Pair of Straight Lines
6.0 Pair of bisectors of lines represented by general equation
6.0 Pair of bisectors of lines represented by general equation
The pair of bisectors of the lines represented by $$a{x^2} + 2hxy + b{y^2} + 2gx + 2fy + c = 0...(1)$$ is $$\frac{{{{(x - \alpha )}^2} - {{(x - \beta )}^2}}}{{(a - b)}} = \frac{{(x - \alpha )(y - \beta )}}{h}$$ where $(\alpha ,\beta )$ be the point of intersection of the pair of straight lines represented by equation $(1)$.
Question 3. Find the value of $\lambda $ for which the equation $$12{x^2} - 10xy + 2{y^2} + 11x - 5y + \lambda = 0$$ represents a pair of straight lines. Find the equations of two lines and the angle between them.
Solution: Comparing the given equation with $$a{x^2} + 2hxy + b{y^2} + 2gx + 2fy + c = 0$$ we get, $a = 12,\;h = - 5,\;b = 2,g = \frac{{11}}{{12}},{\text{ }}f = - \frac{5}{2},{\text{ }}c = \lambda $
If the given equation represents a pair of straight lines, then $$abc + 2fgh - a{f^2} - b{g^2} - c{h^2} = 0$$ or, $$\begin{equation} \begin{aligned} (12 \times 2 \times \lambda ) + \left( {2 \times - \frac{5}{2} \times \frac{{11}}{2} \times - 5} \right) - \left( {12 \times \frac{{25}}{4}} \right) - \left( {2 \times \frac{{121}}{4}} \right) - \left( {\lambda \times 25} \right) = 0 \\ \lambda = 2 \\\end{aligned} \end{equation} $$
Also, $${h^2} - ab = 25 - 24 = 1 > 0$$
Therefore, the given equation will represent a pair of straight lines, if $\lambda = 2$.
Now, To find the equations of two lines, write the given equation as quadratic equation in $x$
$$\begin{equation} \begin{aligned} 12{x^2} + ( - 10y + 11)x + 2{y^2} - 5y + 2 = 0 \\ \therefore x = \frac{{ - ( - 10y + 11) \pm \sqrt {{{( - 10y + 11)}^2} - 48(2{y^2} - 5y + 2)} }}{{24}} \\ 24x = (10y - 11) \pm \sqrt {4{y^2} + 20y + 25} \\ 24x = 12y - 6{\text{ and }}24x = 8y - 16 \\ {\text{or, }}4x - 2y + 1 = 0{\text{ and }}3x - y + 2 = 0 \\\end{aligned} \end{equation} $$
are the required lines.
If $\theta $ is the angle between the lines, then $$\begin{equation} \begin{aligned} \theta = {\tan ^{ - 1}}\left( {\frac{{2\sqrt {({h^2} - ab)} }}{{\left| {a + b} \right|}}} \right) \\ {\text{ }} = {\tan ^{ - 1}}\left( {\frac{{2\sqrt {(25 - 24)} }}{{\left| {12 + 2} \right|}}} \right) \\ {\text{ = }}{\tan ^{ - 1}}\left( {\frac{1}{7}} \right) \\\end{aligned} \end{equation} $$
