(a) The line x + y = 3 intersects the parabola y = x^2 + 1 at points A and B. Determine the coordinates of A and B.
(b) If C is the vertex of the given parabola, determine the area of triangle ABC.
x + y = 3 ⇒ y = 3 - x (1)
y = x^2 + 1 (2)
Equate (1) and (2) and we have that
3 - x = x^2 + 1 rearrange as
x^2 + x - 2 = 0 factor
(x + 2) ( x - 1) = 0 Set each factor to 0 and solve for x and we have that
x = -2 and x = 1
So when x = -2, x + y = 3 implies that y = 5
And when x = 1, x + y = 3 implies that y = 2
So A = ( -2, 5) and B = ( 1, 2)
Second one
The vertex of the parabola is (0, 1)
So A = (-2, 5) B = (1, 2) C = ( 0,1)
We can let AB be the base of the triangle
AB = sqrt [ ( -2 - 1)^2 + (5 - 2)^2 ] = sqrt [ 3^2 + 3^2] = sqrt [18] = 3sqrt (2)
And we can use the formula for the distance that (0, 1) is from the line x + y - 3 = 0.....this will be the height of ABC
So we have
l Ax + By - C l
_____________ =
sqrt ( A^2 + B^2)
l 1(0) + 1(1) - 3 l l -2 l 2
_________________ = _______ = ______ = sqrt (2)
sqrt [ 1^2 + 1^2] sqrt(2) sqrt (2)
So.....the area of ABC = (1/2) AB * height of ABC = (1/2)(3 sqrt (2) ) ( sqrt (2) ) = (1/2) (3) (2) =
3 units^2 = [ ABC ]