# Find the point(s) of intersection of the parabola with equation y = x^2 - 5x + 4 and the line with equation y = 2x - 2

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### 3 Answers

At the point of intersection of the parabola y = x^2 - 5x + 4 and the line y = 2x - 2, the value of x and y is equal.

So we can equate the two and solve for x.

x^2 - 5x + 4 = 2x - 2

=> x^2 - 7x + 6 = 0

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

=> x(x-6) - 1(x - 6) = 0

=> (x - 1)(x -6) = 0

So we have x = 1 and 6

The value of y for x = 1 is 0 and for x = 6, y = 10.

**Therefore the point of intersection are ( 1, 0) and (6, 10)**

Given the parabola y= x^2 - 5x + 4 and the line y= 2x-2

We need to find the intersection points of the parabola and the line.

The intersection points are the values of x and y such that:

the parabola y = the line y

==> x^2 - 5x + 4 = 2x -2

We will combine like terms and solve for x.

==> x^2 - 7x + 6 = 0

Now we will factor.

==> (x-6)(x-1) = 0

==> x1= 6 ==> y1= 2*6-2 = 10

==> x2 = 1==> y2= 2*1 -2 = 0

Then we have two intersection points:

**The intersection points of the parabola and the line are (6, 10) and (1,0)**

Q: Find the point(s) of intersection of the parabola with equation y = x2 - 5x + 4 and the line with equation y = 2x - 2.

A:

y = x^2-5x+4 ...(1).

y = =2x-2.......(2).

The y coordinates at the point of intersection is same.

So we equate the right sides of (1) and (2):

x^2-5x+4 = 2x-2. We subtract x-2 from both sides and get:

=> x^2-5x+4-2x+2 = 0.

=> x^2-7x+6 = 0.

=> (x-6)(x-1) = 0.

So x-2 = 0, or x-1 = 0.

=> x= 6, or x= 1.

Therefore we put x= 6, or x= 1 in equation (2), y = 2x-2. to obtain ordinates :

When x = 6, y = 2*6-2 = 12-2 = 10.

When x= 1, y = 2*1-2 = 0.

**Therefore (6,10) and (1,0) are two points of intersection of the parabola y = x^2-5x+4 and the line y = 2x-2 .**