X - co-ordinates of vertices of a square of unit area are roots of equation x^2 - 3 |x| + 2 = 0 and

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9.Straight Line

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The $x -$ co-ordinates of the vertices of a square of unit area are the roots of the equation $x^2 - 3 |x| + 2 = 0$ and the $y -$ co-ordinates of the vertices are the roots of the equation $y^2 - 3y + 2 = 0$ then the possible vertices of the square is/are :

A

$(1, 1), (2, 1), (2, 2), (1, 2)$

B

$(- 1, 1), (- 2, 1), (- 2, 2), (- 1, 2)$

C

$(2, 1), (1, - 1), (1, 2), (2, 2)$

D

$(A)$ or $(B)$ both

Solution

$x^{2}-3|x|+2=0$

$\Longrightarrow(|x|-1)(|x|-2)=0 \therefore x=-1,1,2,-2$

$y^{2}-3 y+2=0$

$\Longrightarrow(y-1)(y-2)=0 \Rightarrow y=1,2$

The square is of unit radius. Hence, the coordinates of the square can be:

(1,1),(1,2),(2,1),(2,2)

(-1,1),(-1,2),(-2,2),(-2,1)

Standard 11

Mathematics

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