Question

22. Show that the two pair of lines 3x² + 8xy – 3y² = 0 and
3x² +8xy – 3y2 + 2x - 4y-1=0 form a square​

Answer 1

Answer:

the answer is ²34 ⁵67 ⁶⅘

it could have different or difficult tqs

Answer 2

Answer:

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