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
\[\tan \theta = \dfrac{{2\sqrt {{h^2} - ab} }}{{a + b}}\]
As a consequence of this formula, we can conclude that
1. The lines are real and distinct, if \[{h^2} - ab > 0\]
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 , \[a = 1,h = 0,b = - 1\]
\[\theta = {\tan ^{ - 1}}\left( {\dfrac{{2\sqrt {{h^2} - ab} }}{{a + b}}} \right)\]
\[\Rightarrow \theta = {\tan ^{ - 1}}\left( {\dfrac{{2\sqrt {0 - (1)( - 1)} }}{{1 - 1}}} \right)\]
\[\Rightarrow \theta = {\tan ^{ - 1}}\left( \infty \right)\]
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 \[\tan \theta = \infty \] 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.