If \(3 a+5 b+6 c=0\) then the family of lines \(a x+b y+c=0\) pass through the fixed point
If \(3 a+5 b+6 c=0\) then the family of lines \(a x+b y+c=0\) pass through the fixed point
\(\left(\frac{5}{6}, \frac{1}{2}\right)\)
\(\left(\frac{1}{2}, \frac{1}{3}\right)\)
\(\left(\frac{1}{3}, \frac{1}{2}\right)\)
\(\left(\frac{1}{2}, \frac{5}{6}\right)\)
Solution
Given equation of family of lines
\(\begin{aligned}
& a x+b y+c=0 \\
& \because \quad 3 a+5 b+6 c=0, \text { so } a x+b y-\frac{3 a+5 b}{6}=0 \\
& \Rightarrow \quad 3 a(2 x-1)+b(6 y-5)=0
\end{aligned}\)
So, the family of lines passes through the fixed point \(\left(\frac{1}{2}, \frac{5}{6}\right)\).
Hence, option (d) is correct.