Question

For the straight lines $4x+3y–6=0$ and $5x+12y+9=0$, find the Bisector of the angle which contains $(0,0)$

Answer 1

Equations of bisectors of the angles between the given lines are

$\frac{4x+3y-6}{\sqrt{4^2+3^2}}=\pm \frac{5x+12y+9}{\sqrt{5^2+{12}^2}}$

$\Rightarrow \frac{4x+3y-6}{\sqrt{16+9}}=\pm \frac{5x+12y+9}{\sqrt{25+144}}$

$\Rightarrow \frac{4x+3y-6}{\sqrt{25}}=\pm \frac{5x+12y+9}{\sqrt{169}}$

$\Rightarrow \frac{4x+3y-6}{5}=\pm \frac{5x+12y+9}{13}$

$\Rightarrow 52x+39y-78=\pm (25x+60y+45)$

$\Rightarrow 52x+39y-78=25x+60y+45,52x+39y-78=-25x-60y-45$

$\Rightarrow 27x+21y-123=0$ and $77x+99y-33=0$

$\Rightarrow 9x+7y-41=0$ and $7x+9y-3=0$

If $\theta$ is the acute angle between the line $4x+3y–6=0$ and the bisector $9x–7y–41=0,$ then

$\tan \theta =\left|\frac{\frac{-4}{3}-\frac{9}{7}}{1+(-\frac{4}{3})\frac{9}{7}}\right|=\left|\frac{-28-27}{21-36}\right|=\frac{55}{15}=\frac{11}{3}>1$

Hence the bisector of the angle containing the origin is

$\frac{-4x-3y+6}{\sqrt{{(-4)}^2+{(-3)}^2}}=\frac{5x+12y+9}{\sqrt{5^2+{12}^2}}$

or $\frac{-4x-3y+6}{\sqrt{25}}=\frac{5x+12y+9}{\sqrt{169}}$

or $\frac{-4x-3y+6}{5}=\frac{5x+12y+9}{13}$

or $-52x-39y+78=25x+60y+45$

or $-77x-99y+33=0$

or $7x+9y-3=0$