1) To find the point of intersection A, set the two equations equal to each other:
√3x = -√3x + 6
2√3x = 6
x = 3
Plug the value of x back into one of the equations to find y:
y = √3(3) = 3√3
So point A is (3, 3√3).
To find point B, set y = 0 in the equation y = -√3x + 6:
0 = -√3x + 6
√3x = 6
x = 2√3
So point B is (2√3, 0).
2) Use the distance formula to find the distances:
OA = √((3-0)^2 + (3√3-0)^2) = √(9+27) = √36 = 6
OB = √((2√3-0)^2 + (0-0)^2) = √(12) = 2√3
AB = √((3-2√3)^2 + (3√3-0)^2)
3) To find the area of triangle ABO, use the formula for the area of a triangle given the lengths of the sides:
S = 1/2 * base * height
S = 1/2 * AB * OA = 1/2 * √((3-2√3)^2 + (3√3-0)^2) * 6 = 1/2 * √(9-12√3+12 + 27) * 6 = 9√3
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
from Kumon I level Page 189b
1 answer