Image of a line on the projective plane

Question

Suppose the linear fractional transformation as $\lambda \mapsto \frac{a \lambda + b}{c \lambda + d}$. On a projective plane the origin corresponds to line $x = 1$ (so, $y = \lambda x$ is a projective line). But how to find a equation of the image line on this projective plane? So, graphically we have bijection of the origin line with the image line via projective line connection. And map the origin line $x = 1$ via linear operator $A$ where $$A = \begin{bmatrix} a & b\\ c & d \end{bmatrix}$$ and $$ \begin{pmatrix} y \\ x \end{pmatrix} = A^{-1} \begin{pmatrix} y' \\ x' \end{pmatrix}$$ does not produce a result

Answer 1

Let $\lambda' = \frac{a \lambda + b}{c \lambda + d}$. So then $\lambda = \frac{d \lambda' - b}{-c \lambda' + a}$.

We can express a point on the plane in vector form as $ \begin{pmatrix} y \\ x \end{pmatrix} = \begin{pmatrix} -b \\ a \end{pmatrix} + \lambda'\begin{pmatrix} d \\ -c \end{pmatrix} $.

The substitution of $\lambda'$ gives us $ \begin{pmatrix} y \\ x \end{pmatrix} = \begin{pmatrix} -b \\ a \end{pmatrix} + \frac{a \lambda + b}{c \lambda + d}\begin{pmatrix} d \\ -c \end{pmatrix} $. So, there is a first vector component express a starting point on the plane and a unit vector is the second. $\lambda'$ shows how the unit vector scaled to the intersection with the projective line $y = \lambda x$.

In general form the image of the origin line expressed as $c y + d x = ad - bc$.