(9am to 6pm)

Ask Questions, Get Answers

Want help in doing your homework? We will solve it for you. Click to know more.
Home  >>  JEEMAIN and NEET  >>  Mathematics  >>  Class11  >>  Coordinate Geometry

Find the equations of the bisectors of the angles between the straight lines $3x-4y+6=0$ and $12x+5y-3=0$.

$\begin{array}{1 1}(a)\;21x+77y-93=0,99x-77y+63=0\\(b)\;20x+77y-90=0,99x-77y+60=0\\(c)\;21x-77y+93=0,99x+77y+63=0\\(d)\;2x+70y-93=0,99x-70y+63=0\end{array}$

1 Answer

Need homework help? Click here.
$\large\frac{(3x-4y+6)}{\sqrt{3^2+(-4)^2}}$$=\pm \large\frac{(12x+5y-3)}{\sqrt{(12)^2+(5)^2}}$
$\large\frac{3x-4y+6}{5}=$$\pm \large\frac{(12x+5y-3}{13}$
Taking +ve sign
$21x+77y-93=0$(one of the bisector)
Taking -ve sign
Hence (a) is the correct answer.
answered Feb 4, 2014 by sreemathi.v

Related questions