geometry
Suppose all I am given, is the coordinates of two points like the following:
What are some ways I could go about finding the values of every point on this line segment? Like the y-value at 2.3, 2.4, 2.7 etc.
Any suggestions as to how I could go about doing this would be appreciated.
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.
This is the problem of finding the intersection of a straight line and a circle, as commented by J.M.. The more elementary method using analytical geometry, without rotate or translate the coordinate axes (which would make the computation easier$^1$), although not being a compact one, is the following (see sketch).
The equation defined by points $ S(N,J)$ and $T(M,I)$ is given by $$ y-J=m(x-N),\qquad m=\frac{I-J}{M-N}\tag{1}. $$ The equation of the circle centered at $R$ with radius $d=\overline{RQ}$ is $$ (x-L)^{2}+(y-H)^{2}=d^{2}.\tag{2} $$ You need to solve the following system $$ \left\{ \begin{array}{c} y-J=m(x-N) \\ (x-L)^{2}+(y-H)^{2}=d^{2}, \end{array}\tag{3} \right. $$ which is equivalent to $$ \left\{ \begin{array}{c} x=\frac{y-J+mN}{m} \\ \left(\frac{y-J+mN}{m}-L\right)^{2}+(y-H)^{2}=d^{2}.\tag{4} \end{array} \right. $$ Solving the quadratic equation yields (with the help of SWP): $$ y=\frac{1}{ m^{2}+1 }\left( -mN+Lm+J+m^{2}H\pm \sqrt{\Delta}\right), \tag{5} $$ where the discriminant is
$$\begin{eqnarray*} \Delta &=&A+B, \\ \text{with } A &=&-m^{4}N^{2}+m^{4}d^{2}-m^{4}L^{2}-m^{2}J^{2}-m^{2}H^{2}+d^{2}m^{2}-2m^{3}NH, \\ B &=&2Lm^{3}H+2Jm^{2}H+2m^{4}NL-2m^{3}JL+2m^{3}JN.\tag{6} \end{eqnarray*}$$
The information $\overline{RT}<\overline{RQ}<\overline{RS}$ will define the signal of the term $ \pm \sqrt{\Delta}$. The coordinates of $Q$ are $O=x,K=y$.
$^1$By making the translation $X=x-L$ and $Y=y-H$, and computing the new coordinates of the points in this $X,Y$ system, the above formulae simplify (it is equivalent to set $L=H=0$ in them). In the end they should be convert back to the original $x,y$ system.
Best Answer
Here is another way. Choose $t \in [0,1]$, then let $p(t) = (x_1+ t (x_2-x_1), y_1+t(y_2-y_1))$.
Then $p(0) = (x_1,y_1)$, $p(1) = (x_2,y_2)$ and for $t \in (0,1)$, $p(t)$ will be a point in between.
This scheme works even if the two $x$ coordinates are the same.