The equation of a line reflected about another line I need to find the equation of this reflected line:

 A: The General Formula for finding the equation of the line when the line $Ax+By=C$ is reflected at the line $Dx+Ey=F$. 
$$y-h=(\frac{AE^2-AD^2-2BDE}{BE^2-BD^2+2ADE})(x-k)$$
where
$$k=\frac{CE-BF}{AE-BD}$$
$$h=\frac{CD-AF}{BD-AE}$$
$h$ and $k$ is simply the intersection of the lines although if the two lines are parallel then it $h=0, k=0$,
This is easily derived by the use of perpendicular slopes and midpoint formula.
A: The equation for your reflected line can be constructed using the point-slope form, $y=m(x-x_Q)+y_Q$. The point $(x_Q,y_Q)$ is easily obtained as the intersection of your "mirror" line (the blue one) and the line to be reflected (the solid red one). This leaves the problem of the slope. To get you started, recall that if a line has slope $m$ and is at an angle $\phi$ from the $x$-axis, then $m=\tan\,\phi$. You can try using the sum/difference formulae for the tangent to derive the slope of the reflected line (the dashed red one) from the slopes $k_n$ and $k_m$.
A: Let's observe picture below. First of all note that $k_1=\tan\theta_1$ ; $k_2=\tan\theta_2$ and $k_3=\tan\theta_3$
Since dashed red line is reflection of the solid red line we may write next equality: $\alpha=\beta$ $\Rightarrow \theta_2-\theta_1=\theta_3-\theta_2$ . Now, if we apply $\tan$ on the both sides we get following: $\tan(\theta_2-\theta_1)=\tan(\theta_3-\theta_2)$ $\Rightarrow \frac{\tan\theta_2-\tan\theta_1}{1+\tan\theta_2\tan\theta_1}=\frac{\tan\theta_3-\tan\theta_2}{1+\tan\theta_3\tan\theta_2}$$\Rightarrow \frac{k_2-k_1}{1+k_2k_1}=\frac{k_3-k_2}{1+k_3k_2}$, so we can find $k_1$ from the last equation. Since $y_1=k_1x_1+C_1$ and we know that point $Z(x_q,y_q)$ belongs to the dashed red line we may write $y_1=k_1x_1+C_1$ and find $C_1$.

