Prove that the equation $3{\mathrm{x}}^2+7\mathrm{xy}+2{\mathrm{y}}^2+5\mathrm{x}+5\mathrm{y}+2=0$ resents a pair of straight lines and find coordinates of the point of intersection.

Given equation $3{\mathrm{x}}^2+7\mathrm{xy}+2{\mathrm{y}}^2+5\mathrm{x}+5\mathrm{y}+2=0$
Comparing with
$\begin{array}{l}{\mathrm{ax}}^2+2\mathrm{bxy}+{\mathrm{by}}^2+2\mathrm{gx}+2\mathrm{fy}+\mathrm{c}=0\\ \mathrm{a}=3;2\mathrm{h}=7;\mathrm{b}=2;2\mathrm{g}=5;2\mathrm{f}=5;\mathrm{c}=2\\ \mathrm{h}=\frac{7}{2};\mathrm{g}=\frac{5}{2};\mathrm{f}=\frac{5}{2}\\ \mathrm{\Delta }=\mathrm{abc}+2\mathrm{fgh}-{\mathrm{af}}^2-{\mathrm{bg}}^2-{\mathrm{ch}}^2\\ =3\left(2\right)\left(2\right)+\text{⧸}2\left(\frac{5}{\text{⧸}2}\right)\left(\frac{5}{2}\right)\left(\frac{7}{2}\right)\\ -3\left(\frac{25}{4}\right)-2\left(\frac{25}{4}\right)-2\left(\frac{49}{4}\right)\\ =\frac{48+175-75-50-98}{4}\\ =\frac{223-223}{4}\\ \Rightarrow \mathrm{\Delta }=0\end{array}$
$\left(\mathrm{i}\right)Now {\mathrm{g}}^2-\mathrm{ac}=\frac{25}{4}-3\left(2\right)=\frac{25-24}{4}$
$\begin{array}{l}{\mathrm{g}}^2-\mathrm{ac}=\frac{1}{4}>0\\ \Rightarrow {\mathrm{g}}^2>ac\\ {\mathrm{f}}^2-\mathrm{bc}=\frac{9}{4}>0\\ \Rightarrow {\mathrm{f}}^2>\mathrm{bc}\\ \mathrm{and} {\mathrm{h}}^2-\mathrm{ab}=\frac{25}{4}>0\\ \Rightarrow {\mathrm{h}}^2>\mathrm{ab}\end{array}$
$$\begin{gathered} \therefore \mathrm{The} \mathrm{line} \mathrm{represents} \mathrm{a} \mathrm{pair} \mathrm{of} \mathrm{straight} \mathrm{lines}. \\ \text{ P.O.I }=\left(\frac{\mathrm{hf}-\mathrm{bg}}{\mathrm{ab}-{\mathrm{b}}^2},\frac{\mathrm{gh}-\mathrm{af}}{\mathrm{ab}-{\mathrm{b}}^2}\right)=\left(-1,\frac{1}{5}\right) \\ \text{ The required point of intersection }=\left(-1,\frac{1}{5}\right) \end{gathered}$$