homogeneous coordinates I am to use homogeneous coordinates to calculate a standard matrix for a projection onto the line $4x-2y=6$ from the point $(3,10)$.
I'm not sure what homogeneous coordinates are and neither how to use them to calculate my problem? Please believe me, I've been looking for information on various sources!
 A: In 2D points in homogeneous coordinates have the form $P = (x,y,1)$ and lines $L=(a,b,c)$ such that the equation for the line can be found by
$$ L \cdot P = 0 \} a x + b y + c = 0 $$
So in your case, the homogeneous coordinates for the point is $P=(3,10,1)$ and the line $L=(4,-2,-6)$.
The trick with homogeneous coordinates is that if you multiply them with a scalar value, it does not change the underlying geometry. This is powerful be because the coordinates of a point $P=(a,b,c)$ are $(x,y) = (\frac{a}{c},\frac{b}{c})$. To test if a point belongs to a line simply do the dot product $P\cdot L=0$ and check for zero. The coordinates of a line joining two points $P$, $Q$ are
$$\begin{pmatrix} p_x\\p_y\\1 \end{pmatrix} \times \begin{pmatrix} q_x\\q_y\\1 \end{pmatrix} = \begin{pmatrix} p_y-q_y \\ q_x-p_x \\ p_x q_y - p_y q_x \end{pmatrix} $$
and similarly the point where two lines $L=(a,b,c)$ and $M=(u,v,w)$ meet is
$$\begin{pmatrix} a\\b\\c \end{pmatrix} \times \begin{pmatrix} u\\v\\w \end{pmatrix} = \begin{pmatrix} b w - c v \\ c u-a w \\ a v - b u \end{pmatrix} $$
The above has coordinates $(x,y) = \left( \frac{b w -c v}{a v - b u} , \frac{c u-a w}{a v - b u} \right)$ which if you do the calculations with vectors you will end up with the same value.
Futhermore, the minimum distance of the line $L=(a,b,c)$ to the origin is $d = -\frac{c}{\sqrt{a^2+b^2}}$ and the point closest to the origin has coordinates $P=(-a c,-b c, a^2+b^2) \equiv \left(-\frac{a c}{a^2+b^2},-\frac{b c}{a^2+b^2}\right)$
In addition, the minimum distance of the line $L=(a,b,c)$ to the point $P=(u,v,w)$ is
$$ {\rm dist}(L,P) = \frac{a u + b v + c w}{w \sqrt{a^2+b^2}} $$
and the projected point on the line has coordinates
$$ {\rm proj}(L,P) = \left( b( b u-a v)-a c, -a ( b u - a v)- b c, w (a^2+b^2) \right) $$
An affine 3×3 transfromation can be defined for translations and rotations.
$$  \begin{bmatrix} 1 & 0 & t_x \\ 0 & 1 & t_y \\ 0 & 0 & 1 \end{bmatrix} \begin{pmatrix} p_x\\p_y\\1 \end{pmatrix} =  \begin{pmatrix} p_x+t_x\\p_y+t_y\\1 \end{pmatrix}$$
$$  \begin{bmatrix} \cos\theta & -\sin\theta & 0 \\ \sin\theta & \cos\theta & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{pmatrix} p_x\\p_y\\1 \end{pmatrix} =  \begin{pmatrix} p_x \cos\theta-p_y \sin \theta\\p_x \sin\theta + p_y \cos\theta\\1 \end{pmatrix}$$
With these basics you can built what you need from here.
A: Since I don't have practice using this setup, this is probably not the most efficient solution, but here is what I learned by doing it.
Since this is about projecting onto a line not passing through the origin, it's most likely that the intended answer is to use $3\times 3$ matrices operating on homogenous coordinates for the plane.
The usual scheme is to identify the points of the plane with the coordinates of the form $\begin{bmatrix}x\\y\\1\end{bmatrix}$ for $x,y\in \Bbb R$, and have $3\times 3$ matrices that operate on these and perform transformations like translations, rotations, reflections and projections.
Here is the strategy. We will translate everything up 3 units so that the line passes through the origin. Then we will find a usual $2\times 2$ orthogonal projection matrix $P$ to project the matrix down onto the line, and then form the matrix $\begin{bmatrix}P&0\\0&1\end{bmatrix}$ which will be the $3\times 3$ transformation matrix that does this for our homogeneous coordinates. Then we will translate the picture back down $3$ units to restore the original position of the line and the point, and in doing so, we'll have the position of the projection of the point.
Translation matrices are easy: if you have your $2\times 1$ vector $v$ in $2$-space and you want to translate the homogenous picture with it, this is the matrix that accomplishes that:
$\begin{bmatrix}1&0&v_1\\0&1&v_2\\0&0&1\end{bmatrix}$.
So in our picture, the first transformation is given by $A=\begin{bmatrix}1&0&0\\0&1&3\\0&0&1\end{bmatrix}$
Next up is to compute $P$. A unit vector perpendicular to the line is $u=\begin{bmatrix}\frac{2}{\sqrt{5}}\\\frac{-1}{\sqrt{5}}\end{bmatrix}$, and so the projection for this vector is $P=I-u^Tu$.
Finally, find the matrix $C$ which translates everything back down $3$ units. (This is up to you: it's just like the first task.)
Applying these in order, you would use the matrix $CPA$ on the homogeneous coordinates. The question is "where does $\begin{bmatrix}3\\ 10\\ 1\end{bmatrix}$ map to?"
The answer is $\begin{bmatrix}\frac{29}{5}\\ \frac{43}{5}\\ 1\end{bmatrix}$, so that you can check your work.
A: A hint, not a numerical answer:
Imagine representing 2D points with 3D vectors with third component set to 1. In this form, you can write the line as a homogeneous expression $\vec{r}\cdot\vec{n}=0$ where $\vec{n}=(4,-2,6)$. This is a 3D plane through origin. Now you use projection in 3D, which shouldn't be a problem (it's subtraction of projection along the plane's normal). Don't forget the normalized normal.
Transformed points have to be renormalized to have $z=1$ in order to read the 2D coordinates from the first two components (divide the vector by its third component).
