Question

Find the angle between the two straight lines $3x=4y+7$ and $5y=12x+6$ and also the equations to the two straight lines which pass through the point $(4,5)$ and make equal angles with the two given lines.

Answer 1

$3x=4y+74y=3x-7y=\frac{3}{4}x-\frac{7}{4}$

Slope of line ${=m}_1=\frac{3}{4}$

$5y=12x+6y=\frac{12}{5}x+\frac{6}{5}{m}_2=\frac{12}{5}$

Angle between two lines i.e. $\tan \theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|$

$\tan \theta =\left|\frac{\frac{3}{4}-\frac{12}{5}}{1+\frac{3}{4}.\frac{12}{5}}\right|=\left|\frac{\frac{15-48}{20}}{\frac{20+36}{20}}\right|\tan \theta =\left|\frac{-33}{56}\right|=\frac{33}{56}\Rightarrow \theta ={\tan }^{-1}\frac{33}{56}$

Let the slope of other line be $m$ and it makes an angle $\alpha$ with both the given lines

Angle with line having slope $m_1$

$\tan \theta =\left|\frac{m_1-m}{1+m_{1}m}\right|\tan \theta =\left|\frac{\frac{3}{4}-m}{1+\frac{3}{4}m}\right|=\left|\frac{3-4m}{4+3m}\right|.......\left(i\right)$

Angle with line having slope $m_2$

$\tan \alpha =\left|\frac{{m-m}_2}{1+m_{2}m}\right|\tan \alpha =\left|\frac{m-\frac{12}{5}}{1+\frac{12}{5}m}\right|=\left|\frac{5m-12}{5+12m}\right|.......\left(ii\right)$

From $\left(i\right)$ and $\left(ii\right)$

$\left|\frac{3-4m}{4+3m}\right|=\left|\frac{5m-12}{5+12m}\right|\Rightarrow \frac{3-4m}{4+3m}=\frac{5m-12}{5+12m}\Rightarrow 15-20m+36m-48m^2=20m-48+15m^2-36m\Rightarrow 63m^2-32m-63=0\Rightarrow 63m^2-81m+49m-63=0\Rightarrow 9m(7m-9)+7(m-9)=0\Rightarrow (9m+7)(7m-9)=0\Rightarrow m=\frac{9}{7},-\frac{7}{9}$

For $m=-\frac{7}{9}$ equation of line passing through $(4,5)$ is

$y-5=-\frac{7}{9}(x-4)9y-45=-7x+287x+9y-73=0$

For $m=\frac{9}{7}$ equation of line passing through $(4,5)$ is

$y-5=\frac{9}{7}(x-4)7y-35=9x-369x-7y-1=0$