The angle bisector of L_1:a_1x+b_1y+c_1=0 and L_2:a_2x+b_2y+c_2=0 (a_i,b_i,c_i) in R can be found be solving the equation (a_1x+b_1y+c_1)/(sqrt(a^2_1+b^2_1))=+-(a_2x+b_2y+c_2)/(sqrt(a^2_2+b^2_2)) But our teacher told us that the equation of the angle bisector pasing through the region containing the origin can be obtained by solving only the positive case of the equation given that (c_1,c_2)>0.How can we prove this?

Widersinnby7

Widersinnby7

Answered question

2022-11-03

The angle bisector of
L 1 : a 1 x + b 1 y + c 1 = 0
and
L 2 : a 2 x + b 2 y + c 2 = 0
( a i , b i , c i ) R
can be found be solving the equation a 1 x + b 1 y + c 1 a 1 2 + b 1 2 = ±   a 2 x + b 2 y + c 2 a 2 2 + b 2 2
But our teacher told us that the equation of the angle bisector pasing through the region containing the origin can be obtained by solving only the positive case of the equation given that ( c 1 , c 2 ) > 0.How can we prove this?

Answer & Explanation

Geovanni Shelton

Geovanni Shelton

Beginner2022-11-04Added 15 answers

| a 1 x + b 1 y + c 1 | a 1 2 + b 1 2 = | a 2 x + b 2 y + c 2 | a 2 2 + b 2 2
a 1 x + b 1 y + c 1 a 1 2 + b 1 2 = ±   a 2 x + b 2 y + c 2 a 2 2 + b 2 2
Put (0,0) into both sides of your equation:
c 1 a 1 2 + b 1 2 = ±   c 2 a 2 2 + b 2 2
You can see that both sides should yield nonnegative values, so we only choose + in ±.

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