I have 2 points X,Y (for example [5,10] and [20,30]) and I need to interpolate points between these 2 points in order that all this points are spaced by 1 measurement unit.
Let's pretend I am using cm (as my measurement unit) and I have a point at [5,10] and another at [20,30]. How can I know the first point in this interpolation so it's spaced only 1cm from [5,10]? Walking 1cm each step, I would like to know every coordinate of points till I reach the last point [20,30].
3 Answers
Given two points $A$ and $B$, this answer will put points on the line segment between $A$ and $B$ so that the first point is $1$ unit from $A$, the second point is $2$ units from $A$, and so forth until the last point, which is a whole number of units from $A$ and one unit or less from $B$.
Suppose the Cartesian $(x,y)$ coordinates of the points are $A = (x_A, y_A)$ and $B = (x_B, y_B)$. Let $d$ be the distance between these two points; by the Pythagorean Theorem, $$d = \sqrt{(x_A - x_B)^2 + (y_A - y_B)^2}.$$
Since $B$ is at a distance $d$ from $A$, to move $d$ units from $A$ toward $B$ we add $x_B - x_A$ to $x_A$ and $y_B - y_A$ to $y_A$ to get the new $(x,y)$ coordinates. To move just $1$ unit we want to move $\frac1d$ times as far, that is, the point $1$ unit from $A$ is $(x_1,y_1)$ where \begin{align} x_1 = x_A + \frac1d(x_B - x_A),\\ y_1 = y_A + \frac1d(y_B - y_A). \end{align} The next point is at $\frac2d$ of the distance from $A$ to $B$, the next at $\frac3d$ the distance, and so forth. In general the $n$th point that we place along the segment from $A$ to $B$ should be at coordinates $(x_n,y_n)$ where \begin{align} x_n = x_A + \frac nd(x_B - x_A),\\ y_n = y_A + \frac nd(y_B - y_A). \end{align} We do this for each integer $n$ such that $1 \leq n < d$.
The line between $(5,10)$ and $(20,30)$ is $y=\frac{10-30}{5-20}(x-5)+10$ and the distance between the points is $\sqrt{(5-20)^2+(10-30)^2}=25$, so your problem reduces to putting points on the line at $1$ point per unit for $25$ units. Clearly this means putting down $24$ points.
I'll show you how you can find the first point. Since it's $1$ unit away from $(5,10)$, we can call it $(x,y)$ and say $\sqrt{(x-5)^2+(y-10)^2}=1$. We can then use our equation for the line and substitute to say $\sqrt{(x-5)^2+\left[(\frac{4}{3}(x-5)+10)-10)^2\right]}=1$, which you can easily solve for $x$ with a bit of algebra. I'm sure you can do the rest by yourself.
The other answers are fine but I prefer to use more affine formalism.
In an affine space, we can see a point as the result of the external operation point + vector = point. The points you are searching are the chain of results starting from A and adding a vector colinear to XY and , as stated, of length 1.
Then, here, we have to compute the vector XY and to divide it by its norm to get a length of one unit. For convenience, let's rename X and Y : A and B.
$\vec {AB} = \vec {( B_x - A_x, B_y - A_y )}$
vector $norm(\vec {AB} ) = (( B_x - A_x )^2 + ( B_y - A_y )^2)^{1/2}$
Then our unit vector on the line is
$\vec u = \vec {( \frac{B_x - A_x}{(( B_x - A_x )^2 + ( B_y - A_y )^2)^{1/2}}, \frac{B_y - A_y}{(( B_x - A_x )^2 + ( B_y - A_y )^2)^{1/2}} )}$
To get the next point from the current point , you add $\vec u$ until you stop on or else after the target.
Remains a last question : will the target be in these steps or between 2 stop points ? Merely, its depends of the length of AB ( the original XY ) : it will be fine only if the length may be expressed by an integer times the length unit.
Finally, the searched points are $ P_0 = A$ , $P_1 = A + \vec u$ , $P_{12} = A + 12 * \vec u$ , etc
Application with $A_x=5 , Ay=10 , B_x = 20$ and $B_y = 30$ :
$norm(\vec {AB} ) = (( B_x - A_x )^2 + ( B_y - A_y )^2)^{1/2} = (( 20 - 5 )^2 + ( 30 - 10 )^2)^{1/2} = 25$ which is an integer. Fine, the target will be reached exactly.
$\vec u = \vec {( \frac{B_x - A_x}{25}, \frac{B_y - A_y}{25} )} = \vec {( \frac{15}{25}, \frac{20}{25} )} = \vec {( \frac{3}{5}, \frac{4}{5} )}$
and finaly compute the points with the point + vector operation. For example $P_{12} = A + 12 * \vec u = (5,10) + \vec {( \frac{12 \times 3}{5}, \frac{12 \times 4}{5} )} = (5+\frac{12 \times 3}{5},10+\frac{12 \times 4}{5}) $
