Find the value of `lambda` the equation is `5x+y+lambda=0` where `x=1` and `y=5`

Answer

C

Explanation

Given that,
`a(b-c)x^(2)+b(c-a)xy+c(a-b)y^(2)` is perfect square
`:. b^(2)(c-a)^(2)=4a(b-c)*c(a-b)`
`implies b^(2)(c-a)^(2)=4ac(a-b)(b-c)`
`implies[a(b-c)+c(a-b)^(2)]=4ac(a-b)(b-c)`
`" " " "[:.a(b-c)+b(c-a)+c(a-b)=0]`
`implies [a(b-c)-c(a-b)]^(2)=0`
`implies a(b-c)-c(a-b)=0`
`implies ab-ac-ca+bc=0" " implies b(b+c)=2ac`
`implies b=(2ac)/(a+b)`
`implies a,b,c are in HP.