Let the line pair be represented by the equation \[ax^2+2hxy+by^2=0\] Let $y=m_1x$ and $y=m_2x$ be the equations of two lines represented by the above line pair where \[m_1+m_2=-\frac{2h}{b}\text{ and }m_1m_2=\frac{a}{b}\] Let $\theta$ be the angle between the line pair. Then, \[\tan\theta=\pm\frac{m_1-m_2}{1+m_1m_2}\] \[=\pm\frac{\sqrt{(m_1+m_2)^2-4m_1m_2}}{1+m_1m_2}\] \[=\pm\frac{\sqrt{\frac{4h^2}{b^2}-\frac{4a}{b}}}{1+\frac{a}{b}}=\pm\frac{2\sqrt{h^2-ab}}{a+b}\] \[\therefore\tan\theta=\pm\frac{2\sqrt{h^2-ab}}{a+b}\]
The homogeneous equation of degree two i.e. $ax^2+2hxy+by^2=0$ represents
- two real and distinct lines if $h^2-ab>0$
- two real and coincident lines if $h^2-ab=0$
- two imaginary lines if $h^2-ab<0$
Condition of Parallelism
Two lines will be parallel to each other if $\theta=0°$. \[\therefore\tan 0°=\pm\frac{2\sqrt{h^2-ab}}{a+b}\] \[\sqrt{h^2-ab}=0\] \[\therefore h^2=ab\]
Condition of Perpendicularity
Two lines will be perpendicular to each other if $\theta=90°$. \[\therefore\tan 90°=\pm\frac{2\sqrt{h^2-ab}}{a+b}\] \[\cot 90°=\pm\frac{a+b}{2\sqrt{h^2-ab}}\] \[\therefore a+b=0\] \[\therefore\text{coeff. of }x^2+\text{coeff. of }y^2=0\]
Find the angle between the line pair $x^2+9xy+14y^2=0$.
Here, \[x^2+9xy+14y^2=0\] \[\begin{array}{c} \therefore a=1, & 2h=9 &\text{and}& b=14 \\ a=1, & h=\frac{9}{2} &\text{and}& b=14\end{array}\] Let $\theta$ be the angle between the line pair. Then, \[\tan\theta=\pm\frac{2\sqrt{h^2-ab}}{a+b}\] \[=\pm\frac{2\sqrt{\frac{81}{4}-14}}{1+14}=\pm\frac{2×5}{2×15}=\pm\frac{1}{3}\] \[\therefore\theta=\tan^{-1}\frac{1}{3}\]
Find the angle between the line pair $x^2-5xy+4y^2+x+2y-2=0$.
Given line pair is, \[x^2-5xy+4y^2+x+2y-2=0\] Taking homogeneous part only, \[x^2-5xy+4y^2=0\] \[\begin{array}{c} \therefore a=1, & 2h=-5 &\text{and}& b=4 \\ a=1, & h=\frac{-5}{2} &\text{and}& b=4\end{array}\] Let $\theta$ be the angle between the line pair. Then, \[\tan\theta=\pm\frac{2\sqrt{h^2-ab}}{a+b}\] \[=\pm\frac{2\sqrt{\frac{25}{4}-4}}{1+4}=\pm\frac{2×\frac{3}{2}}{5}=\pm\frac{3}{5}\] \[\therefore\theta=\tan^{-1}\frac{3}{5}\]
Prove that the two lines are parallel in the line pair $x^2+6xy+9y^2+12y-5=0$.
Given line pair is, \[x^2+6xy+9y^2+4x+12y-5=0\] Taking homogeneous part only, \[x^2+6xy+9y^2=0\] \[\begin{array}{c} \therefore a=1, & 2h=6 &\text{and}& b=9 \\ a=1, & h=3 &\text{and}& b=9\end{array}\] The two lines will be parallel if \[h^2=ab\] \[3^2=1×9\] \[9=9\] Hence, the two lines are parallel.
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