Question

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 distances OA, OB and AB.

Answer 1

To find the coordinates of point A where the two curves intersect, we can 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

Plugging x=3 back into the equation y=sqrt(3x) or y=-sqrt(3x)+6, we get y=3.

Therefore, point A is (3,3).

To find the coordinates of point B where the curve y=-sqrt(3x)+6 intersects the x-axis, we set y=0 in the equation:

0 = -sqrt(3x)+6
sqrt(3x) = 6
3x = 36
x = 12

Therefore, point B is (12,0).

Now we can find the distances OA, OB, and AB using the distance formula:

OA = sqrt((3-3)^2 + (3-0)^2) = sqrt(0 + 9) = sqrt(9) = 3
OB = sqrt((12-0)^2 + (0-0)^2) = sqrt(144) = 12
AB = sqrt((12-3)^2 + (0-3)^2) = sqrt(81 + 9) = sqrt(90) = 3*sqrt(10)

Therefore, the distances are OA = 3, OB = 12, and AB = 3*sqrt(10).