Question

For the straight lines $4x+3y-6=0$ and $5x+12y+9=0$ the equation of the bisector of the obtuse angle between them is

Answer 1

The equations of the given straight lines are

$4x+3y-6=0$ .......(1)

$5x+12y+9=0$ .......(2)

The equation of the bisectors of the angles between lines (1) and (2) are

$\frac{4x+3y-6}{\sqrt{4^2+3^2}}=\pm \frac{5x+12y+9}{\sqrt{5^2+{12}^2}}or\frac{4x+3y-6}{5}=\pm \frac{5x+12y+9}{13}$

Taking the positive sign we have $\frac{4x+3y-6}{5}=\frac{5x+12y+9}{13}$

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

$27x-21y-123=0$

or $9x-7y-41=0$

Taking the negative sign we have $\frac{4x+3y-6}{5}=-\frac{5x+12y+9}{13}$

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

$77x+99y-33=0or7x+9y-3=0$

Hence the equation of the bisectors are

$9x-7y-41=0$ ........(3)

and $7x+9y-3=0$ ........(4)

Noe slope of line (1) = $-\frac{4}{3}$ and slope of the bisector (3) = $\frac{9}{7}$

If $\theta$ be the acute angle between the line (1) and the bisector (3) then

$\tan \theta =\left|\frac{m_1+m_2}{1+m_1{m}_2}\right|=\left|\frac{27+28}{21-36}\right|=\left|\frac{55}{-15}\right|=\frac{11}{3}>1$

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

$\theta >{45}^{\circ }$

Hence $9x-7y-41=0$ is the bisector of the obtuse angle between the given lines (1) and (2)