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. To find the point of intersection A, set the two equations equal to each other:
√3x = -√3x +6
Solving for x, we get x = 1. Substituting x back into y=√3x, we get y = √3. Therefore, the point of intersection A is (1, √3).
To find the point of intersection B, set y = 0 in the equation y = -√3x + 6:
0 = -√3x + 6
Solving for x, we get x = 2. Therefore, the point of intersection B is (2, 0).
2. To find the distances OA, OB, and AB, we can use the distance formula:
OA = √((2-1)^2 + (0-√3)^2) = √2
OB = 2
AB = √((2-1)^2 + (0-√3)^2) = √2
3. To find the area of triangle ABO, we can use Heron's formula since we know all three side lengths.
s = (OA + OB + AB)/2
s = (2 + √2 + √2)/2
s = (4 + 2√2)/2
s = 2 + √2
Plugging into Heron's formula, we get:
S = √(s(s-OA)(s-OB)(s-AB))
S = √((2 + √2)(2 + √2 - 2)(2 + √2 - √2)(2 + √2 - √2))
S = √(2 + √2)(2 - √2)(2)
S = √(4 - 2)(2)
S = √(4 - 2)(2)
S = √2
Therefore, the area of triangle ABO is √2.