Fixed Point of a Family

The fixed point of the family is the intersection of L₁ = 0 and L₂ = 0, independent of λ.

The family $L_1 + \lambda L_2 = 0$ passes through the intersection $P = L_1 \cap L_2$ for all values of $\lambda$. This point $P$ is the fixed point of the family.

Finding the fixed point

Solve $L_1 = 0$ and $L_2 = 0$ simultaneously. The solution is $P$, which is independent of $\lambda$.

Standard JEE form

A family is often given as a single equation with a parameter:

$$(2 + 3\lambda)x + (1 - \lambda)y + (5 + 2\lambda) = 0$$

Regroup by $\lambda$:

$$(2x + y + 5) + \lambda(3x - y + 2) = 0$$

Fixed point: solve $2x + y + 5 = 0$ and $3x - y + 2 = 0$ simultaneously.

Adding: $5x + 7 = 0 \Rightarrow x = -7/5$. Then $y = -5 - 2x = -5 + 14/5 = -11/5$.

Fixed point: $\left(-\dfrac{7}{5}, -\dfrac{11}{5}\right)$.

Every member of the family — for every value of $\lambda$ — passes through this point.