Equation of the given line is $ax+by+c-1=0$

$\Rightarrow 20ax+20by+20c-20=0$     .....(i)

From the given relation, substituting value of $c$, we get,
$\mathrm{20 }ax+20by+t-5a-4b-20=0$

$\Rightarrow 20a\left(x-\frac{1}{4}\right)+20b\left(y-\frac{1}{5}\right)+\left(t-20\right)=0$

Clearly for $t=20,$ the given line will pass through the point $\left(\frac{1}{4},\frac{1}{5}\right)$ for all the values of $a \& b$