Question

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

• A. $0$
• B. $20$
• C. $30$
• D. None of these

Answer 1

Answer: (B)

Equation of the given line is $ax+by+c-1=0$
$\Rightarrow 20ax+20by+20c-20=0\dots \left(1\right)$
From the given relation, substituting value of $c$ , we get,
$20ax+20by+t-5a-4b-20=0$
$\Rightarrow 20a\left(x-\frac{1}{4}\right)+20b\left(y-\frac{1}{5}\right)+\left(t-20\right)=0$
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 $a\&b$