Mathematics - Straight Lines Question with Solution | TestHub

First, find the intersection point of $L_1$ and $L_2$. Multiply $L_1$ by 4: $12x-16y+20=0$. Subtract this from $L_2$: $(12x+5y-13)-(12x-16y+20)=0⟹21y-33=0⟹y=11/7$. Substitute $y=11/7$ into $L_1$: $3x-4(11/7)+5=0⟹3x-9/7=0⟹x=3/7$. The intersection point is $P(3/7,11/7)$.

Next, find the angle bisector containing the origin. For the origin to be in the angle, the constant terms of the lines must have the same sign after normalizing the coefficients. $L_1:3x-4y+5=0$. Constant term is $+5$. $L_2:12x+5y-13=0$. To make its constant term positive, we write it as $-12x-5y+13=0$. The equation of the angle bisector containing the origin is:$$\frac{3x-4y+5}{\sqrt{3^2+(-4{)}^2}}=\frac{-12x-5y+13}{\sqrt{(-12{)}^2+(-5{)}^2}}$$ $$\frac{3x-4y+5}{5}=\frac{-12x-5y+13}{13}$$ $13(3x-4y+5)=5(-12x-5y+13)⟹39x-52y+65=-60x-25y+65⟹99x-27y=0⟹11x-3y=0$. This is the equation of the diagonal. We can verify that $P(3/7,11/7)$ lies on this line: $11(3/7)-3(11/7)=33/7-33/7=0$.