Could anyone point me to some equation to find the point p(s1,s2), which is the point of intersection of a line segment with points Q(x1,y1), which is outside the square and the center of the Square C(x2,y2).
Help appreciated.
Here's the image for reference
Best Answer
If $|x_1 - x_2| > |y_1 - y_2|$, then the line will intersect the square in one of the sides, left or right. Then if $x_1 > x_2$, the point of intersection is on the left, so the point will be some $(x,y)$ with $x = x_2 - 15$. Since it is on the line $(x_2,y_2) + \lambda (x_2 - x_1, y_2 - y_1)$ we know that for some $\lambda$, $x_2 - 15 = x_2 + \lambda(x_2 - x_1)$ and $y = y_2 + \lambda(y_2 - y_1)$. The first equality gives you $\lambda = 15/(x_1 - x_2)$ so that the second equality gives you $y$, i.e. $y = y_2 - 15\frac{y_1 - y_2}{x_1 - x_2}$. So the point of intersection is $(x_2 - 15, y_2 - 15\frac{y_1 - y_2}{x_1 - x_2})$.
You can do the same for $x_1 < x_2$, when you get $x = x_2 + 15$, while for $|x_1 - x_2| < |y_1 - y_2|$ you know that the intersection is either at the top or at the bottom of the square, giving a similar analysis.