1. To find the point of intersection A, we can set the two equations equal to each other:
√3x = -√3x +6
2√3x = 6
x = 6/(2√3) = 3√3
Substitute x back into one of the equations to find y:
y = √3(3√3) = 3√9 = 3(3) = 9
Therefore, point A is (3√3, 9).
To find the point of intersection B, we set the equation y = -√3x +6 equal to 0 (since it intersects the x-axis):
0 = -√3x +6
x = 2
Therefore, point B is (2, 0).
2. Using the distance formula, we can calculate the distances:
OA = √[(3√3 - 0)^2 + (9 - 0)^2] = √(27 + 81) = √108 = 6√3
OB = √[(2 - 0)^2 + (0 - 0)^2] = √4 = 2
AB = √[(3√3 - 2)^2 + (9 - 0)^2] = √(27 + 1) = √28 = 2√7
3. To find the area of triangle ABO, we can use the formula for the area of a triangle given three side lengths:
S = √[s(s-OB)(s-OA)(s-AB)] where s is the semi-perimeter = (6√3 + 2 + 2√7)/2 = (4 + 6√3 + 2√7)/2 = 2 + 3√3 + √7
Plugging values in:
S = √[(2 + 3√3 + √7)(2 + 3√3 + √7 - 2)(2 + 3√3 + √7 - 6√3)(2 + 3√3 + √7 - 2√7)]
S = √[(2 + 3√3 + √7)(3√3 + √7)(2 - 3√3)(3 - √7)]
S = √[(2 + 3√3 + √7)(6 + 3√3 - 3√3 - 7)]
S = √[(12 - 7)]
S = √5
Therefore, the area of triangle ABO is √5.
When lines y=√3x and y=-√3x +6 intersect at point A, and line y=-√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 answer