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
question from Kumon I math level page 189b
1) To find the point of intersection A, we set y=√3x and y = -√3x+6 equal to each other:
√3x = -√3x+6
2√3x = 6
x = 3
Plugging x=3 back into y=√3x, we get y=√3(3)=3√3. Therefore, point A is (3,3√3).
To find the point of intersection B, we set y=-√3x+6 equal to 0 (since it intersects the x-axis):
-√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: √((x2-x1)^2 + (y2-y1)^2).
OA = √((3-0)^2 + (3√3-0)^2) = √(9 + 27) = √36 = 6
OB = √((2-0)^2 + (0-0)^2) = √4 = 2
AB = √((2-3)^2 + (0-3√3)^2) = √(1 + 27) = √28
3) To find the area of triangle ABO, we can use the formula for the area of a triangle: 1/2 * base * height.
The base AB = 2 and the height from A to the x-axis is 3√3. Therefore, the area S of triangle ABO is:
S = 1/2 * 2 * 3√3 = 3√3.