Q.Prove that the straight lines joining the origin to the points of intersection of $3x^2 + 5xy - 3y^2 + 2x + 3y = 0$ and $3x - 2y - 1 = 0$ are at right angles.

The lines joining the origin to the two points where $S=0$ meets $L=0$ can be found, without ever solving for the intersection points, by homogenizing $S$ using $L$ (written so its constant term is $1$) — the result is exactly the combined equation of those two lines through the origin.

Step 1. Write the line with constant term $1$.

Call $L=3x-2y$, so $L=1$ on the line.

Step 2. Homogenize the curve $S:3x^2+5xy-3y^2+2x+3y=0$. The quadratic terms are already degree $2$ and are left alone; the linear terms $2x+3y$ are degree $1$, so we multiply them by $L^{2-1}=L$ (i.e. replace the implicit "$1$" that would multiply a constant, using $L=1$) to raise them to degree $2$:

(There is no constant term in $S$, so nothing else needs adjusting.)

Step 3. Expand $(2x+3y)(3x-2y)$.

Step 4. Add to the quadratic part of $S$. …

• Homogenizing with $L=3x-2y-1$ (constant $-1$, not $+1$) instead of first rewriting the line as $3x-2y=1$ …