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
3) Find the area S of triangle ABO
To find the area of triangle ABO, we first need to find the coordinates of points A and B.
1. Find point A:
Since lines y=√3x and y = -√3x+6 intersect at point A, we can set the two equations equal to each other:
√3x = -√3x+6
2√3x = 6
x = 3/√3
x = √3
Plugging x = √3 into y = √3x, we get:
y = √3(√3)
y = 3
Therefore, point A is (√3, 3).
2. Find point B:
Since line y = -√3x + 6 intersects the x-axis at point B, we can set y = 0 and solve for x:
0 = -√3x + 6
√3x = 6
x = 6/√3
x = 2√3
Therefore, point B is (2√3, 0).
Now that we have the coordinates of points A and B, we can find the lengths of the sides of triangle ABO and then calculate the area using the formula for the area of a triangle (1/2 * base * height).
The base AB = distance between points A and B:
AB = √[(√3 - 2√3)^2 + (3 - 0)^2]
AB = √[(√3)^2 + 3^2]
AB = √(3 + 9)
AB = √12
AB = 2√3
The height of the triangle is the y-coordinate of point A:
Height = 3
Now we can calculate the area of triangle ABO:
S = 1/2 * AB * Height
S = 1/2 * 2√3 * 3
S = 3√3
Therefore, the area of triangle ABO is 3√3.