Question

If $(b-c)x^2+(c-a)xy+(a-b)y^2=0$ is a perfect square, then the quantities $a,b,c$ are in

• A. $AP$
• B. $GP$
• C. $HP$
• D. $noneofthese$

Answer 1

The correct option is B $GP$

The given equation is,

$(b-c)x^2+(c-a)xy+(a-b)y^2=0$.

Now the expression on the right hand side will be perfect square if the discriminant of the equation is $0$.

Then we get,

$(c-a{)}^2-4(a-b\left)\right(b-c)$

Or, $(c^2+a^2-2ac)-4(ab-ac-b^2+bc)=0$

Or, $a^2+4b^2+c^2-4ab-4bc+2ca=0$

Or, $(a+c{)}^2-2.(a+c).2b+(2b{)}^2=0$

Or, $(a+c-2b{)}^2=0$

Or, $a+c-2b=0$

Or, $b-a=c-b$.

So $a,b,c$ are in A.P.