Question

For what value of $\lambda$ does the equation $12x^2-10xy+2y^2+11x-5y+\lambda =0$ represent a pair of straight lines? Find their equations, point of intersection, acute angle between them and pair of angle bisector.

Answer 1

$12x^2-10xy+2y^2+11x-5y+\lambda =0$

For pair of straight line

$a{x}^2+2hxy+b{y}^2+2gx+2fy+c=0$

$\left[\begin{array}{ccc}a& h& g\\ h& b& f\\ g& f& c\end{array}\right]=0$

Here, $a=12,h=\cfrac=-{10}{2}=-5,b=2,g=\cfrac{11}{2}

,\cfrac=-{5}{2}& c=\lambda$

$\left[\begin{array}{ccc}12& -5& \frac{11}{2}\\ -5& 2& -\frac{5}{2}\\ \frac{11}{2}& -\frac{5}{2}& \lambda \end{array}\right]=0$

$=12(12\lambda -\frac{25}{4})+5(-5\lambda +\frac{55}{4})+\frac{11}{2}(\frac{25}{2}-11)=0$

$or,12(8\lambda -25)-100\lambda +275+33=096\lambda -300-100\lambda +308=0or,-4x+8=0\lambda =2$

Equation is

$12x^2-10xy+2y^2+11x-5y+2=012x^2-4xy+8x+2y^2-6xy-4y+3x-y+2=04x(3x-y+2)-2y(-y+3x+2)+(4x-2y+1)=0$

Two eqations are

$3x-y+2=0\times 2\to \left(1\right)4x-2y+1=0\times 1\to \left(2\right)6x-2y+4=04x-2y+1=0--------2x+3=0x=-\frac{3}{2}$

Putting value of $x$ in $\left(2\right)$

$4(-\frac{3}{2})-2y+1=0-6-2y=1=0y=-\frac{5}{2}$

So point of intersection is

$(-\frac{3}{2},\frac{5}{2})y=3x=2$ here$m_1=3m_1=3y=\frac{4x+1}{2}\&m_2=\frac{4}{2}=2$

So, angle between them

$={\tan }^{-1}(\pm \frac{m_1-m_2}{1+m_1{m}_2})={\tan }^{-1}\left(\frac{3-2}{1+6}\right)\&{\tan }^{-1}(-\frac{3-2}{1+6})={\tan }^{-1}\left(\frac{1}{7}\right)\&{\tan }^{-1}(-\frac{1}{7})$

and other angls are

$=\pi -{\tan }^{-1}\left(\frac{1}{7}\right)\&\pi -{\tan }^{-1}(-\frac{1}{7})$

Pair of angle bisector

$\frac{a_{1}x+b_{1}y+c_1}{\sqrt{a_1^2+b_1^2}}=\pm \frac{a_{2}x+b_{2}y+c_2}{\sqrt{a_2^2+b_2^2}}$

where equations are $a_{1}x+b_{1}y+c_1=0$ & $a_{2}x+b_{2}y+c_2=0$

So, $\frac{3x-y+2}{\sqrt{3^2+1^2}}=\pm \frac{4x-2y+1}{\sqrt{4^2+2^2}}\frac{3x-y+2}{\sqrt{10}}=\pm \frac{4x-2y+1}{\sqrt{20}}$

or, $3x-y+2=\pm \frac{4x-2y+1}{\sqrt{2}}$