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
2) Find the distance OA, OB and AB.
3) Find the area S of triangle ABO
2) To find the distance OA, we first need to find the coordinates of point A where the two lines intersect. Setting y=√3x equal to -√3x+6, we get:
√3x = -√3x + 6
2√3x = 6
x = 3
Plugging x=3 into y=√3x, we get y=√(3*3) = √9 = 3. Therefore, point A is (3,3).
The distance OA is the length of the line connecting points O(0,0) and A(3,3), which can be found using the distance formula:
OA = √((3-0)^2 + (3-0)^2) = √(3^2 + 3^2) = √18 = 3√2
To find the coordinates of point B, we set y=-√3x+6 equal to 0:
-√3x + 6 = 0
-√3x = -6
x = 2
Therefore, point B is (2,0). The distance OB is the length of the line connecting points O(0,0) and B(2,0), which is simply OB = 2.
Lastly, the distance AB is the length of the line connecting points A(3,3) and B(2,0), which can be found using the distance formula:
AB = √((3-2)^2 + (3-0)^2) = √(1 + 9) = √10
3) To find the area of triangle ABO, we can use the formula for the area of a triangle given its side lengths a, b, and c:
S = √(s(s-a)(s-b)(s-c))
where s is the semi-perimeter of the triangle given by s = (a + b + c)/2.
In this case, a = OB = 2, b = OA = 3√2, and c = AB = √10. Plugging these values into the formula, we get:
s = (2 + 3√2 + √10)/2
S = √((2 + 3√2 + √10)(3√2 - 2)(√10 - 2))
After calculating the expression, we will obtain the area of triangle ABO.