When lines y=square root of 3x and y=-square root of 3x +6 intersect at point A, and line y=- square root of 3x +6 and the x axis intersect at point B, solve the following problems. Find the distances OA, OB and AB.
First, let's find the coordinates of points A and B:
1. To find point A, we need to solve for the intersection of the two equations y = √3x and y = -√3x + 6. Setting the two equations equal to each other, we get:
√3x = -√3x + 6
2√3x = 6
x = 3
Plugging x = 3 back into y = √3x, we get y = √9 = 3. Therefore, point A is (3, 3).
2. To find point B, we need to solve for the intersection of the equation y = -√3x + 6 and x-axis (y = 0). Setting y = 0 in the equation y = -√3x + 6, we get:
0 = -√3x + 6
√3x = 6
x = 2√3
Therefore, point B is (2√3, 0).
Now, we can find the distances OA, OB, and AB using the distance formula:
OA = √((3-0)^2 + (3-0)^2) = √(9 + 9) = √18 = 3√2
OB = √((2√3-0)^2 + (0-0)^2) = √(12) = 2√3
AB = √((3-2√3)^2 + (3-0)^2) = √(9 + 12) = √21
Therefore, OA = 3√2, OB = 2√3, and AB = √21.