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 distance OA, OB and AB.
3) Find the area S of triangle ABO
1)
To find the point of intersection A, we set the two equations equal to each other:
√3x = -√3x + 6
2√3x = 6
x = 3
Substitute x back into y=√3x or y=-√3x+6 to find the y-coordinate of point A:
y = √3(3) = 3√3
Therefore, point A is at (3, 3√3).
To find the point of intersection B, we set y=-√3x+6 equal to y=0 (since it intersects the x-axis):
-√3x + 6 = 0
√3x = 6
x = 2√3
Therefore, point B is at (2√3, 0).
2)
To find the distance OA, we use the distance formula:
OA = √((3-0)^2 + (3√3-0)^2)
OA = √(9 + 27)
OA = √36
OA = 6
To find the distance OB, we use the distance formula:
OB = √((2√3-0)^2 + (0-0)^2)
OB = √(12)
OB = 2√3
To find the distance AB, we use the distance formula:
AB = √((3-2√3)^2 + (3√3-0)^2)
AB = √(9 - 12√3 + 12 + 27)
AB = √(39 - 12√3)
AB = √(39 - 12√3)
3)
To find the area S of triangle ABO, we can use the formula for the area of a triangle given its three vertices:
S = 0.5 |x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)|
Substitute the coordinates of A, B, and O into the formula:
S = 0.5|(3[(0) - (0)] + 2√3(3√3 - 0) + 0(3 - 3√3)]|
S = 0.5|(0 + 18 + 0)|
S = 0.5 * 18 = 9
Therefore, the area of triangle ABO is 9.