The combined equation of bisector of angles between the lines $L_1$ and $L_2$ is
$$2x^2-3xy-2y^2-x+7y-3=0$$
$P(4,-3)$ is a point on $L_1$. If the equation of obtuse angle bisector is $ax+by-3=0$, then choose the correct option:
(A) $|a-b|=9$
(B)$a^2+b^2=45$
(C)$a^2-b^2=27$
(D) None of these
One method would be to find separate equations of line by comparing $(ax+by+1)(px+qy+1)=\frac{2x^2-3xy-2y^2-x+7y-3}{-3}=0$
But is there any better approach?
Notice, the line $L_1$ passes through the intersection point $(1, 1)$ of angle bisectors & the point $(4, -3)$ hence the equation of line $L_1$ $$y-1=\frac{1-(-3)}{1-4}(x-1)$$ $$4x+3y-7=0$$ plot the line $L_1$ & the angle bisectors: $x-2y+1=0$ & $2x+y-3=0$. One can easily find that the angle between line $L_1$ & bisector: $x-2y+1=0$ is greater than the angle between line $L_1$ & bisector: $2x+y-3=0$ hence, $x-2y+1=0$ is the obtuse angle bisector which can be re-written in intercept form $$\color{red}{\frac{x}{-1}+\frac{y}{\frac{1}{2}}=1}$$ comparing the above equation with $ax+by-3=0$ or $\color{blue}{\frac{x}{\frac{3}{a}}+\frac{y}{\frac{3}{b}}=1}$ one should find $$\frac{3}{a}=-1, \ \frac{3}{b}=\frac 12$$ or$$ \ \ a=-3, \ b=6$$ Now, one can easily check the given options by substituting values of $a$ & $b$ $$|a-b|=|-3-6|=9$$ & $a^2+b^2=(-3)^2+(6)^2=45$
both the options (A) & (B) are correct.