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|$$

Substituting $m_1=\mathrm{cosec}\theta +\cot \theta$ and $m_2=\mathrm{cosec}\theta -\cot \theta$: $$\tan \alpha =\left|\frac{(\mathrm{cosec}\theta +\cot \theta )-(\mathrm{cosec}\theta -\cot \theta )}{1+(\mathrm{cosec}\theta +\cot \theta )(\mathrm{cosec}\theta -\cot \theta )}\right|$$ $$\tan \alpha =\left|\frac{2\cot \theta }{1+{\mathrm{cosec}}^2\theta -{\cot }^2\theta }\right|$$

Using the identity ${\mathrm{cosec}}^2\theta =1+{\cot }^2\theta$: $$\tan \alpha =\left|\frac{2\cot \theta }{1+(1+{\cot }^2\theta )-{\cot }^2\theta }\right|$$ $$\tan \alpha =\left|\frac{2\cot \theta }{2}\right|$$ $$\tan \alpha =\cot \theta$$

Since $\tan \alpha =\cot \theta$, we have: $$\alpha =90^{\circ }-\theta$$

Thus, the angle between the lines is $90^{\circ }-\theta$.

Final Answer:

The separate equations of the pair of lines are: $$(1+\cos \theta )x-y\sin \theta =0$$ $$(1-\cos \theta )x-y\sin \theta =0$$

The angle between the lines is $90^{\circ }-\theta$.