find acute bisector between 2 lines

3x+4y=11,12x-5y=2

3x + 4y = 11, 12x - 5y = 2
Here we have to make the constant term positive.
So -3x -4y +11 = 0 and -12x + 5y +2 = 0
Now we have to check a1a2 + b1b2
So a1a2 + b1b2  = 36 -20 = 16 > 0
So the equation of bisector between two lines is
-3x -4y +119+16 =--12x +5y +2144+2513(-3x-4y+11) =-5(-12x + 5y +2)11x +3y -17 = 0Hence this is the required equation of bisector.

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