As a little bit of background, I am self studying Introduction to Calculus and Analysis I by Courant and John. In the first chapter of the book he discusses the concept of considering functions as maps.
He provides several geometric examples. One is the projection from a point on a line, $x$, to a point on a parallel line, $y$, from another point that is in the same plane as the other two lines. Then he provides the map as $y = ax+b$. This result is simple to derive.
The second example is a map for a similar situation except the lines are not parallel. This map is provided as $y=\frac{ax+b}{cx+d}$. I have been trying to figure out how this result was arrived at but, I have been unsuccessful.
Case 1: Parallel lines
Case 2: Lines not parallel
These figures are exactly as provided in the book. So the assignment of the points a, b, c, and d are left up to the reader.
I have looked at the fact that at the point of observation you will get equal angles that are opposite to each other at the intersection of the lines. However, I cannot found a use for that since the lines are not parallel.
In the book they refer to it as a "perspective" mapping which, led me to look into projective geometry. Where the point of intersection would be the view of two parallel lines from infinity. Implying that the points on the line would be getting infinitely closer to each other as you approach the point of intersection. However, I am unsure of how to apply it to this problem or if it is even necessary.
Thank you for any assistance.
Let $m$, $n$ be the distances of point $O$ from lines $x$, $y$. Let then $x_0$, $y_0$ be the coordinates of points $H$, $K$ and $x$, $y$ the coordinates of corresponding points $P$, $Q$ (see picture). Then we have: $$ x-x_0=m\tan\theta,\quad y-y_0=n\tan(\theta+\alpha)=n{\tan\theta+\tan\alpha\over1-\tan\alpha\tan\theta}. $$ Substitute now $\tan\theta=(x-x_0)/m$ into the second formula above, to get: $$ y-y_0=n{x-x_0+m\tan\alpha\over m-\tan\alpha(x-x_0)}. $$ Rearranging this gives the desired result.