Family of Lines Through Intersection of Two Lines

For any value of λ, this represents a line passing through the intersection of L₁: a₁x+b₁y+c₁=0 and L₂: a₂x+b₂y+c₂=0.

Let $L_1 \equiv a_1x+b_1y+c_1=0$ and $L_2 \equiv a_2x+b_2y+c_2=0$ intersect at $P$.

Why $L_1 + \lambda L_2 = 0$ passes through $P$ for all $\lambda$

At $P$: $L_1(P) = 0$ and $L_2(P) = 0$. Therefore:

$$L_1(P) + \lambda L_2(P) = 0 + \lambda \cdot 0 = 0$$

for every value of $\lambda$. So every member of the family passes through $P$.

What the family covers

$(a_1+\lambda a_2)x + (b_1+\lambda b_2)y + (c_1+\lambda c_2) = 0$ is linear in $x$ and $y$ for all $\lambda$ — it is always a straight line. As $\lambda$ varies over $\mathbb{R}$, every line through $P$ is obtained except $L_2$ itself (which corresponds to $\lambda \to \infty$).

Finding a specific member

To find the member satisfying an extra condition, write $L_1 + \lambda L_2 = 0$ and impose the condition to get one equation in $\lambda$. Solve, substitute back. The intersection $P$ need never be computed explicitly.

Fixed point of a parametric family

If a family is given as $(p + \lambda q)x + (r + \lambda s)y + (u + \lambda v) = 0$, regroup:

$$(px + ry + u) + \lambda(qx + sy + v) = 0$$

This is $L_1 + \lambda L_2 = 0$ with $L_1: px+ry+u=0$ and $L_2: qx+sy+v=0$. The fixed point is $L_1 \cap L_2$ — independent of $\lambda$.