Point of Intersection of the Pair

Point of intersection of the two lines represented by the general equation, obtained by solving ∂S/∂x = 0 and ∂S/∂y = 0 simultaneously. Valid when ab − h² ≠ 0 (lines are not parallel).

For $S \equiv ax^2+2hxy+by^2+2gx+2fy+c=0$ representing a pair of lines, the intersection point satisfies both lines simultaneously.

Method using partial derivatives: The point of intersection $(x_0, y_0)$ is where both lines meet. Differentiating $S$ partially:

$$\frac{\partial S}{\partial x} = 2ax+2hy+2g = 0 \implies ax+hy+g = 0$$ $$\frac{\partial S}{\partial y} = 2hx+2by+2f = 0 \implies hx+by+f = 0$$

Solving these two linear equations simultaneously:

$$\frac{x}{hf-bg} = \frac{y}{gh-af} = \frac{1}{ab-h^2}$$ $$x_0 = \frac{hf-bg}{ab-h^2}, \quad y_0 = \frac{gh-af}{ab-h^2}$$

Why partial derivatives? The equation $S = 0$ is homogeneous of degree 2 in $(x-x_0, y-y_0)$ when shifted to the intersection point. At the intersection, both first-order terms (in $x-x_0$ and $y-y_0$) must vanish, giving exactly $\partial S/\partial x = 0$ and $\partial S/\partial y = 0$.

Verification: Substitute $(x_0, y_0)$ back into $S = 0$. Since the intersection point lies on both lines, $S(x_0, y_0) = 0$ should hold — this provides an independent check.

When $ab = h^2$: The lines are parallel, $ab-h^2 = 0$, and the system has no solution (or infinitely many, for coincident lines). The intersection is at infinity.