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Question - $L_1⇒2x+y-1=0$ , $L_2⇒2x-y+3=0$. Find the equation of angle bisector passing through quadrant containing $(2,3).$

Effort $⇒$ Found out first angle bisector $b_1⇒y=2$ , second angle bisector $b_2⇒-\dfrac{1}{2}$

Parity check for $L_1 ⇒2(2)+3-1=6>0$

Parity check for $L_2⇒2(2)-(3)+3=4>0$

Therefore $(2,3)$ is in $(+,+)$ with respect to $L_1$and $L_2$.

What to do next? I am stuck here , can someone help?I only know that we check $(0,0)$ after this. But how to get angle bisector after that? Thanks.

Aleph
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1 Answers1

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Start with finding the point of intersection: $y=1-2x, y=2x+3$. Point of intersection is $(-\frac{1}{2},2)$. Note that the slopes are opposite so one bisector is vertical and one is horizontal. The required bisector is $y=2$.

Vasili
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  • That's the correct answer. It's easy to see it in a plot. But can you prove it in a general case? – Andrei Feb 04 '22 at 20:58
  • Since slopes are tangents of angles between axis $x$ and lines, opposite slopes means that one angle is $\alpha$ and the other one is $\pi-\alpha$. Now let's consider a triangle formed by the lines and axis $x$. It will have two angles equal to $\alpha$ so the bisector will be perpendicular to axis $x$ – Vasili Feb 04 '22 at 21:34
  • Can you find the answer using parity check because for complicated values your method might not help – Aleph Feb 05 '22 at 04:21