Show that the equation 2x² – 13xy – 7y² + x +23y – 6 = 0 represents a pair of straight lines. Also find the angle between them and the coordinates of the point of intersection of the lines.

Given equation is
$2{\mathrm{x}}^2-13\mathrm{xy}-7{\mathrm{y}}^2+\mathrm{x}+23\mathrm{y}-6=0$
comparing the equation with we get
$\mathrm{a}=2,\begin{array}{c}2\mathrm{h}=-13\\ \Rightarrow \mathrm{h}=\frac{-13}{2}\end{array},\mathrm{b}=-7,\begin{array}{c}2\mathrm{g}=1\\ \Rightarrow \mathrm{g}=\frac{1}{2}\end{array},\begin{array}{c}2\mathrm{f}=23\\ \Rightarrow \mathrm{f}=\frac{23}{2}\end{array},\mathrm{c}=-6$
$\text{ Consider i) }\mathrm{abc}+2\mathrm{fgh}-{\mathrm{af}}^2-{\mathrm{bg}}^2-{\mathrm{ch}}^2$
$\begin{array}{l}=2(-7)(-6)+2\left(\frac{23}{2}\right)\left(\frac{1}{2}\right)\left(\frac{-13}{2}\right)-2{\left(\frac{23}{2}\right)}^2-(-7){\left(\frac{1}{2}\right)}^2-(-6){\left(\frac{-13}{2}\right)}^2\\ \\ =\frac{84}{1}-\frac{299}{4}-2\times \frac{529}{4}+7\cdot \frac{1}{4}+6\times \frac{169}{4}\\ =\frac{336-299-1058+7+1014}{4}=0\\ \therefore \mathrm{abc}+2\mathrm{fgh}-{\mathrm{af}}^2-{\mathrm{bg}}^2-{\mathrm{ch}}^2=0\end{array}$
$\text{ Now ii) }{\mathrm{h}}^2-\mathrm{ab}={\left(\frac{-13}{2}\right)}^2-\left(2\right)(-7)>0\Rightarrow {\mathrm{h}}^2>\mathrm{ab}$
$\begin{array}{l}{\mathrm{g}}^2-\mathrm{ac}={\left(\frac{1}{2}\right)}^2-2(-6)>0\Rightarrow {\mathrm{g}}^2>\mathrm{ac}\\ {\mathrm{f}}^2-\mathrm{bc}={\left(\frac{23}{2}\right)}^2-(-7)(-6)>0\Rightarrow {\mathrm{f}}^2>\mathrm{bc}\end{array}$
From (i) (ii) the given equation represents a pair of lines
$\mathrm{Point} \mathrm{of} \mathrm{intersection}=\left(\frac{\mathrm{hf}-\mathrm{bg}}{\mathrm{ab}-{\mathrm{h}}^2},\frac{\mathrm{gh}-\mathrm{af}}{\mathrm{ab}-{\mathrm{h}}^2}\right)$
$=\left(\frac{\frac{-13}{2}\times \frac{23}{2}-(-7)\left(\frac{1}{2}\right)}{2(-7)-{\left(\frac{-13}{2}\right)}^2},\frac{\frac{1}{2}\times \frac{-13}{2}-2\times \frac{23}{2}}{2(-7)-{\left(\frac{-13}{2}\right)}^2}\right)$
$\begin{array}{l}=\left(\frac{\frac{-299}{4}+\frac{7}{2}}{-14-\frac{169}{4}},\frac{\frac{-13}{4}-23}{-14-\frac{169}{4}}\right)\\ =\left(\frac{-299+14}{-56-169},\frac{-13-92}{-56-169}\right)\\ =\left(\frac{-285}{-225},\frac{-105}{-225}\right)=\left(\frac{19}{15},\frac{7}{15}\right)\end{array}$
If $\mathrm{\theta }$ is the acute angle between the lines then
$\begin{array}{l}\cos \mathrm{\theta }=\frac{|\mathrm{a}+\mathrm{b}|}{\sqrt{(\mathrm{a}-\mathrm{b}{)}^2+4{\mathrm{h}}^2}}\\ =\frac{|2-7|}{\sqrt{(2+7{)}^2+4{\left(\frac{-13}{2}\right)}^2}}\\ =\frac{|-5|}{\sqrt{(9{)}^2+4\left(\frac{169}{4}\right)}}\\ =\frac{|-5|}{\sqrt{81+169}} \\ =\frac{|-5|}{\sqrt{250}} \\ =\frac{5}{5\sqrt{10}}\\ \therefore \mathrm{\theta }={\cos }^{-1}\left(\frac{1}{\sqrt{10}}\right)\end{array}$