Question

If the parabola $y=(a-b) x^{2}+(b-c) x+(c-a)$ touches the $x$-axis then the line $a x+b y+c=0$

• A. always passes through a fixed point
• B. represents the family of parallel lines
• C. is always perpendicular to $x$-axis
• D. always has negative slope

Answer 1

Answer: (A)

Solving equation of parabola with $x$-axis $(y=0)$, we get
$(a-b) x^{2}+(b-c) x+(c-a)=0$
which should have two equal values of $x$, as $x$-axis touches the parabola.
$\therefore (b-c)^{2}-4(a-b)(c-a)=0$
$\Rightarrow (b+c-2 a)^{2}=0$
$\Rightarrow (b+c-2 a)^{2}=0 $
$\Rightarrow-2 a+b+c=0$
Thus, $a x+b y+c=0$ always passes through $(-2,1)$.