Question

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 distance OA, OB and AB.

Answer 1

To find the intersection point of the two lines y = √3x and y = -√3x + 6, we can set the two equations equal to each other:

√3x = -√3x + 6
2√3x = 6
x = 3

Plugging this x value back into either equation gives us the y value at point A:

y = √3(3) = 3√3

So point A is at (3, 3√3).

Next, to find the intersection point of the line y = -√3x + 6 and the x-axis, we can set y equal to 0:

0 = -√3x + 6
√3x = 6
x = 2√3

So point B is at (2√3, 0).

Now we can calculate the distances:

Distance OA:
Using the distance formula: √((3-0)^2 + (3√3-0)^2) = √(9 + 27) = √36 = 6

Distance OB:
Using the distance formula: √((2√3-0)^2 + (0-0)^2) = √(12) = 2√3

Distance AB:
Using the distance formula: √((3-2√3)^2 + (3√3-0)^2)
= √(9 - 12√3 + 12 + 27)
= √(21 - 12√3)

Therefore, the distances are:
- OA = 6
- OB = 2√3
- AB = √(21 - 12√3)