find acute bisector between 2 lines

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 a

_{1}a

_{2}+ b

_{1}b

_{2}

So a

_{1}a

_{2}+ b

_{1}b

_{2}= 36 -20 = 16 > 0

So the equation of bisector between two lines is

$\frac{-3x-4y+11}{\sqrt{9+16}}=-\frac{-12x+5y+2}{\sqrt{144+25}}\phantom{\rule{0ex}{0ex}}\Rightarrow 13(-3x-4y+11)=-5(-12x+5y+2)\phantom{\rule{0ex}{0ex}}\Rightarrow 11x+3y-17=0\phantom{\rule{0ex}{0ex}}Hencethisistherequiredequationofbisector.$

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