Question

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

Answer 1

1)
To find the point of intersection A, we need to set the two equations equal to each other:
√3x = -√3x + 6
2√3x = 6
x = 3

Plugging in x = 3 into y=√3x, we get:
y = √3(3) = 3√3

Therefore, point A is at (3, 3√3).

To find the point of intersection B, we need to set y = 0 in the equation y = -√3x + 6:
0 = -√3x + 6
√3x = 6
x = 2

Therefore, point B is at (2, 0).

2)
To find the distances OA, OB, and AB, we can use the distance formula:

OA = √[(3-0)^2 + (3√3-0)^2] = √[9 + 27] = √36 = 6

OB = √[(2-0)^2 + (0-0)^2] = √[4] = 2

AB = √[(3-2)^2 + (3√3-0)^2] = √[1 + 27] = √28

3)
To find the area S of triangle ABO, we can use the formula for the area of a triangle given the lengths of two sides and the included angle:

S = (1/2) * OA * OB * sin(θ)

Where θ is the angle between OA and OB. Since point A and B are on the x-axis, the angle is 90 degrees. Therefore, sin(90) = 1 and the area becomes:

S = (1/2) * 6 * 2 * 1 = 6