To find the area of triangle ABO, we need to first find the coordinates of points A and B.
Let's first find the coordinates of point A, where the two lines y=sqrt(3x) and y=-sqrt(3x)+6 intersect.
Setting the two equations equal to each other:
sqrt(3x) = -sqrt(3x) + 6
2sqrt(3x) = 6
sqrt(3x) = 3
3x = 9
x = 3
Substitute x=3 into either equation to find y:
y = sqrt(3 * 3) = sqrt(9) = 3
Therefore, point A is at (3, 3).
Next, let's find the coordinates of point B, where the line y=-sqrt(3x)+6 intersects the x-axis.
Setting y=0 in the equation y=-sqrt(3x)+6:
0 = -sqrt(3x) + 6
sqrt(3x) = 6
3x = 36
x = 12
Therefore, point B is at (12, 0).
Now, we can find the area of the triangle ABO using the coordinates of the three points:
S = 1/2 * |(x1(y2-y3) + x2(y3-y1) + x3(y1-y2))|
S = 1/2 * |(3(0-6) + 12(3-0) + 12(6-3))|
S = 1/2 * |(-18 + 36 + 36)|
S = 1/2 * |(54)|
S = 27
Therefore, the area of triangle ABO is 27 square units.
When lines y=square root by 3x and y=-square root by 3x +6 intersect at point A, and line y=- square root by 3x +6 and the x axis intersect at point B, solve the following problems.
Find the area S of triangle ABO
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