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 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 and solve for x:

√3x = -√3x + 6
2√3x = 6
x = 3√3

Now plug x back into either equation to find the y-coordinate of point A:

y = √3(3√3) = 3√3

So, point A is (3√3, 3√3).

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

0 = -√3x + 6
√3x = 6
x = 2√3

So, point B is (2√3, 0).

2) To find the distances:
Distance OA = √[(3√3)^2 + (3√3)^2] = √(27 + 27) = √54 = 3√6
Distance OB = 2√3
Distance AB = √[(3√3 - 0)^2 + (3√3 - 0)^2] = √(27 + 27) = √54 = 3√6

3) To find the area S of triangle ABO, use the formula for the area of a triangle given the lengths of the sides:

S = √[s(s-a)(s-b)(s-c)]

where s = (3√6 + 2√3 + 3√6)/2 = 4.5√6

Plugging in the values:

S = √[4.5√6(4.5√6-3√6)(4.5√6-2√3)(4.5√6-3√6)]

S = √[4.5√6(1.5√6)(2.5√6)(1.5√6)]

S = √[4.5(1.5)(2.5)(1.5)(6)(6)]

S = √[1012.5]

S = 31.824 units squared

Therefore, the area of triangle ABO is 31.824 square units.