show that the following pair of lines
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Given equations for the pairs of lines are :
6 x² -5 x y - 6 y² = 0
=> (3x + 2y)(2x - 3y) = 0
L1 : 3 x + 2y =0 Slope: -3/2
L2: 2 x - 3y = 0 Slope: 2/3
L1 and L2 are perpendicular clearly.
Their point of intersection is: Origin O (0,0).
Other pair of lines: 6 x² - 5xy - 6y² + x + 5y - 1 = 0
Clear the terms with x², xy and y² are same. It means that this pair of lines is parallel to the first pair of lines.
Factorize it to: (3x + 2y - 1) (2x - 3 y + 1) = 0
So L3 : 3x + 2y - 1 = 0 Slope: -3/2
L4 : 2x - 3y + 1 = 0 Slope = 2/3
Their point of intersection is: C(1/13, 5/13)
Since the distance between parallel lines, L1 & L3 = 1
the distance between parallel lines L2 and L4 = 1
Hence they form a square of side length = 1 unit.
6 x² -5 x y - 6 y² = 0
=> (3x + 2y)(2x - 3y) = 0
L1 : 3 x + 2y =0 Slope: -3/2
L2: 2 x - 3y = 0 Slope: 2/3
L1 and L2 are perpendicular clearly.
Their point of intersection is: Origin O (0,0).
Other pair of lines: 6 x² - 5xy - 6y² + x + 5y - 1 = 0
Clear the terms with x², xy and y² are same. It means that this pair of lines is parallel to the first pair of lines.
Factorize it to: (3x + 2y - 1) (2x - 3 y + 1) = 0
So L3 : 3x + 2y - 1 = 0 Slope: -3/2
L4 : 2x - 3y + 1 = 0 Slope = 2/3
Their point of intersection is: C(1/13, 5/13)
Since the distance between parallel lines, L1 & L3 = 1
the distance between parallel lines L2 and L4 = 1
Hence they form a square of side length = 1 unit.
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