If the parabola `y=(a-b)x^2+(b-c)x+(c-a)` touches x- axis then the line `ax+by+c=0`

Options

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

Answer

A

Explanation

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.
`:. (b-c)^(2) -4(a-b) (c-a) =0`
`rArr (b+c -2a)^(2) =0`
`rArr (b+c -2a)^(2) =0`
`rArr -2a +b + c=0`
Thus, `ax +by +c =0` always passes through `(-2,1)`.