If x/b+c-a=y/b-c-a=z/a+b-c=k then what is x² + y² + z² - 2xy - 2yz + 2zx equal to ?

Correct Answer is : k2(a+b+c)2k2(a + b + c)²k2(a+b+c)2

Correct Answer is : k2(a+b+c)2k2(a + b + c)²k2(a+b+c)2

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UPSC CDS 2 2021 Math Paper

Question : 25 of 100

Marks: +1, -0

\frac{x}{b+c-a}=\frac{y}{b-c-a}=\frac{z}{a+b-c}=k

x^{2} + y^{2} + z^{2} - 2xy - 2yz + 2zx

equal to ?

Solution:

Concept:

Recognize the expression

x^{2}+y^{2}+z^{2}-2xy-2yz+2zx

as the square of

(x-y+z)

. Factor it as

(x-y+z)^{2}

Explanation:

Given:

\frac{x}{b+c-a}=\frac{y}{b-c-a}=\frac{z}{a+b-c}=k

x = k(b+c-a)

y = k(b-c-a)

z = k(a+b-c)

We need

x^{2}+y^{2}+z^{2}-2xy-2yz+2zx

Observe:

x^{2}+y^{2}+z^{2}-2xy-2yz+2zx = (x-y+z)^{2}

Substitute the expressions:

x-y+z = k[(b+c-a) - (b-c-a) + (a+b-c)]

Simplify inside brackets:

(b+c-a) - (b-c-a) + (a+b-c) = b+c-a -b +c +a + a+b-c = a+b+c

x-y+z = k(a+b+c)

Hence

(x-y+z)^{2} = k^{2}(a+b+c)^{2}

Thus

x^{2}+y^{2}+z^{2}-2xy-2yz+2zx = k^{2}(a+b+c)^{2}

Answer:

k^{2}(a+b+c)^{2}

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