perspective projection transformation matrix In one text,  the  derivation for  perspective  projection  goes like  this:   

If $x',y',z'$ represent any point along  the  projection line,and $x_{prp},y_{prp},z_{prp}$ are the projection reference point, then:

$$x'=x-(x-x_{prp})u$$
 $$y'=y-(y-y_{prp})u$$
$$z'=z-(z-z_{prp})u$$
where $0\le u \le 1$  

Now,   finding the value of u from the third equation above and  replacing in the first and  second  equations,   we get at  $z_vp$:

$$ x_p=x'= x.\frac{z_{vp}-z_{prp}}{z-z_prp}+x_{prp}.\frac{z-z_{vp}}{z-z_{prp}}$$
$$ y_p=y'= y.\frac{z_{vp}-z_{prp}}{z-z_prp}+y_{prp}.\frac{z-z_{vp}}{z-z_{prp}}$$

and there is no projection transformation matrix  given.  

In the second text,  the  derivation is using  similar triangles:

comparing the  similar triangles  we  get:
$$x_p=\frac{-x_c}{\frac{z_c}{d}}$$
$$y_p=\frac{y_c}{\frac{z_c}{d}}$$

and  the projection transformation for  homogeneous coordinates x,y,z is  given by:
$$
\begin{bmatrix}
-1&amp;&amp;0&amp;&amp;0&amp;&amp;0\\
0&amp;&amp;1&amp;&amp;0&amp;&amp;0\\
0&amp;&amp;0&amp;&amp;1&amp;&amp;0\\
0&amp;&amp;0&amp;&amp;\frac{1}{d}&amp;&amp;0
\end{bmatrix}
$$    

so,  my questions are:


How do I find the projection matrix of coordinates as found from 
derivation 1?
Both are given in the perspective projection area of
these different texts,  are both transformations the  same? If not, 
how  are  they different?
Given your equations (which i just assume are correct):
$$
\begin{align}
x_p &amp;= x \frac{z_{vp}-z_{prp}}{z-z_{prp}}+x_{prp} \frac{z-z_{vp}}{z-z_{prp}}\\
y_p &amp;= y \frac{z_{vp}-z_{prp}}{z-z_{prp}}+y_{prp} \frac{z-z_{vp}}{z-z_{prp}}
\end{align},
$$

if we take $(x_{prp}, y_{prp}, z_{prp}) = (0, 0, 0)$, then we have

$$
\begin{align}
x_p &amp;= x\frac{z_{vp}}{z}\\
y_p &amp;= y\frac{z_{vp}}{z}
\end{align}
$$

Assuming the projection plane is at distance $d$ from the origin, and, by your drawing, that the projection plane is on the negative side of the Z axis, we have $z_{vp} = -d$, which leads to

$$
\begin{align}
x_p &amp;= - x\frac{d}{z}\\
y_p &amp;= - y\frac{d}{z}
\end{align}
$$

Now we can follow the same reasoning (homogeneous coordinates) as in my answer to your other question (using column notation) to write

$$
\begin{align}
\begin{bmatrix}
x\\y\\z\\ \frac{-z}{d}
\end{bmatrix}
=
\begin{bmatrix}
1&amp;0&amp;0&amp;0\\
0&amp;1&amp;0&amp;0\\
0&amp;0&amp;1&amp;0\\
0&amp;0&amp;\frac{-1}{d}&amp;0
\end{bmatrix}
\begin{bmatrix}
x\\y\\z\\1
\end{bmatrix}
\end{align}
$$

which after the division by the last coordinate will give the projected point. Notice that because we are writing it in column notation, the perspective matrix is the transpose of the matrix in your other question. Also, the negative sign is because the camera reference frame is a right-handed one.

As for the second textbook that you are reading, I cannot understand why you would have a negative sign only in the x axis without knowing more about the camera reference frame. By the diagram you posted, there should not be a negative sign.