Hello, mrwelcam!
I would like to know if its possible to find a equation for a curve
if 3 points were given, say: .A(0,0), B(5,2), C(10,0)
If so, how would i go about solving this?Code:| | B | o | . . o - - - - - - - o - - A C
With only three points, there is an infinite number of equations.
The first idea would be a down-opening parabola.
But there are an infinite number of cubics, quartics, etc. that contain those points
. . as well as zillions of circles, ellipses, sine/cosine functions . . .
Hi. I'm found this on the web:
A quadratic equation has the form:
Y = A * X^2 + B * X + C
Since we have three points, we have three sets of values for X and Y. That gives us three equations and three unknowns: A, B, and C.
Plugging in the points (X1, Y1), (X2, Y2), and (X3, Y3) gives:
Y1 = AX1^2 + BX1 + C Y2 = AX2^2 + BX2 + C Y3 = AX3^2 + BX3 + C
Subtracting equation 1 from the other two gives:
Y2 - Y1 = A(X2^2 - X1^2) + B(X2-X1) Y3 - Y1 = A(X3^2 - X1^2) + B(X3-X1)
Multiplying the first equation by (X1-X3) and the second by (X2-X1) and adding them gives:
(Y2-Y1)(X1-X3) = (X1-X3)A(X2^2 - X1^2) + B(X2-X1)(X1-X3) (Y3-Y1)(X2-X1) = (X2-X1)A(X3^2 - X1^2) + B(X3-X1)(X2-X1)
Or
(Y2-Y1)(X1-X3) + (Y3-Y1)(X2-X1) = A[(X1-X3)(X2^2-X1^2) + (X2-X1)(X3^2-X1^2)]
Solving for A gives:
A = [(Y2-Y1)(X1-X3) + (Y3-Y1)(X2-X1)]/[(X1-X3)(X2^2-X1^2) + (X2-X1)(X3^2-X1^2)]
Plugging into the previous equation gives:
B = [(Y2 - Y1) - A(X2^2 - X1^2)] / (X2-X1)
Plugging into the original equation gives:
C = Y1 - AX1^2 - BX1
This formula seems to be working well for me. How would i go about expanding this formula to allow for 5 points.
Thanks for your time.
The same applies but obviously, more complicated. You could also try Cramer's Rule
Cramer's rule - Wikipedia, the free encyclopedia