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Fixed Point of ax+by+c=0 from Parabola Touching x-axis | JEE

JEE Maths question with a full step-by-step solution.

Question
If the parabola y=(a−b)x2+(b−c)x+(c−a)y = (a - b)x^2 + (b - c)x + (c - a) touches the xx-axis, then the line ax+by+c=0ax + by + c = 0
Aalways passes through a fixed pointcorrect
Brepresents the family of parallel lines
Cis always perpendicular to the xx-axis
Dalways has a negative slope
Solution
Step 1: Touching the xx-axis means equal roots, so the discriminant vanishes.
(b−c)2−4(a−b)(c−a)=(b+c−2a)2=0⇒b+c−2a=0.(b - c)^2 - 4(a - b)(c - a) = (b + c - 2a)^2 = 0 \Rightarrow b + c - 2a = 0.
 Step 2: The constraint −2a+b+c=0-2a + b + c = 0 has coefficients (−2,1,1)(-2, 1, 1), so ax+by+c=0ax + by + c = 0 always passes through the fixed point (−2,1)(-2, 1). Correct answer: (1)
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