Question
Prove that the family of lines represented by $\text{x}(1+\lambda)=\text{y}(2-\lambda)+5=0,$ $\lambda$ being arbitrary, pass through a fixed point. Also, find that point.

Answer

$\text{x}(1+\lambda)+\text{y}(2-\lambda)+5=0$
$\Rightarrow\text{x}+\text{x}\lambda+2\text{y}-\lambda\text{y}+5=0$
$\Rightarrow\lambda(\text{x}-\text{y})+(\text{x}+2\text{y}+5)=0$
$\Rightarrow(\text{x}+2\text{y}+5)+\lambda(\text{x}-\text{y})=0$
This is of the form $\text{L}_1+\lambda\text{L}_2=0$
So it represents a line passing through the intersection of x - y = 0 and x + 2y = -5.
Solving the two equations, we get $\Big(\frac{-5}{3},\frac{-5}{3}\Big)$ which is the fixed point through which the given family of lines passes for any value of $\lambda.$

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