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Equation of the bisector of the acute angle between the lines  \mathrm{3x-4y+7=0}

and  \mathrm{12x+5y-2=0}  is 

Option: 1

\mathrm{21 x+77 y-101=0}


Option: 2

\mathrm{21 x-77 y+101=0}


Option: 3

\mathrm{11 x-3 y+9=0}


Option: 4

\mathrm{11 x+3 y-9=0}


Answers (1)

best_answer

c

The given equations can be rewritten as 

\mathrm{3 x-4 y+7=0 \text { and }-12 x-5 y+2=0}

(making constant terms positive)

Since  \mathrm{a_1 a_2+b_1 b_2=-36+20<0},

\mathrm{\therefore }  positive sign in equation of bisectors gives the bisector of acute angle.

Hence acute angle bisector is

\mathrm{\frac{3 x-4 y+7}{\sqrt{\left(3^2+4^2\right)}}=\frac{-12 x-5 y+2}{\sqrt{\left(12^2+5^2\right)}}}

or  \mathrm{11 x-3 y+9=0}, which is given in (c).

Posted by

Ritika Kankaria

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