Find the values of k, if the lines joining the origin to the points of intersection of the curve 2x² – 2xy + 3y² + 2x – y – 1 = 0 and the line x + 2y = k are mutually perpendicular. 

The given curve is $2{\mathrm{x}}^2-2\mathrm{xy}+3{\mathrm{y}}^2+2\mathrm{x}-\mathrm{y}-1=0 ...\left(1\right)$

And given line is x + 2y = k
$\Rightarrow \frac{\mathrm{x}+2\mathrm{y}}{\mathrm{k}}=1$ ...(2)
Let the given curve (1) and the given line (2) intersect at A and B and join $\overline{\mathrm{OA}} \mathrm{and} \overline{\mathrm{OB}}$
By homogenising the curve (1) with the line (2) we get
$\begin{array}{l}2{\mathrm{x}}^2-2\mathrm{xy}+3{\mathrm{y}}^2+2\mathrm{x}\left(1\right)-\mathrm{y}\left(1\right)-1(1{)}^2=0\\ \Rightarrow 2{\mathrm{x}}^2-2\mathrm{xy}+3{\mathrm{y}}^2+2\mathrm{x}\left(\frac{\mathrm{x}+2\mathrm{y}}{\mathrm{k}}\right)-\mathrm{y}\left(\frac{\mathrm{x}+2\mathrm{y}}{\mathrm{k}}\right)-1{\left(\frac{\mathrm{x}+2\mathrm{y}}{\mathrm{k}}\right)}^2=0\\ \\ \Rightarrow 2{\mathrm{x}}^2-2\mathrm{xy}+3{\mathrm{y}}^2+\frac{2{\mathrm{x}}^2+4\mathrm{xy}}{\mathrm{k}}-\left(\frac{\mathrm{xy}+2{\mathrm{y}}^2}{\mathrm{k}}\right)-\left(\frac{{\mathrm{x}}^2+4{\mathrm{y}}^2+4\mathrm{xy}}{{\mathrm{k}}^2}\right)=0\\ \\ \Rightarrow 2{\mathrm{k}}^2{\mathrm{x}}^2-2{\mathrm{k}}^2\mathrm{xy}+3{\mathrm{k}}^2{\mathrm{y}}^2+2{\mathrm{kx}}^2+4\mathrm{kxy}-\mathrm{kxy}-2{\mathrm{ky}}^2-{\mathrm{x}}^2-4{\mathrm{y}}^2-4\mathrm{xy}=0\\ \\ \Rightarrow {\mathrm{x}}^2\left(2{\mathrm{k}}^2+2\mathrm{k}-1\right)+\mathrm{xy}\left(-2{\mathrm{k}}^2+3\mathrm{k}-4\right)+\left(3{\mathrm{k}}^2-2\mathrm{k}-4\right){\mathrm{y}}^2=0\dots \left(3\right)\\ \end{array}$
Now (3) represents the pair of lines $\overline{\mathrm{OA} },\overline{\mathrm{OB}}$
$\text{ Since the lines }\overline{\mathrm{OA}},\overline{\mathrm{OB}}\text{ are perpendicular then a+b=0}$
$\begin{array}{l} \\ \Rightarrow 2{\mathrm{k}}^2+2\mathrm{k}-1+3{\mathrm{k}}^2-2\mathrm{k}-4=0\\ \Rightarrow 5{\mathrm{k}}^2=5\\ \Rightarrow {\mathrm{k}}^2=1\\ \therefore \mathrm{k}=\pm 1\end{array}$