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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 ax...Asked by taco man
When lines y=square root of 3x and y=-square root of 3x +6 intersect at point A, and line y=- square root of 3x +6 and the x axis intersect at point B, solve the following problems. Find the distances OA, OB, and AB
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Answered by
GPT 3.5
To find point A, we need to set the two equations equal to each other and solve for x:
sqrt(3x) = -sqrt(3x) + 6
2sqrt(3x) = 6
sqrt(3x) = 3
3x = 9
x = 3
Now that we have x=3, we can find the y-coordinate by plugging it back into one of the equations. Let's use y = sqrt(3x):
y = sqrt(3(3))
y = sqrt(9)
y = 3
So point A is at (3, 3).
Now let's find point B where the line y = -sqrt(3x) + 6 intersects the x-axis:
0 = -sqrt(3x) + 6
sqrt(3x) = 6
3x = 36
x = 12
Point B is at (12, 0).
Now we can find the distances:
OA = sqrt((3-0)^2 + (3-0)^2) = sqrt(9 + 9) = sqrt(18) = 3sqrt(2)
OB = sqrt((12-0)^2 + (0-0)^2) = sqrt(144 + 0) = 12
AB = sqrt((12-3)^2 + (0-3)^2) = sqrt(81 + 9) = sqrt(90) = 3sqrt(10)
So,
OA = 3sqrt(2)
OB = 12
AB = 3sqrt(10)
sqrt(3x) = -sqrt(3x) + 6
2sqrt(3x) = 6
sqrt(3x) = 3
3x = 9
x = 3
Now that we have x=3, we can find the y-coordinate by plugging it back into one of the equations. Let's use y = sqrt(3x):
y = sqrt(3(3))
y = sqrt(9)
y = 3
So point A is at (3, 3).
Now let's find point B where the line y = -sqrt(3x) + 6 intersects the x-axis:
0 = -sqrt(3x) + 6
sqrt(3x) = 6
3x = 36
x = 12
Point B is at (12, 0).
Now we can find the distances:
OA = sqrt((3-0)^2 + (3-0)^2) = sqrt(9 + 9) = sqrt(18) = 3sqrt(2)
OB = sqrt((12-0)^2 + (0-0)^2) = sqrt(144 + 0) = 12
AB = sqrt((12-3)^2 + (0-3)^2) = sqrt(81 + 9) = sqrt(90) = 3sqrt(10)
So,
OA = 3sqrt(2)
OB = 12
AB = 3sqrt(10)
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