Help needed over here :S

[HUSH1994]

this is a question from a past paper,im not sure which year but it is pure one,i need a way by co-ordinate geometry equations to solve it,not differentiation.
Q.The equation of the curve is y=x^2-4x+7 and the equation of the line is y+3x=9. The curve and the line intersect at the points A and B.
(i)the mid-point of AB is M,show that the co-ordinates of M are (0.5,7.5)  [not the problem,its easy ]

(ii) Find the coordinates of the point Q on the curve at which the tangent is parallel to the line y+3x=9. [this is the real pain]

thanks,please help as soon as possible,as i have an exam on saturday

[elemis]

2 minutes.

[elemis]

Whether you like or not differentiation is necessary to solve part (b).

The tangent to the curve is parallel to the y+3x=9 right ? So the gradient of the tangent will be 3 as well.

Differentiating the curve equation gives : 2x - 4

We must now find the point on the curve at which dy/dx = 3

Hence, 2x-4 = 3

x = 3.5

Corresponding y coordinate is : -1.5

Therefore, the coordinates of Q are : (3.5,-1.5)