Question 156379: find the distance between the poionts of intersection of the line with equation y=2x+1 and the parabola with equation y=x^2-4x+6. Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website! find the distance between the poionts of intersection of the line with equation y=2x+1 and the parabola with equation y=x^2-4x+6.
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First find the 2 points.
y = x^2-4x+6 = 2x+1 (both = y so they're equal)
x^2-4x+6 = 2x+1
x^2-6x+5 = 0
Quadratic equation (in our case ) has the following solutons:
For these solutions to exist, the discriminant should not be a negative number.
First, we need to compute the discriminant : .
Discriminant d=16 is greater than zero. That means that there are two solutions: .
Quadratic expression can be factored:
Again, the answer is: 5, 1.
Here's your graph:
The graph shown is not of the original parabola. Only the solutions matter,
x = 1 and x = 5
These are the values of x where the original parabola intersects the line.
Sub the values of x into either function (the line is easier) to get the 2 points:
(1,3) and (5,11)
Now find the distance s between the 2 points.
s^2 = (x2-x1)^2 + (y2-y1)^2 (Pythagorean theorem)
s^2 = (5-1)^2 + (11-3)^2
s^2 = 16 + 64 = 80
s = sqrt(80) = 4*sqrt(5)
