Question

find an equation of the line that bisects the acute angle formed by the graphs of 3x+5y+2=0 and 5x+3y-2=0

Answer 1

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

$\frac{|A1x+B1y+C1|}{\sqrt{A1^2+B1^2+C1^2}} = \frac{|A2x+B2y+C2|}{\sqrt{A2^2+B2^2+C2^2}}$      .........................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

$\frac{|3x+5y+2|}{\sqrt{3^2+5^2+2^2}} = \frac{|5x+3y-2|}{\sqrt{3^2+5^2+(-2)^2}}$    

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