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I have a list of points $S$ in the form of $(p, q)$:

$$begin{align}
S = &(43, 58), (44, 60), (40, 60), (41, 61), \
&(46, 60), (40, 57), (53, 62), (50, 61)
end{align}$$

And I wish to center them on the origin $(0, 0)$. I would do this by subtracting from them the midpoints ($(bar{p}, bar{q})$) for each dimension:

$$begin{align}
bar{p} &= frac{p_1 + p_2 + dots + p_n}{n} \
bar{q} &= frac{q_1 + q_2 + dots + q_n}{n}
end{align}$$

I find $bar{p} = 44.625$ and $bar{q} = 59.875$. I find my new $S$ to be:

$$begin{align}
S_{text{new}} = &(-1.625, -1.875), (-0.625, 0.125), (-4.625, 0.125), (-3.625, 1.125), \
&(1.375, 0.125), (-4.625, -2.875), (8.375, 2.125), (5.375, 1.125)
end{align}$$

Using linear regression, I've found the line of best fit for this data set which crosses the origin to be $y = 0.26x + 0$. This is the line in which I want to project points of data onto from right angles.

My question is, how do I find these projected points (marked as red dots)? Taking point $(1.375, 0.125)$, I can make a triangle with vertices at the origin, the point, and the projected point like so:

I know the slope of $c$ ($0.26$), the position of vertex $ba$ ($(1.375, 0.125)$), and position of vertex $ca$ ($(0, 0)$), but how do I find the position of vertex $cb$?

This is for principal component analysis. To find the eigenvalue, I need the sum of squared distances from projected points to the origin. I've already found the eigenvector to be $begin{bmatrix}0.96 \ 0.25end{bmatrix}$.

One way to do this is by calculating the euclidean vector of the blue line, in this case it is $begin{bmatrix} 1 \ 0.26 end{bmatrix}$, you want its norm to be $1$ so you divide it by its norm to get: $v = begin{bmatrix} 0.97 \ 0.25 end{bmatrix}$.

Then see every point as vector and to get the coordinates of point A one the blue line you just have to calculate $(A cdot v) cdot v$.

For example, for the point $(-1.625, −1.875)$, you would find:

$$left( begin{bmatrix} -1.625 \ -1.875 end{bmatrix} cdot begin{bmatrix} 0.97 \ 0.25end{bmatrix}right) cdot begin{bmatrix} 0.97 \ 0.25 end{bmatrix} = begin{bmatrix} -1.98 \ -0.51 end{bmatrix}$$

The other answer to totally correct, but I would like to add a bit more explanation. What you would like to do is to transform the original data to a new coordinate frame. Let the new coordinate frame be represented by the vectors $textbf{v}$ and $textbf{w}$, as shown in the image below. Because $textbf{v}$ and $textbf{w}$ are perpendicular, their dot product is zero: $textbf{v} cdot textbf{w} = 0$. Furthermore, I assume that $textbf{v}$ is a unit vector (which is not the case in the image due to my bad drawing skills...), so $textbf{v} cdot textbf{v}=1$.

Any datapoint $textbf{x}$ can be represented in the new coordinate frame:
$$
textbf{x} = a textbf{v} + b textbf{w}.
$$

As you are interested in the position on the line, you are only interested in $a$. To get $a$, we multiply the equation on both sides with $textbf{v}$:
$$
textbf{x} cdot textbf{v} = a textbf{v} cdot textbf{v} + b textbf{w} cdot textbf{v} = a,
$$
where I used $textbf{v} cdot textbf{w} = 0$ and $textbf{v} cdot textbf{v}=1$.

The only thing you now need to do is to get the position on the line in the original frame by simply multiplying $a$ with $textbf{v}$, i.e., $(textbf{x} cdot textbf{v}) textbf{v}$.

In your case, $textbf{v} = frac{1}{sqrt{1+0.26^2}} begin{bmatrix} 1 \ 0.26end{bmatrix} approx begin{bmatrix} 0.96 \ 0.25 end{bmatrix}$, and this results in:

-1.9787   -0.5145

-0.5550   -0.1443

-4.3017   -1.1184

-3.1215   -0.8116

 1.3184    0.3428

-5.0323   -1.3084

 8.3622    2.1742

 5.3086    1.3802