Answer
See explanation
Work Step by Step
\[
P=\left[\begin{array}{cccc}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & -1 & 0
\end{array}\right]
\]
Because in the previous exercise, the matrix $\mathrm{P}$ that performs
the perspective projection is given in left
\[
S=\left[\begin{array}{ccc}
9 & 12 & 1.8 \\
3 & 8 & 2.7 \\
-5 & 2 & 1 \\
1 & 1 & 1
\end{array}\right]
\]
The homogeneous coordinates of the vertices of the triangle may be written as $(9,3,-5,1),(12,8,2,1),$ and $(1.8,2.7,1,1),$ therefor the data matrix for S is as given
$D=\left[\begin{array}{cccc}1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & -1 & 0\end{array}\right]\left[\begin{array}{ccc}9 & 12 & 1.8 \\ 3 & 8 & 2.7 \\ -5 & 2 & 1 \\ 1 & 1 & 1\end{array}\right]=\left[\begin{array}{ccc}9 & 12 & 1.8 \\ 3 & 8 & 2.7 \\ 0 & 0 & 0 \\ 1.5 & 0.8 & 0.9\end{array}\right]$
data matrix for the transformed triangle
$(9,3,0,1.5) \rightarrow(9 / 1.5,3 / 1.5,0 / 1.5)=(6,2,0)$
$(12,8,0,0.8) \rightarrow(12 / 0.8,8 / 0.8,0 / 0.8)=(15,10,0)$
$(1.8,2.7,0,0.9) \quad \rightarrow(1.8 / 0.9,2.7 / 0.9,0 / 0.9)=(2,3,0)$
the columns of this matrix may be converted from homogeneous coordinates by dividing by the final coordinate:
So the coordinates of the vertices of the transformed triangle are $(6,2,0),(15,10,0),$ and (2,3,0)
