Question

Show that the straight lines given by
$x(a+2b)+y(a+3b)=a+b$.
For different values of $a$ and $b$ pass through a fixed point.

Answer 1

The given equation can be put in the form
(a) $a(x+y-1)+b(2x+3y-1)=0$
or $(x+y-1)+\lambda (2x+3y-1)=0$
where $\lambda =b/a$
Above is of the form $P+\lambda Q$ and as such it represents a line through the intersection of the lines $P=0$ and $Q=0$
i.e. $x+y-1=0$ and $2x+3y-1=0$.
Solving the above, we get the point $(2,-1)$ which is fixed.