# mathWhat is the intercepting point of the line y=5-x and the parabola y=(13-x^2)^1/2?

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At the point of intersection the x and y coordinates are the same.

To find the point of intersection we solve y=5-x and y=(13-x^2)^1/2

substitute y = 5 - x in y=(13-x^2)^1/2

=> 5 - x = (13 - x^2)^(1/2)

=> (5 - x)^2 = 13 - x^2

=> 25 + x^2 - 10x = 13 - x^2

=> 2x^2 - 10x + 12 = 0

=> 2x^2 - 6x - 4x + 12 = 0

=> 2x(x - 3) - 4(x - 3) = 0

=> (2x - 4)(x - 3) = 0

=> x = 2 and x = 3

y = 3 , 2

**The points of intersection are (2, 3) and (3, 2)**

We'll solve the simultaneous equations to find out the intercepting points.

We'll use the sum and the product to solve it.

We'll note x+y = S and xy = P.

We'll re-write the second equation as:

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

x^2 + y^2 = S^2 - 2P

We'll re-write the system in S and P.

S = 5

S^2 - 2P = 13

We'll susbtitute S in the second equation:

25 - 2P = 13

-2P = 13 - 25

-2P = -12

P = 6

Now, we'll form the quadratic knowing the sum and the product:

x^2 - 5x + 6 = 0

x1 = [5+sqrt(25 - 24)]/2

x1 = (5+1)/2

x1 = 3

x2 = (5-1)/2

x2 = 2

**The requested intercepting points are the solutions of the symmetric system and they are represented by the pairs: {2 ; 3} or {3 ; 2}.**