Equation To The Pair Of Angle Bisectors

Question

Find the equation to the pair of angle bisectors of the pair of lines $(ax+by)^2=3(bx-ay)^2$.

Efforts: $$(ax+by)^2=3(bx-ay)^2$$ After simplifying, I got: $$x^2(a^2-3b^2)+8abxy+y^2(b^2-3a^2)=0$$ Now, what should I do next?

Answer 1

Taking the square root on both sides, we obtain $$ax+by=-\sqrt3(bx-ay) \quad \text{and} \quad ax+by=\sqrt3(bx-ay)$$ and the two lines are $$ (a+\sqrt3b)x + (b-\sqrt3 a)y = 0\quad\text{and}\quad (a-\sqrt3b)x + (b+\sqrt3 a)y = 0. $$ The equation for the bisectors is known to be $$ \frac{|(a+\sqrt3b)x + (b-\sqrt3 a)y|}{\sqrt{(a+\sqrt3b)^2+(b-\sqrt3 a)^2}}=\frac{|(a-\sqrt3b)x + (b+\sqrt3 a)y|}{\sqrt{(a-\sqrt3b)^2+(b+\sqrt3 a)^2}}, $$ which in fact is equivalent to $$ |(a+\sqrt3b)x + (b-\sqrt3 a)y|=|(a-\sqrt3b)x + (b+\sqrt3 a)y|. $$ This gives $$ bx -ay=0 \quad\text{or}\quad ax + by=0. $$

Answer 2

You can simplify and try the hard way. Here's what I will do: Here: $$(ax + by)^2 - 3*(bx − ay)^2 = 0$$ this is of the form $$a^2-b^2 = (a+b)(a-b)$$ => the two lines are $$(ax + by + sqrt(3)*bx -sqrt(3)*ay = 0)$$ and $$(ax + by -sqrt(3)*bx + sqrt(3)*ay = 0)$$