Question

The acute angle bisector between the lines $3x-4y=0,5x+12y-26=0$ is

Answer 1

We have,

Equation of lines are

$3x-4y=0......\left(1\right)$

$5x+12y-26=0......\left(2\right)$

From equation (1) to,

$3x-4y=0$

$\Rightarrow 3x=4y$

$\Rightarrow 4y=3x$

$\Rightarrow y=\frac{3}{4}x$

On comparing that,

$y=m_{1}x$

Such that,

$m_1=\frac{3}{4}$

Now,

From equation (2) to,

$5x+12y=26$

$\Rightarrow 12y=26-5x$

$\Rightarrow y=\frac{-5}{12}x+\frac{26}{12}$

On comparing that,

$y=m_{2}x+C$

Now,

$m_2=\frac{-5}{12}$

So,

Using property of angle between two lines are

$\tan \theta =\frac{m_1\sim m_2}{1+m_1{m}_2}$

$\tan \theta =\frac{\frac{3}{4}+\frac{5}{12}}{1-\frac{3}{4}\times \frac{5}{12}}$

$\tan \theta =\frac{\frac{36+20}{48}}{\frac{48-15}{48}}$

$\tan \theta =\frac{56}{33}$

$\theta ={\tan }^{-1}\frac{56}{33}$

Acute angle is above.

Hence, this is the answer.