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Question: The angle between the pair of straight lines \[{x^2} - {y^2} - 2y - 1 = 0\] is 1\. \[{90^ \circ }...

The angle between the pair of straight lines x2−y2−2y−1=0{x^2} - {y^2} - 2y - 1 = 0 is
1. 90∘{90^ \circ }
2. 60∘{60^ \circ }
3. 75∘{75^ \circ }
4. 36∘{36^ \circ }

Explanation

Solution

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 ax2+2hxy+by2=0a{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, ax2+2hxy+by2=(lx+my)(px+qy)a{x^2} + 2hxy + b{y^2} = (lx + my)(px + qy)where lp=alp = a , mq=bmq = b and lq+mp=2hlq + mp = 2h.
Also, the separate equations of lines are lx+my=0lx + my = 0 and px+qy=0px + qy = 0.
So, the angle between the lines is given by
tan⁡θ=2h2−aba+b\tan \theta = \dfrac{{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 h2−ab>0{h^2} - ab > 0
2. The lines are real and coincident, if h2−ab=0{h^2} - ab = 0
3. The lines are not real (imaginary), if h2−ab<0{h^2} - ab < 0
Given the equation of pair of straight lines
x2−y2−2y−1=0{x^2} - {y^2} - 2y - 1 = 0
x2−(y2+2y+1)=0{x^2} - \left( {{y^2} + 2y + 1} \right) = 0
x2−(y+1)2=0{x^2} - {\left( {y + 1} \right)^2} = 0
Equating with the standard equation ax2+2hxy+by2=0a{x^2} + 2hxy + b{y^2} = 0
We have , a=1,h=0,b=−1a = 1,h = 0,b = - 1
θ=tan⁡−1(2h2−aba+b)\theta = {\tan ^{ - 1}}\left( {\dfrac{{2\sqrt {{h^2} - ab} }}{{a + b}}} \right)
⇒θ=tan⁡−1(20−(1)(−1)1−1)\Rightarrow \theta = {\tan ^{ - 1}}\left( {\dfrac{{2\sqrt {0 - (1)( - 1)} }}{{1 - 1}}} \right)
⇒θ=tan⁡−1(∞)\Rightarrow \theta = {\tan ^{ - 1}}\left( \infty \right)
Therefore θ=90∘\theta = {90^ \circ }.
Therefore option(1) is the correct answer.

Note: Two lines are coincident if tan⁡θ=0\tan \theta = 0 i.e. if h2−ab=0{h^2} - ab = 0. Two lines are perpendicular if tan⁡θ=∞\tan \theta = \infty i.e. if a+b=0a + 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 ax2+2hxy+by2=0a{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=0xy = 0. Therefore h=0h = 0. The two bisectors are always perpendicular.