Let $x,y,z$ be coordinates in $\Bbb R^3$ and $\pi :(z=1) , \pi': (y=1)$ two hyperplanes. Write down the perspectivity $\phi : \pi \rightarrow \pi '$ from $ O =(0,0,0)$ in terms of coordinates $(x,y)$ on $\pi$ and $(x,z)$ on $\pi'$. Find and describe the points of $\pi$ where $\phi$ is not defined. Prove that $\phi$ takes a line $L \subset \pi$ to a line $L'= \phi(L) \subset \pi'$( with a single exception).
So I introduced points $P, Q$, with $P=( \alpha, \beta, 1) \in \pi$ and $Q=(\gamma,1, \mu) \in \pi'$. And I wanted to find a line through $(\alpha, \beta, 1) and (0,0,0)$. $PO$={$ratio's (\lambda \alpha: \lambda \beta: \lambda) $for each $\lambda\neq 0$}. Which is the same ratio as$( \alpha: \beta: 1)$. So PO intersects $\pi'$ when $\beta=1$. I don't really know if this is correct, and if so where to go from here
Let $(x,y,1)$ be an arbitrary point on $\pi$. The direction vector from $(0,0,0)$ to this point is $(x,y,1)$. So, the equation of a line through $(0,0,0)$ and $(x,y,1)$ is
$$r(t)=(xt,yt,t).$$
This line punches $\pi'$ if $yt=1$. That is, if $t=\frac1y$. If $y=0$ then there will not be a point but in the infinity.
Finally the mapping is
$$(x,y,1)\rightarrow \ \left(\frac xy,1,\frac1y\right)$$ if $y\not=0.$
If we map $(x,ax+b,1)$, then the lines through the origin and through this straight line will be in a plane (determined by this line and the origin). The intersection of this plane with $\pi'$ is a straight line.