Given pair of lines are x² + 4xy + y² = 0
⇒ (y/x) ² + 4 y/x + 1 = 0
⇒ y/x = -4±2√3/2 = -2±√3,
∴ The lines y = (-2 + √3) x and y = (-2 - √3) x and x - y = 4 forms an equilateral triangle
Clearly the pair of lines x² + 4xy +y² = 0 intersect at origin,
The perpendicular distance form origin to x - y = 4 is the height of the
h = 2 √ 2
∵ Area of triangle = h²/√3 = 8/√3
Answer:
x^2-4x+y^2+6y-108=0
Step-by-step explanation:
Answer:
x=0 y=-11 (0,-11)
Step-by-step explanation:
1. take what y equals (3x-11) and plug it in for y in the second equation
Should look like this: O= 6x-2(3x-11)-22
2. Then distribute O=6x-6x+22-22
3. Then combine like terms O= 0 the 22's cancel out as well as the x's
x= 0
4. plug 0 in for x in the first equation y=3(0)-11
5. y= -11
6. x=0 y=-11 (0,-11)
The surface area is 6075.96 in^3
