The equations of the bisectors of the lines 3 x-4 y-7-0 and 12 x-5 y-8-0 are-A- 21 x -27 y -131-0B- 21 x -27 y -131-0C- 99 x -77 y -51-0D- 99 x -77 y -51-0

The equations of the bisectors are : 3x 4y+732 nbsp;+ nbsp;( 4)2=±12x 5y 8122+( 5)2⇒3x 4y+75=±12x 5y 813⇒39x 52y+91=±60x 25y 40⇒21x+27y 131=0 nbsp; nbsp ...

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Standard XII

Mathematics

Bisectors of Angle between Two Lines

The equations...

Question

The equations of the bisectors of the lines 3x-4y+7=0 and 12x-5y-8=0 are:

21x+27y-131=0

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21x-27y+131=0

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99x-77y+51=0

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99x+77y-51=0

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Solution

The correct options are
A 21x+27y-131=0
C 99x-77y+51=0
The equations of the bisectors are :
$\frac{3 x - 4 y + 7}{\sqrt{3^{2} + {(- 4)}^{2}}} = \pm \frac{12 x - 5 y - 8}{\sqrt{12^{2} + {(- 5)}^{2}}} \Rightarrow \frac{3 x - 4 y + 7}{5} = \pm \frac{12 x - 5 y - 8}{13} \Rightarrow 39 x - 52 y + 91 = \pm (60 x - 25 y - 40) \Rightarrow 21 x + 27 y - 131 = 0 or 99 x - 77 y + 51 = 0$

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The equations of the bisectors of the lines 3x-4y+7=0 and 12x-5y-8=0 are:

The correct options are A 21x+27y-131=0 C 99x-77y+51=0The equations of the bisectors are : 3x−4y+7√32 + (−4)2=±12x−5y−8√122+(−5)2⇒3x−4y+75=±12x−5y−813⇒39x−52y+91=±(60x−25y−40)⇒21x+27y−131=0 or 99x−77y+51=0