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

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Correct answer: (B) 20. Equation of the given line is ax+by+c−1=0 ⇒20 ax+20 by+20 c−20=0 .....(i) From the given relation, substituting value of c, we get, 20 ax+20 by+t−5a−4b−20=0 ⇒20 a( x−1/4 )+20 b( y−1/5 )+( t−20 )=0 Clearly for t=20, the given line will pass through the point ( 1/4 , 1/5 ) for all the values of a & b

If $5 a + 4 b + 20 c = t$ , then the value of $t$ for which the line $a x + b y + c - 1 = 0$ always passes through a fixed point is

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