When lines y=square root by 3x and y= negative square root by 3x +6 intersect at point A, and line y= negative square root by 3x +6 and the x axis intersect at point B, solve the following problems.
1. Find the points of intersection A and B
2. Find the distances OA,OB and AB
3. Find the area S of triangle ABO
1.
To find the point of intersection A, we set the two equations equal to each other:
√(3x) = -√(3x) + 6
√(3x) + √(3x) = 6
2√(3x) = 6
√(3x) = 3
3x = 9
x = 3
Substitute x back into one of the equations to find the y-coordinate:
y = -√(3(3)) + 6
y = -√9 + 6
y = -3 + 6
y = 3
Therefore, point A is (3, 3).
To find the point of intersection B, we set the equation equal to zero:
-√(3x) + 6 = 0
√(3x) = 6
3x = 36
x = 12
Therefore, point B is (12, 0).
2.
To find the distances OA and OB, we use the distance formula:
Distance OA = √[(x₂ - x₁)² + (y₂ - y₁)²]
Distance OA = √[(3 - 0)² + (3 - 0)²]
Distance OA = √[9 + 9]
Distance OA = √18
Distance OB = √[(12 - 0)² + (0 - 0)²]
Distance OB = √[144]
Distance OB = 12
3.
To find the area of triangle ABO, we can use the formula for the area of a triangle given its three vertices:
S = 0.5 * |x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)|
S = 0.5 * |3(0 - 0) + 12(3 - 3) + 0(3 - 0)|
S = 0.5 * |0 + 0 + 0|
S = 0
Therefore, the area of triangle ABO is 0.