Answer:
x+ y = 0
and
-2x + 5y + 2 = 0
Step-by-step explanation:
given line
3x+5y+2=0
5x+3y-2=0
solution
when two line bisect each other then line equation of bisector is express as
[tex]\frac{|A1x+B1y+C1|}{\sqrt{A1^2+B1^2+C1^2}} = \frac{|A2x+B2y+C2|}{\sqrt{A2^2+B2^2+C2^2}}[/tex] .........................1
and here
A1 = 3
B1 = 5
C1 = 2
and
A2 = 5
B2 = 3
C2 = -2
so now put value in equation 1 we get
[tex]\frac{|3x+5y+2|}{\sqrt{3^2+5^2+2^2}} = \frac{|5x+3y-2|}{\sqrt{3^2+5^2+(-2)^2}}[/tex]
solve it we get
-2x + 5y + 2 = 0 ..........1
and
3x+5y+2 = - ( 5x+3y-2 )
solve it we get
8x + 8y = 0
x + y = 0 ................2
