Show that the pair of straight lines
i) 6x² – 5xy – 6y² = 0 and 6x²–5xy – 6y²+ x + 5y–1= 0 forms a square.
ii) 3x² + 8xy – 3y² = 0 and 3x²+ 8xy – 3y² + 2x – 4y – 1 = 0 form a square

Given pair of lines are
$$\begin{gathered} 6{\mathrm{x}}^2-5\mathrm{xy}-6{\mathrm{y}}^2=0 ...\left(1\right) \\ 6{\mathrm{x}}^2-5\mathrm{xy}-6{\mathrm{y}}^2+\mathrm{x}+5\mathrm{y}-1=0 ...\left(2\right) \end{gathered}$$
In (1) and (2) coefficient of x² + coefficient y²
= 6 – 6 = 0
$\therefore$Equation (1) represents a pair of perpendicular lines passing through origin & equation (2) represents a pair of perpendicular  lines parallel to (1)
$\therefore \left(1\right)\text{ and (2) form a rectangle.let }\mathrm{OABC}\text{ be the }$$\text{ rectangle such that }\left(1\right)\text{ represents }\overline{\mathrm{OA}},\overline{\mathrm{OC}}$ and (2)$\text{ represents }\overline{\mathrm{BA}},\overline{\mathrm{BC}}$
$\text{ Equation of }\overline{\mathrm{AC}}\text{ is }\left(2\right)-\left(1\right)=0$
i.e x + 5y – 1 = 0
$\text{ slope of }\overline{\mathrm{AC}}\text{ is }-\frac{1}{5}\left(\because \mathrm{m}=-\frac{\mathrm{a}}{\mathrm{b}}\right)$
comparing equation (2)
with ${\mathrm{ax}}^2+2\mathrm{hxy}+{\mathrm{by}}^2+2\mathrm{gx}+2\mathrm{by}+\mathrm{c}=0$
we get
$$\begin{gathered} \mathrm{a}=6,\mathrm{b}=-6,\mathrm{c}=-1 \\ \begin{array}{c}and 2\mathrm{h}=-5\\ \Rightarrow \mathrm{h}=-\frac{5}{2}\end{array}, \\ \begin{array}{c}2\mathrm{g}=1\\ \Rightarrow \mathrm{g}=\frac{1}{2}\end{array}, \\ \begin{array}{c}2\mathrm{f}=5\\ \Rightarrow \mathrm{f}=\frac{5}{2}\end{array} \end{gathered}$$
Now B = point of intersection of (2)
$\begin{array}{l}=\left(\frac{\mathrm{hf}-\mathrm{bg}}{\mathrm{ab}-{\mathrm{h}}^2},\frac{\mathrm{gh}-\mathrm{af}}{\mathrm{ab}-{\mathrm{h}}^2}\right)\\ \mathrm{B}=\left(\frac{-\frac{5}{2}\times \frac{5}{2}-(-6)\left(\frac{1}{2}\right)}{6(-6)-(-5/2{)}^2},\frac{\frac{1}{2}\times \frac{-5}{2}-6\times \frac{5}{2}}{6(-6)-(-5/2{)}^2}\right)\\ =\left(\frac{-\frac{25}{4}+3}{-36-\frac{25}{4}},\frac{\frac{-5}{4}-15}{-36-{\displaystyle \frac{25}{4}}}\right.)\\ =\left(\frac{-25+12}{-144-25},\frac{-5-60}{-144-25}\right)=\left(\frac{-13}{-169},\frac{-65}{-169}\right)\\ \Rightarrow \mathrm{B}=\left(\frac{1}{13},\frac{5}{13}\right)\end{array}$
$\text{ Slope of }\overline{\mathrm{OB}}\text{ is }=\frac{5/13-0}{1/13-0}=5$
$$\begin{gathered} \text{ (Slope of }\overline{\mathrm{AC}}\text{ ) (slope of }\overline{\mathrm{OB}}\text{ ) }=-\frac{1}{5}\times 5=-1 \\ \therefore \overline{\mathrm{AC}}\perp \overline{\mathrm{OB}} \end{gathered}$$
$$\begin{gathered} \Rightarrow \mathrm{diagonals} \mathrm{of} \mathrm{the} \mathrm{rectangle} \mathrm{OABC} \mathrm{are} \mathrm{perpendicular} \\ \Rightarrow \mathrm{OABC} \mathrm{is} \mathrm{a} \mathrm{square} \end{gathered}$$
ii) Given pair of lines are $3{\mathrm{x}}^2+8\mathrm{xy}-3{\mathrm{y}}^2=0$ ...(1)
In (1) and (2) coefficient of x² + coefficient of y²
= 3 – 3 = 0
Equation (1) represents a pair of perpendicular lines passing through origin and equation (2) represents a pair perpendicular of lines parallel to (1)
$\therefore \text{ (1) and (2) forms a rectangle }$
Let OABC be the rectangle such that (1) represents $\overline{\mathrm{OA}},\overline{\mathrm{OC}} , \left(2\right) \mathrm{represents} \overline{\mathrm{BA}},\overline{\mathrm{BC}}$
$\text{ Equation to }\overline{\mathrm{AC}}\text{ is }\left(2\right)-\left(1\right)=0$
2x – 4y – 1 = 0
$\text{ Slope of }\overline{\mathrm{AC}}=\frac{-2}{-4}=\frac{1}{2}[\because \text{ slope }=-\mathrm{a}/\mathrm{b}]$
Comparing equation (2) with
${\mathrm{ax}}^2+2\mathrm{hxy}+{\mathrm{by}}^2+2\mathrm{gx}+2\mathrm{fy}+\mathrm{c}=0$
$\text{ we get }\mathrm{a}=3,\mathrm{b}=-3,\mathrm{c}=-1,2\mathrm{h}=8\Rightarrow \mathrm{h}=4$
$2\mathrm{g}=2\Rightarrow \mathrm{g}=1;2\mathrm{f}=-4\Rightarrow \mathrm{f}=-2$
Now B = point of intersection of (2)
$\begin{array}{l}=\left(\frac{\mathrm{hf}-\mathrm{bg}}{\mathrm{ab}-{\mathrm{h}}^2},\frac{\mathrm{gh}-\mathrm{af}}{\mathrm{ab}-{\mathrm{h}}^2}\right)\\ =\left(\frac{4\times -2-(-3)\left(1\right)}{3(-3)-(4{)}^2},\frac{1\left(4\right)-3(-2)}{3(-3)-(4{)}^2}\right)\\ =\left(\frac{-8+3}{-9-16},\frac{4+6}{-9-16}\right)=\left(\frac{-5}{-25},\frac{10}{-25}\right)=\left(\frac{1}{5},\frac{-2}{5}\right)\end{array}$
$\text{ Slope of }\mathrm{OB}\text{ is }=\frac{\left(-\frac{2}{5}-0\right)}{1/5-0}=-2$
$\left(\text{ slope of }\mathrm{AC}\right)\left(\text{ slope of }\mathrm{OB}\right)=\frac{1}{2}\times -2=-1$
$\therefore \mathrm{AC}\perp \mathrm{OB}$
$$\begin{gathered} \therefore \mathrm{diagonals} \mathrm{of} \mathrm{the} \mathrm{rectangle} \mathrm{OABC} \mathrm{are} \mathrm{perpendicular} \\ \therefore \mathrm{the} \mathrm{given} \mathrm{lines} \mathrm{form} \mathrm{a} \mathrm{square} \end{gathered}$$