MHT CET · Maths · Straight Lines
If \(2 a+b+3 c=0\), then the line \(a x+b y+c=0\) passes through the fixed point that is
- A \(\left(\frac{2}{3}, \frac{1}{3}\right)\)
- B \((0,1)\)
- C \(\left(\frac{2}{3}, 0\right)\)
- D None of these
Answer & Solution
Correct Answer
(A) \(\left(\frac{2}{3}, \frac{1}{3}\right)\)
Step-by-step Solution
Detailed explanation
Given, \(2 a+b+3 c+0\)...(i)
and line, \(a x+b y+c=0\)
\(\Rightarrow 3 a x+3 b y+3 c=0\)...(ii)
On subtracting Eq. (i) from Eq. (ii), we get \((3 x-2) a+(3 y-1) b=0 \cdot a+0 \cdot b\)
On comparing both sides, we get
\(3 x-2=0 \Rightarrow x=\frac{2}{3}\)
and \(3 y-1=0 \Rightarrow y=\frac{1}{3}\)
\(\therefore\) Line, \(a x+b y+c=0\) passes through the fixed point \(\left(\frac{2}{3}, \frac{1}{3}\right)\).
and line, \(a x+b y+c=0\)
\(\Rightarrow 3 a x+3 b y+3 c=0\)...(ii)
On subtracting Eq. (i) from Eq. (ii), we get \((3 x-2) a+(3 y-1) b=0 \cdot a+0 \cdot b\)
On comparing both sides, we get
\(3 x-2=0 \Rightarrow x=\frac{2}{3}\)
and \(3 y-1=0 \Rightarrow y=\frac{1}{3}\)
\(\therefore\) Line, \(a x+b y+c=0\) passes through the fixed point \(\left(\frac{2}{3}, \frac{1}{3}\right)\).
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