show that the lines x^-4xy+y^=0 and x+y=1 form an equilateral triangle and find its area
To show that the lines x^2 - 4xy + y^2 = 0 and x + y = 1 form an equilateral triangle, we need to use a geometric property of equilateral triangles.
The property we will use is that in an equilateral triangle, the distance between any two points on the triangle is the same. So, we need to find the distance between the points of intersection of the given lines.
First, we need to find the points of intersection by solving the system of equations formed by the two lines:
x^2 - 4xy + y^2 = 0 --- (1)
x + y = 1 --- (2)
We can solve equation (2) for x and substitute it into equation (1), or vice versa. Let's solve equation (2) for x:
x = 1 - y.
Now we substitute x = 1 - y into equation (1):
(1 - y)^2 - 4(1 - y)y + y^2 = 0
Simplifying this equation gives us:
1 - 2y + y^2 - 4y + 4y^2 + y^2 = 0
6y^2 - 6y + 1 = 0.
This is a quadratic equation in terms of y. We can solve this equation to find the values of y.
Using the quadratic formula, we have:
y = (-(-6) ± √((-6)^2 - 4(6)(1))) / (2(6))
y = (6 ± √(36 - 24)) / 12
y = (6 ± √12) / 12
y = (6 ± 2√3) / 12
y = (3 ± √3) / 6.
Now substitute these values of y back into equation (2) to find the corresponding x values:
For y = (3 + √3) / 6:
x = 1 - (3 + √3) / 6
x = (6 - (3 + √3)) / 6
x = (3 - √3) / 6.
For y = (3 - √3) / 6:
x = 1 - (3 - √3) / 6
x = (6 - (3 - √3)) / 6
x = (3 + √3) / 6.
So, the two points of intersection are:
A: (x₁, y₁) = ((3 - √3) / 6, (3 + √3) / 6)
B: (x₂, y₂) = ((3 + √3) / 6, (3 - √3) / 6).
Now, we need to find the distance between these two points to check if it is equal to the distance between the other two pairs of intersection points.
The distance between two points (x₁, y₁) and (x₂, y₂) is found using the distance formula:
d = √((x₂ - x₁)^2 + (y₂ - y₁)^2).
Let's calculate the distance between points A and B:
d = √(((3 + √3) / 6 - (3 - √3) / 6)^2 + ((3 - √3) / 6 - (3 + √3) / 6)^2)
d = √((2√3 / 6)^2 + (-2√3 / 6)^2)
d = √((4 * 3 / (6^2)) + (4 * 3 / (6^2)))
d = √((12 / 36) + (12 / 36))
d = √(24 / 36)
d = √(2/3).
Now we need to verify if this distance is equal to the distance between points A and C, and between points B and C.
To find the coordinates of point C, we need to find the intersection of line x = 1 - y and line x + y = 1. Substituting x = 1 - y into x + y = 1:
(1 - y) + y = 1
1 - y + y = 1
1 = 1.
This equation is true for any value of y. Therefore, the line x = 1 - y and the line x + y = 1 are coincident, and point C is the point of intersection of these two lines.
Hence, points A, B, and C are the vertices of the equilateral triangle.
To find the area of the equilateral triangle, we can use the formula:
Area = (sqrt(3) / 4) * s^2.
Since the triangle is equilateral, all three sides have the same length as the distance between any two points.
So, the area of the equilateral triangle formed by the lines x^2 - 4xy + y^2 = 0 and x + y = 1 is:
Area = (sqrt(3) / 4) * (2/3)^2
Area = (sqrt(3) / 4) * (4/9)
Area = sqrt(3) / 9 square units.
If that's x^2 - 4xy + y^2 = 0, that means that
y = (2+√3)x and y = (2-√3)x
Find where those lines intersect y = -x + 1
A: (2+√3)x = 1-x
x = 1/(3+√3)
y = 1-x = (2+√3)/(3+√3)
B:(2-√3)x = 1-x
x = 1/(3-√3)
y = 1 - x = (2-√3)/(3-√3)
After a little algebra, we find that
|OA|2 = 2/3
|OB|2 = 2/3
|AB|^2 = √6/3
The altitude = 1/√2, so the area of OAB = 1/2 * 1/√2 * √6/3 = √3/6