Question

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

Answer 1

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

$\Rightarrow x+x\lambda +2y-\lambda y+5=0$

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

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

This is of the form $L_1+\lambda L_2=0$

So it represents a line passing through the intersenction of $x-y=0$ and $x+2y=-5$

Solving the two equations,

we get $(\frac{-5}{3},\frac{-5}{3})$ which is the fixed point through which the given family of lines passes for any value of $\lambda .$