Quadratic Equation, Plz help

As the slope and the y-intercept of L are m and 1 respectively,
so, let the equation of L be y = mx + 1
Sub. it into the equation y = x^2 - 2x + 2, we get,
mx + 1 = x^2 - 2x + 2
==> x^2 - (m + 2)x + 1 = 0
If x1, x2 are the roots of the above equation, then
x1 + x2 = m + 2
==> (x1 + x2)/2 = (m + 2)/2

(1) Let the coordinates of the points M, N be (x1, y1), (x2, y2).
Then the coordinates of the mid-point of MN is ((x1 + x2)/2, (y1 + y2)/2),
ie. the x-coordinate of the mid-point is (m + 2)/2
so the y-coordinate of the mid-point is :
(y1 + y2)/2
= m(m + 2)/2 + 1
= (m^2 + 2m + 2)/2

(2) (m^2 + 2m + 2)/2
= [{m + 1)^2 + 1]/2
>= 1/2
That is, the y-coordinates of the mid-point MN is always greater than zero,
so it cannot lie on the x-axis.