When lines y=√3x and y= negative√3x +6 intersect at point A, and line y= negative√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 points of intersection, we set the two equations equal to each other:
√3x = -√3x + 6
2√3x = 6
x = 3
Substitute x back into the first equation to find y:
y = √3(3) = 3√3
So point A is (3, 3√3)
Next, to find point B, we set y = 0 in the equation y = -√3x + 6:
0 = -√3x + 6
√3x = 6
x = 2
So point B is (2, 0)
2. To find the distances OA, OB, and AB, we use the distance formula:
Distance OA = √((3-0)^2 + (3√3-0)^2)
Distance OA = √(9 + 27)
Distance OA = √36
Distance OA = 6
Distance OB = √((2-0)^2 + (0-0)^2)
Distance OB = √4
Distance OB = 2
Distance AB = √((3-2)^2 + (3√3-0)^2)
Distance AB = √(1 + 27)
Distance AB = √28
3. To find the area of triangle ABO, we can use the formula for the area of a triangle given by two sides and the included angle:
S = 0.5 * OA * OB * sin(θ)
Where θ is the angle between sides OA and OB. Since we have not been given an angle, we can instead use the formula for the area of a triangle given by its three sides, also known as the Heron's formula:
S = √(s(s - OA)(s - OB)(s - AB))
Where s = 0.5 * (OA + OB + AB)
Plugging in the values:
s = 0.5 * (6 + 2 + √28) = 4 + 0.5√28
S = √((4 + 0.5√28)(4 + 0.5√28 - 6)(4 + 0.5√28 - 2)(4 + 0.5√28 - √28))
S = √((4 + 0.5√28)(2 + 0.5√28)(2 - 0.5√28)(4 + 0.5√28 - √28))
S = √((16 - 14)(4 - 0.25√28)(4 + 0.5√28 - √28))
S = √(2 * 4 * (3.75)(1.5))
S = √(30)
Therefore, the area of triangle ABO is √30 units squared.