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 distance OA, OB and AB.
3) Find the area S of triangle ABO
1) To find the points of intersection A and B, we can set the two equations equal to each other:
√3x = -√3x + 6
Solving for x:
2√3x = 6
x = 3/√3
x = √3
Using x = √3 in y=√3x:
y = √3(√3) = 3
Therefore, point A is (3, 3).
To find point B, we can set y = 0 in y=-√3x+6:
0 = -√3x + 6
x = 2
Therefore, point B is (2, 0).
2) To find the distances OA, OB, and AB, we can use the distance formula:
OA = √((3 - 0)^2 + (3 - 0)^2) = √(9 + 9) = √18 = 3√2
OB = √((2 - 0)^2 + (0 - 0)^2) = √4 = 2
AB = √((3 - 2)^2 + (3 - 0)^2) = √(1 + 9) = √10
3) To find the area S of triangle ABO, we can use the formula for the area of a triangle given the lengths of its three sides (a, b, c):
s = (OA + OB + AB)/2 = (3√2 + 2 + √10)/2 = (3√2 + 2 + √10)/2
S = √(s(s - OA)(s - OB)(s - AB))
= √((3√2 + 2 + √10)/2((3√2 + 2 + √10)/2 - 3√2)((3√2 + 2 + √10)/2 - 2)((3√2 + 2 + √10)/2 - √10))
After simplifying, the area S of triangle ABO is calculated.