Show that two pairs of lines 3x2 + 8xy - 3y2 = 0 and 3x2 + 8xy - 3y2 + 2x - 4y -1= 0 forms a square
Answers
Given : two pair of lines 3x² + 8xy – 3y² = 0 and 3x² + 8xy – 3y² + 2x – 4y - 1 = 0
To Find : Show these lines form a square
Solution:
3x² + 8xy – 3y² = 0
=> 3x² + 9xy -xy – 3y² = 0
=> 3x(x + 3y) - y(x + 3y) = 0
=> (3x - y)(x + 3y) = 0
3x - y = 0
x + 3y = 0
3x² + 8xy – 3y² + 2x – 4y - 1 = 0
=> (3x - y - 1) (x + 3y + 1) = 0
3x - y - 1 = 0
x + 3y + 1 = 0
3x - y = 0 and x + 3y = 0
=> ( 0, 0)
3x - y - 1 = 0 x + 3y + 1 = 0
(0.2 , -0.4)
3x - y = 0 and x + 3y + 1 = 0
=> (-0.1 , -0.3)
x + 3y = 0 and 3x - y - 1 = 0
(0.3 , -0.1)
3x - y = 0 , x + 3y = 0 Slopes are 3 , -1/3 ( hence perpendicular )
3x - y - 1 = 0 , x + 3y + 1 = 0 Slopes are 3 , -1/3 ( hence perpendicular )
( 0, 0) , (0.3 , -0.1) , (0.2 , -0.4) , (-0.1 , -0.3)
Using distance formula
Each side length = √0.1
Hence it forms a square
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