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Question: If \((b - c){x^2} + (c - a)xy + (a - b){y^2} = 0\) is a perfect square, then the quantities a,b,c ar...

If (b−c)x2+(c−a)xy+(a−b)y2=0(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. None of these

Explanation

Solution

We have given an equation given that the equation is a perfect square, we know that the quadratic equation that are perfect square or we can say that has equal roots has its determinant equal to zero. Using this we’ll solve the equation to find a relation in a,b,c to find which type of progression they are forming.

Complete step by step answer:

Given data: (b−c)x2+(c−a)xy+(a−b)y2=0(b - c){x^2} + (c - a)xy + (a - b){y^2} = 0
We know that the determinant of a perfect square is equal to zero
If we take the quadratic equation on ‘x’
Then the determinant of the given equation will be
D=((c−a)y)2−4(b−c)(a−b)y2D = {\left( {(c - a)y} \right)^2} - 4(b - c)(a - b){y^2}
And we know that D=0D = 0
⇒((c−a)y)2−4(b−c)(a−b)y2=0\Rightarrow {\left( {(c - a)y} \right)^2} - 4(b - c)(a - b){y^2} = 0
⇒(c−a)2y2−4(b−c)(a−b)y2=0\Rightarrow {(c - a)^2}{y^2} - 4(b - c)(a - b){y^2} = 0
Dividing the whole equation by y2{y^2}
⇒(c−a)2−4(b−c)(a−b)=0\Rightarrow {(c - a)^2} - 4(b - c)(a - b) = 0
Now, using (A−B)2=A2+B2−2AB{\left( {A - B} \right)^2} = {A^2} + {B^2} - 2AB
⇒c2+a2−2ac−4(ab−b2−ac+cb)=0\Rightarrow {c^2} + {a^2} - 2ac - 4(ab - {b^2} - ac + cb) = 0
Simplifying the brackets
⇒c2+a2−2ac−4ab+4b2+4ac−4cb=0\Rightarrow {c^2} + {a^2} - 2ac - 4ab + 4{b^2} + 4ac - 4cb = 0
⇒c2+a2+2ac−4ab+4b2−4cb=0\Rightarrow {c^2} + {a^2} + 2ac - 4ab + 4{b^2} - 4cb = 0
Using A2+B2+2AB=(A+B)2{A^2} + {B^2} + 2AB = {\left( {A + B} \right)^2}
⇒(c+a)2−4b(a+c)+4b2=0\Rightarrow {(c + a)^2} - 4b(a + c) + 4{b^2} = 0
⇒(c+a)2−2×2b(a+c)+(2b)2=0\Rightarrow {(c + a)^2} - 2 \times 2b(a + c) + {(2b)^2} = 0
Using A2+B2−2AB=(A−B)2{A^2} + {B^2} - 2AB = {\left( {A - B} \right)^2}
⇒(c+a−2b)2=0\Rightarrow {(c + a - 2b)^2} = 0
Taking the square root
⇒c+a−2b=0\Rightarrow c + a - 2b = 0
⇒c+a=2b\Rightarrow c + a = 2b
From this, we can say that a,b,ca,b,care in AP as the sum of the first and third term is equal to the twice of the second term
Option(A) is correct.

Note: In the above solution we have taken the quadratic equation in ‘x’ but if take the quadratic equation in ‘y’ answer will remain the same as
If we take the quadratic equation on ‘y’
Then the determinant of the given equation will be
D=((c−a)x)2−4(b−c)(a−b)x2D = {\left( {(c - a)x} \right)^2} - 4(b - c)(a - b){x^2}
And we know that D=0D = 0
⇒((c−a)x)2−4(b−c)(a−b)x2=0\Rightarrow {\left( {(c - a)x} \right)^2} - 4(b - c)(a - b){x^2} = 0
⇒(c−a)2x2−4(b−c)(a−b)x2=0\Rightarrow {(c - a)^2}{x^2} - 4(b - c)(a - b){x^2} = 0
Dividing the whole equation by x2{x^2}
⇒(c−a)2−4(b−c)(a−b)=0\Rightarrow {(c - a)^2} - 4(b - c)(a - b) = 0
Now, using (A−B)2=A2+B2−2AB{\left( {A - B} \right)^2} = {A^2} + {B^2} - 2AB
⇒c2+a2−2ac−4(ab−b2−ac+cb)=0\Rightarrow {c^2} + {a^2} - 2ac - 4(ab - {b^2} - ac + cb) = 0
Simplifying the brackets
⇒c2+a2−2ac−4ab+4b2+4ac−4cb=0\Rightarrow {c^2} + {a^2} - 2ac - 4ab + 4{b^2} + 4ac - 4cb = 0
⇒c2+a2+2ac−4ab+4b2−4cb=0\Rightarrow {c^2} + {a^2} + 2ac - 4ab + 4{b^2} - 4cb = 0
Using A2+B2+2AB=(A+B)2{A^2} + {B^2} + 2AB = {\left( {A + B} \right)^2}
⇒(c+a)2−4b(a+c)+4b2=0\Rightarrow {(c + a)^2} - 4b(a + c) + 4{b^2} = 0
⇒(c+a)2−2×2b(a+c)+(2b)2=0\Rightarrow {(c + a)^2} - 2 \times 2b(a + c) + {(2b)^2} = 0
Using A2+B2−2AB=(A−B)2{A^2} + {B^2} - 2AB = {\left( {A - B} \right)^2}
⇒(c+a−2b)2=0\Rightarrow {(c + a - 2b)^2} = 0
Taking the square root
⇒c+a−2b=0\Rightarrow c + a - 2b = 0
⇒c+a=2b\Rightarrow c + a = 2b
From this, we can say that a,b,ca,b,cis in AP as the sum of the first and third term is equal to the twice of the second term.