Prove that: `|a b a x+b y b c b x+c y a x+b y b x+c y0|=(b^2-a c)(a x^2+2b x y+c y^2)`

Options

A. zero
B. positive
C. negative
D. `b^(2) + ac`

Answer

C

Explanation

We have,
`Delta = |(a,b,ax + by),(b,c,bx + cy),(ax + by,bx + cy,0)|`
`rArr Delta = |(a,b,ax + by),(b,c,bx + cy),(0,0,-(ax^(2) + 2bxy + cy^(2)))| " " [("Applying " R_(3) rarr),(R_(3) - xR_(1) - yR_(2))]`
`rArr Delta = (b^(2) - ac) (ax^(2) + 2bxy + cy^(2))`
Now, `b^(2) - ac lt 0 and a gt 0`
`rArr` Discriminant of `ax^(2) + 2bxy + cy^(2)` is negative and `a gt 0`
`rArr ax^(2) + 2bxy+ cy^(2) gt 0` for all x,`y in R`
`rArr Delta = (b^(2) -ac) (ax^(2) + 2bxy + cy^(2)) lt 0`