Hi, I would like to know if its possible to find a equation for a curve if 3 points were given. Say
1. 0, 0
2. 5, 2
3. 10, 0
If so, how would i go about solving this.
Thanks all
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 . . .
The quadratic is probably the easiest.
$\displaystyle
\text{At}\; (0,0)\;\; 0 = 0 a + 0 b + c
$
$\displaystyle
\text{At}\; (5,2)\;\; 2 = 25a + 5b + c
$
$\displaystyle
\text{At}\; (10,0) \;\; 0 = 100 a + 10 b + c
$
Solving gives $\displaystyle a = - \frac{2}{25},\;\; b = \frac{4}{5},\;\; c= 0$
and so we have
$\displaystyle
y =- \frac{2}{25} x^2 + \frac{4}{5} x.
$
or
$\displaystyle
y = \frac{2}{25} x (10-x)
$
Other possibilities
$\displaystyle y = \frac{2}{5^{p+q}} x^p(10-x)^q$ where $\displaystyle p, q \in \mathbb{R^+}$.
A trig function
$\displaystyle
y = 2 \sin \frac{\pi x}{10}
$
An ellipse
$\displaystyle
\frac{(x-5)^2}{25} + \frac{y^2}{4} = 1
$
Here's a circle
$\displaystyle
(x-5)^2 + \left(y + \frac{21}{4} \right)^2 = \left(\frac{29}{4}\right)^2
$.
Like Soroban said - the list is exhaustive.
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