If the equation ${\mathrm{\lambda x}}^2-10\mathrm{xy}+12{\mathrm{y}}^2+5\mathrm{x}-16\mathrm{y}-3=0$ represents a pair of straight lines then find ’$\mathrm{\lambda }$’ and also find angle between lines and point of intersetion of the lines for this value of ’$\mathrm{\lambda }$’

The given equation is
${\mathrm{\lambda x}}^2-10\mathrm{xy}+12{\mathrm{y}}^2+5\mathrm{x}-16\mathrm{y}-3=0 ...\left(1\right)$
Comparing the given equation with
$\begin{array}{l}{\mathrm{ax}}^2+2\mathrm{hxy}+{\mathrm{by}}^2+2\mathrm{gx}+2\mathrm{fy}+\mathrm{c}=0\text{ we get }\\ \mathrm{a}=\mathrm{\lambda },\begin{array}{c}2\mathrm{h}=-10\\ \mathrm{h}=-5\end{array},\mathrm{b}=12,\begin{array}{c}2\mathrm{g}=5\\ \mathrm{g}=5/2\end{array},\begin{array}{c}2\mathrm{f}=-16\\ \mathrm{f}=-8\end{array},\mathrm{c}=-3\end{array}$
Since the given equation represents a pair of lines
$\begin{array}{l}\Rightarrow \mathrm{abc}+2\mathrm{fgh}-{\mathrm{af}}^2-{\mathrm{bg}}^2-{\mathrm{ch}}^2=0\\ \Rightarrow \mathrm{\lambda }\left(12\right)(-3)+2(-8)\left(\frac{5}{2}\right)(-5)-\mathrm{\lambda }(-8{)}^2-12{\left(\frac{5}{2}\right)}^2-(-3\left)\right(-5{)}^2=0\\ \\ \Rightarrow -36\mathrm{\lambda }+200-\mathrm{\lambda }\left(64\right)-12\times \frac{25}{4}+3\times 25=0\\ \Rightarrow -100\mathrm{\lambda }+200=0\Rightarrow 100\mathrm{\lambda }=200\\ \Rightarrow 100\mathrm{\lambda }=200\\ \therefore \mathrm{\lambda }=2\end{array}$
$\text{ Let ' }\mathrm{\theta }\text{ 'be the angle between the lines then }$
$\begin{array}{l}\cos \mathrm{\theta }=\frac{\mathrm{a}+\mathrm{b}}{\sqrt{(\mathrm{a}-\mathrm{b}{)}^2+4{\mathrm{h}}^2}}\\ =\frac{2+12}{\sqrt{(2-12{)}^2+4(-5{)}^2}}\\ =\frac{14}{\sqrt{100+100}}\\ =\frac{14}{10\cdot \sqrt{2}}\\ \Rightarrow \mathrm{\theta }={\cos }^{-1}\left(\frac{7}{5\sqrt{2}}\right)\end{array}$
Point of intersection of the lines is $\left(\frac{\mathrm{hf}-\mathrm{bg}}{\mathrm{ab}-{\mathrm{h}}^2},\frac{\mathrm{gh}-\mathrm{af}}{\mathrm{ab}-{\mathrm{h}}^2}\right)$
$\begin{array}{l}=\left(\frac{-5\times -8-12\times 5/2}{2\times 12-(-5{)}^2};\frac{5/2\times -5-2\times -8}{2\times 12-(-5{)}^2}\right)\\ =\left(\frac{40-30}{24-25},\frac{-\frac{25}{2}+16}{24-25}\right)=\left(\frac{10}{-1},\frac{-25+32}{-2}\right)\\ =\left(-10,\frac{-7}{2}\right)\end{array}$