Let $l_1$ and $l_2$ be the representations in homogeneous coordinates of two lines in the plane. How could you express the fact that these two lines are orthogonal?
Deduce that, in general, the image by a homography of two orthogonal lines are not orthogonal
$Idea :$ So in 2d a line is represented as $ax +by +c = 0$ so I'm thinking that, in homogeneous coordinates I can express that fact as
$$ l^T p = 0 $$
for $l = (a, b, c)$ and $p = (x, y , 1)$ and use the fact that, in 2d the orthogonality condition is that the slopes multiply $-1$ but I'm stacking here.
Lines in the projective plane $\mathbb{RP}^2$ are planes through $0$ in $\mathbb{R}^3$. The homogeneous coordinates of such a plane are given by $ax + by+cz = 0$ and the vector $(a,b,c) \in \mathbb{R}^3$ is a normal vector of the plane.
Two lines in $\mathbb{RP}^2$ are orthogonal if and only if the corresponding planes in $\mathbb{R}^3$ are orthogonal if and only if the normal vectors of the planes are orthogonal. Two vectors in $\mathbb{R}^3$ are orthogonal if their scalar product is $0$.
So the projective lines $a_1 x + b_1 y + c_1 z = 0$ and $a_2 x + b_2 y + c_2 z = 0$ are orthogonal if and only if $a_1a_2 + b_1b_2 + c_1c_2 = 0$.
For a homography that does not preserve orthogonality one can take a linear map whose matrix is not orthogonal, such as $$ A = \left(\begin{matrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}\right). $$ The orthogonal projective lines given by $x=0$ and $y=0$ with normal vectors $(1,0,0)$ and $(0,1,0)$ are then sent to $x=0$ and $x+y = 0$ with normal vectors $(1,0,0)$ and $(1,1,0)$. These are not orthogonal anymore.