When lines y=√3x and y=-√3x +6 intersect at point A, and line y=-√3x +6 and the x axis intersect at point B, solve the following problems.
2. Find the distances OB and AB
To find the distance OB, we need to find the x-intercept of the line y=-√3x +6. This occurs when y=0, so we set y=0 in the equation:
0 = -√3x + 6
√3x = 6
x = 6/√3
x = 2√3
Therefore, the x-intercept (point B) is at (2√3, 0).
To find the distance OB, we can use the distance formula:
OB = √((x2 - x1)^2 + (y2 - y1)^2)
OB = √((2√3 - 0)^2 + (0 - 0)^2)
OB = √(12)
OB = 2√3
Next, we need to find the coordinates of point A where the two lines intersect. Setting the two equations equal to each other, we can solve for x:
√3x = -√3x + 6
2√3x = 6
x = 3/√3
x = √3
Substitute this x value into one of the equations to find y:
y = √3(√3)
y = 3
Therefore, the point of intersection A is at (√3, 3).
Finally, to find the distance AB, we can use the distance formula again:
AB = √((√3 - 2√3)^2 + (3 - 0)^2)
AB = √((-√3)^2 + 3^2)
AB = √(3 + 9)
AB = √12
AB = 2√3
So, the distances OB and AB are both equal to 2√3.