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If $2 a+b+3 c=0$, then the line $a x+b y+c=0$ passes through the fixed point that is
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Verified Answer
The correct answer is:
$\left(\frac{2}{3}, \frac{1}{3}\right)$
Given, $2 a+b+3 c+0$...(i)
and line, $\quad a x+b y+c=0$
$\Rightarrow \quad 3 a x+3 b y+3 c=0$...(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 $\quad 3 y-1=0 \Rightarrow y=\frac{1}{3}$
$\therefore$ Line, $a x+b y+c=0$ passes through the fixed point $\left(\frac{2}{3}, \frac{1}{3}\right)$.
and line, $\quad a x+b y+c=0$
$\Rightarrow \quad 3 a x+3 b y+3 c=0$...(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 $\quad 3 y-1=0 \Rightarrow y=\frac{1}{3}$
$\therefore$ Line, $a x+b y+c=0$ passes through the fixed point $\left(\frac{2}{3}, \frac{1}{3}\right)$.
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