Answer:
First, we need to find the coordinates of the intersection points between the curve and the line. We can substitute y = 2(1-x) into the equation of the curve and simplify to get:
3x^2 + x(2(1-x)) - (2(1-x))^2 + 4(2(1-x)) - 3 = 0
Simplifying this expression, we get:
3x^2 - 10x + 5 = 0
We can solve for x by using the quadratic formula:
x = (-b ± sqrt(b^2 - 4ac)) / 2a
where a = 3, b = -10, and c = 5. Substituting these values, we get:
x = (-(-10) ± sqrt((-10)^2 - 4(3)(5))) / 2(3)
x = (10 ± sqrt(40)) / 6
x = (5 ± sqrt(10)) / 3
Now we can substitute these values of x back into the equation y = 2(1-x) to find the corresponding values of y:
When x = (5 + sqrt(10)) / 3, y = 2(1 - (5 + sqrt(10)) / 3) = -(4/3) - (2/3)sqrt(10)
When x = (5 - sqrt(10)) / 3, y = 2(1 - (5 - sqrt(10)) / 3) = -(4/3) + (2/3)sqrt(10)
Therefore, the coordinates of point A are ((5 + sqrt(10)) / 3, -(4/3) - (2/3)sqrt(10)), and the coordinates of point B are ((5 - sqrt(10)) / 3, -(4/3) + (2/3)sqrt(10)).