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.

Question

1. Find the points of intersection A and B
2. Find the distances OA,OB and AB
3. Find the area S of triangle ABO

Answer 1

1. To find the point of intersection A, set the two equations equal to each other:
√3x = -√3x + 6
2√3x = 6
x = 3

Substitute x back into the first equation to find y:
y = √3(3) = 3√3

Therefore, point A is (3, 3√3)

To find the point of intersection B, set y = 0 in the equation y = -√3x + 6:
0 = -√3x + 6
√3x = 6
x = 2

Therefore, point B is (2, 0)

2. Calculating the distances OA, OB, and AB:

OA = √((3-0)^2 + (3√3 - 0)^2) = √(9 + 27) = √36 = 6

OB = 2

AB = √((3-2)^2 + (3√3 - 0)^2) = √(1 + 27) = √28

3. Using Heron's formula to find the area of triangle ABO:

S = √(s(s-a)(s-b)(s-c)) where s=(a+b+c)/2

S = √((6 + √28)/2)((6 - √28)/2)(√28/2)(2/2)
S = √((6 + √28)(6 - √28) * 28)
S = √((6^2 - 28) * 28)
S = √((36 - 28) * 28)
S = √(8 * 28)
S = √224
S = 4√14

Therefore, the area of triangle ABO is 4√14.