AP ICET 2019 Shift 1 Paper

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Question : 78 of 200

Marks: +1, -0

If

5(x^2 + y^2 + z^2) = 4(xy + yz + zx)

then

x :y :z =

8:1:4
2:1:8
4:1:8
4:8:1
No answer

Solution:

5 x^{2}+5 y^{2}+5 z^{2}-4 x y-4 y z-4 z x=0

x^{2}-4 x y+4 y^{2}+y^{2}-4 y z+4 z^{2}+4 x^{2}-4 z x+z^{2}=0

(x-2y)^2 + (y-2z)^2 +(2x-z)^2 = 0

x-2y = 0; y-2z = 0; 2x-z = 0

x = 2y; y = 2z; 2x = z

{x}/{y}=2

....(1)

{y}/{z}=2

....(2)

{x}/{z}=1/2

...(3)
(1)×(2)

⇒x/y×y/z=2×2

⇒x/z=4

which is in contradiction to 3
∴ data is inconsistent.

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