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Q.
If $5a+4b+20c=t$ , then the value of $t$ for which the line $ax+by+c-1=0$ always passes through a fixed point is
NTA AbhyasNTA Abhyas 2022
Solution:
Equation of the given line is $ax+by+c-1=0$
$\Rightarrow 20 \, ax+20 \, by+20 \, c-20=0\ldots \left(1\right)$
From the given relation, substituting value of $c$ , we get,
$20ax+20 \, by+t-5a-4b-20=0$
$\Rightarrow 20a\left(x - \frac{1}{4}\right)+20b\left(y - \frac{1}{5}\right)+\left(t - 20\right)=0$
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 $a\&b$