Question

Find the equations to the straight lines which pass through the point $(h,k)$ and are inclined at an angle ${\tan }^{-1}m$ to the straight line $y=mx+c$.

Answer 1

$y=mx+c$

Slope of line $=m$

Other line passes through $(h,k)$, let its slope be $m^{\prime }$

Angle between two lines i.e. $\tan \theta =\left|\frac{m_1-m_2}{1{+m}_1{m}_2}\right|$

$\tan \theta =\pm \frac{m-m^{\prime }}{1+m{m}^{\prime }}$

$\theta ={\tan }^{-1}(\pm \frac{m-m^{\prime }}{1+m{m}^{\prime }})$

${\tan }^{-1}m={\tan }^{-1}(\pm \frac{m-m^{\prime }}{1+m{m}^{\prime }})$

$m=\pm \frac{m-m^{\prime }}{1+m{m}^{\prime }}$

If $m=\frac{m-m^{\prime }}{1+m{m}^{\prime }}$

$m+m^2{m}^{\prime }=m-m^{\prime }$

$m^2{m}^{\prime }+m^{\prime }=0$

$m^{\prime }(m^2+1)=0$

$\Rightarrow m^{\prime }=0$

Equation of line with slope $m^{\prime }=0$ is

$y-k=0(x-h)y-k=0y=k.....\left(i\right)$

Also $m=-\frac{m-m^{\prime }}{1+m{m}^{\prime }}$

$m+m^2{m}^{\prime }=-m+m^{\prime }$

$2m=(1-m^2)m^{\prime }$

$\Rightarrow m^{\prime }=\frac{2m}{1-m^2}$

Equation of line with slope $m^{\prime }$ is

$y-k=\frac{2m}{1-m^2}(x-h)(1-m^2)(y-k)=2m(x-h)$

So, the equation of lines are $y=k$ and $(1-m^2)(y-k)=2m(x-h)$