Question

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

1)Find the points of intersection A and B.

2) Find the distance OA, OB and AB.

3) Find the area S of triangle ABO

Answer 1

[AI-generated answer — Bot GPT 3.5]

1) To find the points of intersection A and B, we first need to set the two equations equal to each other:

√3x = -√3x+6

Solving for x:

2√3x = 6
x = 3/√3
x = √3

Now plug this value of x into the first equation to find the y-coordinate of point A:

y = √3(√3)
y = 3

Therefore, Point A is (3,3).

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

0 = -√3x+6
√3x = 6
x = 6/√3
x = 2√3

So, point B is (2√3, 0).

2) To find the distances OA, OB, and AB:

OA = √[(3-0)^2 + (3-0)^2] = √(9 + 9) = √18 = 3√2
OB = √[(2√3-0)^2 + (0-0)^2] = √(12) = 2√3
AB = √[(3-2√3)^2 + (3-0)^2] = √(9-12√3+12 + 9) = √(21-12√3) = 3

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

Let angle A be the right angle at point B. Then angle BAO = 90 degrees.

Area S = 0.5 * OA * OB = 0.5 * 3√2 * 2√3 = 3√6.