Equation of pair of straight lines is $$7x^2+4xy+4y^2=0$$ What will be new equation of straight lines if the coordinate axes become the angle-bisector of above two lines.
My Try:
$$7x^2+4xy+4y^2=0$$ $$4\left(\frac yx\right)^2+4\left(\frac yx\right)+7=0$$ Solving above quadratic equation for $y/x$ we get $$\frac yx=\frac{-4\pm\sqrt{4^2-4\cdot4\cdot 7}}{2\cdot 4}$$ $$\frac yx=\frac{-1\pm i\sqrt{6}}{2}$$ I get imaginary roots of above equation which doesn't represent a pair of real straight lines. I do not know where is my mistake. Please help me solve this problem.
The problem is that the equation $$ 7x^2+4xy+4y^2=0 $$ does not represent a pair of straight line in $\mathbb R$, but just a point.
Infact you can write: \begin{gather} 7x^2+4xy+4y^2=0\\ \left(\sqrt 7 x + \frac{2}{\sqrt 7}y\right)^2 + \left(2\frac{\sqrt 6}{\sqrt 7} y\right)^2 = 0 \end{gather} And only the point $(0,0)$ satisfies this equation.
Your argument in infact correct when $x\neq 0$ and the result is coherent: there are no points in $\mathbb R^2\ \backslash \ \{(0,0)\}$ that satisfies the equation.