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.
Find the distances OA,OB and AB using point A (3, 3√3) and point B (2√3,0)
To find the distances OA, OB, and AB using the given points A(3, 3√3) and B(2√3, 0), we can use the distance formula:
1. Distance OA:
OA = √((x2 - x1)^2 + (y2 - y1)^2)
OA = √((3 - 0)^2 + (3√3 - 0)^2)
OA = √(9 + 27)
OA = √36
OA = 6
2. Distance OB:
OB = √((x2 - x1)^2 + (y2 - y1)^2)
OB = √((2√3 - 0)^2 + (0 - 0)^2)
OB = √(12)
OB = 2√3
3. Distance AB:
AB = √((x2 - x1)^2 + (y2 - y1)^2)
AB = √((2√3 - 3)^2 + (0 - 3√3)^2)
AB = √(3 + 3*3)
AB = √12
AB = 2√3
Therefore, the distances are:
OA = 6
OB = 2√3
AB = 2√3