Question

The angle between the pair of straight lines $$x^2-y^2-2y-1=0$$ is
1. $${90}^{\circ }$$
2. $${60}^{\circ }$$
3. $${75}^{\circ }$$
4. $${36}^{\circ }$$

Answer 1

Hint: A straight line is a line which is not curved or bent. So, a line that extends to both sides till infinity and has no curves is called a straight line. The equations of two or more lines can be expressed together by an equation of degree higher than one. As we see that a linear equation in x and y represents a straight line, the product of two linear equations represent two straight lines, that is a pair of straight lines.

Complete step-by-step solution:
We know that the equation $$a{x}^2+2hxy+b{y}^2=0$$ represents a pair of straight lines passing through origin and hence it can be written as product of two linear factors, $$a{x}^2+2hxy+b{y}^2=(lx+my)(px+qy)$$where $$lp=a$$ , $$mq=b$$ and $$lq+mp=2h$$.
Also, the separate equations of lines are $$lx+my=0$$ and $$px+qy=0$$.
So, the angle between the lines is given by
$$\tan \theta ={\displaystyle \frac{2\sqrt{h^2-ab}}{a+b}}$$
As a consequence of this formula, we can conclude that
1. The lines are real and distinct, if $$h^2-ab>0$$
2. The lines are real and coincident, if $$h^2-ab=0$$
3. The lines are not real (imaginary), if $$h^2-ab<0$$
Given the equation of pair of straight lines
$$x^2-y^2-2y-1=0$$
$$x^2-\left(y^2+2y+1\right)=0$$
$$x^2-{\left(y+1\right)}^2=0$$
Equating with the standard equation $$a{x}^2+2hxy+b{y}^2=0$$
We have , $$a=1,h=0,b=-1$$
$$\theta ={\tan }^{-1}\left({\displaystyle \frac{2\sqrt{h^2-ab}}{a+b}}\right)$$
$$\Rightarrow \theta ={\tan }^{-1}\left({\displaystyle \frac{2\sqrt{0-(1)(-1)}}{1-1}}\right)$$
$$\Rightarrow \theta ={\tan }^{-1}\left(\mathrm{\infty }\right)$$
Therefore $$\theta ={90}^{\circ }$$.
Therefore option(1) is the correct answer.

Note: Two lines are coincident if $$\tan \theta =0$$ i.e. if $$h^2-ab=0$$. Two lines are perpendicular if $$\tan \theta =\mathrm{\infty }$$ i.e. if $$a+b=0$$. If two pairs of straight lines are equally inclined to one another, then both must have the same pair of bisectors. If the lines given by the equation $$a{x}^2+2hxy+b{y}^2=0$$ are equally inclined to axes, then the coordinate axes are the bisectors, i.e. the equation of pair of bisector must be $$xy=0$$. Therefore $$h=0$$. The two bisectors are always perpendicular.