Reflection of $x=1$ about $x+y=1$

Question

A ray of light travels along the line $x=1$ and gets reflected by a mirror on $x+y=1$. Find the equation of the reflected ray.

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I am to solve this problem using only $\tan\theta=|\dfrac{m_1-m_2}{1+m_1m_2}|$ where $m_1$ and $m_2$ are the slopes of the lines between which the angle $\theta$ is subtended.$$$$

I know one way of solving it is to find any general point on $x=1$, find its image about $x+y=1$ and find the equation of the reflected ray using the 2 point form(once we have the point of intersection of $x=1$ and $x+y=1$. However, this method is not allowed by my teacher.

$$$$I would be grateful for any assistance. Many thanks.

Answer 1

Depending on the direction of the ray we get $(-\infty,1]$ or $[1,\infty)$:

Answer 2

Not quite sure if this is the answer you're looking for, but perhaps you could approach it this way. Intuitively we know that the reflection would be the perpendicular of the line $x=1$. So we shall aim to show that algebraically.

For convenience, we shall rewrite $x+y=1$ as $y=-x+1$.

Let $\theta$ be the angle between $y=-x+1$ and $x=1$, and let $m_1, m_2$ be the slopes of the lines $y=-x+1$ and $x=1$ respectively. Now we define a slope $m_3$, which will become the slope of the reflected line. Since angle of incidence equals angle of reflection, we have the following equation necessarily.

\begin{align} \tan \theta &= \left \lvert \frac{m_1 - m_2}{1+ m_1 m_2} \right \rvert = \left \lvert \frac{m_1 - m_3}{1+ m_1 m_3} \right \rvert \end{align}

Now with $m_1 = -1$, we equate the following: \begin{align} \left \lvert \frac{(-1)- m_2}{1- m_2} \right \rvert = \left \lvert \frac{(-1)- m_3}{1-m_3} \right \rvert \end{align}

After some algebraic manipulation, we have $m_2 \cdot m_3 = -1$, verifying that the reflected line is the perpendicular of $x=1$. Since it passes through the point $(1,0)$, the equation of the reflected line must be $y=0$.