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I need to find the equation of this reflected line:

enter image description here

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Use orthogonal projection matrices and reflection matrices. Hint: the orthogonal projection of a point $x$ onto a vector $v$ is $\frac{vv^T}{v^Tv}x$. The reflection of the point $x$ about the vector $v$ is $2\frac{vv^T}{v^Tv}x-x$. – Shiyu Sep 24 '11 at 5:26

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 $k_m$.

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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 $C_1$.

enter image description here

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