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If \(3 a+5 b+6 c=0\) then the family of lines \(a x+b y+c=0\) pass through the fixed point
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2287 Upvotes
Verified Answer
The correct answer is:
\(\left(\frac{1}{2}, \frac{5}{6}\right)\)
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.
\(\begin{aligned}
& a x+b y+c=0 \\
& \because \quad 3 a+5 b+6 c=0, \text { so } a x+b y-\frac{3 a+5 b}{6}=0 \\
& \Rightarrow \quad 3 a(2 x-1)+b(6 y-5)=0
\end{aligned}\)
So, the family of lines passes through the fixed point \(\left(\frac{1}{2}, \frac{5}{6}\right)\).
Hence, option (d) is correct.
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