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  • Equation of the bisector of the acute angle between the lines <img alt="\mathrm{3x - 4y + 7 = 0 }" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B3x%20-

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


Option: 2

\mathrm{21 \mathrm{x}-77 \mathrm{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

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

 

 

Posted by

Rishabh

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