# Calculate the coordinates of the intercepting point of the line x-2 and the curve x^2-4x+4.

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f(x) = x^2 - 4x + 4

g(x) = x-2

The intercepting point is when f(x) = g(x)

x^2 - 4x + 4 = x-2

Group similar terms:

==> x^2 - 5x + 6 = 0

Now factor:

==> (x-3)(x-2) = 0

==> x1= 3 ==> y1= 3-2 = 1

==> x2= 2 ==> y2= 2-2 = 0

Then the curve f(x) and the line y are intercepting in two point :

**Point of intercepting are: A(3, 1) and B(2, 0)**

First, we'll note the line y = x-2 and the curve as y = x^2-4x+4.

To determine the coordinates of the intercepting point of the line and curve, we'll have to solve the system:

y = x-2 (1)

y = x^2-4x+4 (2)

x-2 = x^2-4x+4 (3)

We notice that the equation (2) is the result of the square expanding:

(x-2)^2 = x^2-4x+4

We'll re-write (3):

(x-2) = (x-2)^2

We'll subtract (x-2) both sides:

(x-2)^2 - (x-2) = 0

We'll factorize by (x-2):

(x-2)[(x-2)-1] = 0

We'll set each factor as 0:

x-2 = 0

We'll add 2 both sides;

**x = 2**

From (1) => y = x-2 => **y = 2-2 = 0**

x-3 = 0

**x = 3**

From (1) => y = x-2 =>** y = 3-2 = 1**

**The coordinates of the intercepting points are: (2 , 0) and (3 , 1).**

To find the intercepting ponits of x^2-4x+4 and x-2.

Solution:

Let y = x^2-4x+4 and y = x-2 be the curve and the line.

Since the y coordinates at the intersection is same for both curve and the line,

x^2-4x+4 = x-2

(x-2)^2 = (x-2)

(x-2)^2 - (x-2 ) = 0

(x-2)(x-2-1) = 0

(x-2) (x-3) = 0

So x= 2 or x=3 are the x coordinates of the intersections.

At x =2, y = (x-2) = 2-2 = 0

At x=3, y = x-2 = 3-2 = 1.

So (2,0) and (3, 1) are coordinates of intersection of x^2-4x+4 and x-2.