show that; given any three non-collinear points (x1,y1), (x2,y2), (x3,y3) there is a parabola p(x)=Ax^2+ Bx + c , containing the three points.
3 equations, three unknowns
if they are co linear though, one equation will be a constant times another and there will be no unique solution.
okay, i'm just not exactly sure where to start in trying to prove this though.
y = a x ^ 2 + b * x + c
You must solve system of equations :
a * x1 ^ 2 + b * x1 + c = y1
a * x2 ^ 2 + b * x2 + c = y2
a * x3 ^ 2 + b * x3 + c = y3
then put values of a , b and c in equation :
y = a x ^ 2 + b * x + c
To show that a parabola exists containing any three non-collinear points (x1, y1), (x2, y2), (x3, y3), we can use the method of finding the equation of a parabola given three points.
The standard form of a quadratic equation is y = ax^2 + bx + c. We need to find the values of a, b, and c that satisfy the conditions of the given points.
Step 1: Substitute the x and y coordinates of the three points into the equation.
For the first point (x1, y1):
y1 = a(x1)^2 + b(x1) + c
y1 = a(x1)^2 + b(x1) + c --- (Equation 1)
For the second point (x2, y2):
y2 = a(x2)^2 + b(x2) + c
y2 = a(x2)^2 + b(x2) + c --- (Equation 2)
For the third point (x3, y3):
y3 = a(x3)^2 + b(x3) + c
y3 = a(x3)^2 + b(x3) + c --- (Equation 3)
Step 2: Solve the system of three equations to find the values of a, b, and c.
To solve this system of equations, we can subtract Equation 1 from both sides of Equation 2 and Equation 3.
Equation 2 - Equation 1:
(y2 - y1) = a(x2^2 - x1^2) + b(x2 - x1)
(y2 - y1) = a(x2 + x1)(x2 - x1) + b(x2 - x1) --- (Equation 4)
Equation 3 - Equation 1:
(y3 - y1) = a(x3^2 - x1^2) + b(x3 - x1)
(y3 - y1) = a(x3 + x1)(x3 - x1) + b(x3 - x1) --- (Equation 5)
Step 3: Calculate the differences in the x-coordinates and y-coordinates.
Compute the differences in the x-coordinates and y-coordinates:
Δx = x2 - x1
Δy = y2 - y1
Step 4: Substitute the differences into Equation 4 and Equation 5.
Equation 4 becomes:
Δy = a(Δx + x1)(Δx) + b(Δx) --- (Equation 6)
Equation 5 becomes:
(y3 - y1) = a(x3 + x1)(x3 - x1) + b(x3 - x1)
Step 5: Simplify Equation 6 and solve for a.
Δy = a(Δx + x1)(Δx) + b(Δx)
Simplifying further:
Δy = Δx(aΔx + ax1) + bΔx
Δy = a(Δx^2) + ax1(Δx) + b(Δx) --- (Equation 7)
Since Δx ≠ 0 (the points are non-collinear), we can divide both sides of Equation 7 by Δx:
Δy/Δx = aΔx + ax1 + b
Step 6: Calculate the slope of the line passing through the two points (x1, y1) and (x2, y2).
The slope of the line passing through (x1, y1) and (x2, y2) is given by:
m = (y2 - y1) / (x2 - x1)
Step 7: Substitute the value of m into the equation obtained in Step 6.
Δy/Δx = aΔx + ax1 + b
m = aΔx + ax1 + b
Step 8: Solve for b.
b = m - aΔx - ax1
Step 9: Substitute the values of a and b into any of the three original equations.
Let's use Equation 1:
y1 = a(x1)^2 + b(x1) + c
Substituting the values of a and b:
y1 = a(x1)^2 + (m - aΔx - ax1)(x1) + c
Step 10: Simplify the equation and solve for c.
y1 = a(x1)^2 + mx1 - aΔx(x1) - a(x1)^2 + c
y1 = mx1 - aΔx(x1) + c
Since Δx ≠ 0, we can divide both sides by Δx:
(y1 - mx1) / Δx = -a(x1) + c
Finally, solving for c:
c = (y1 - mx1) / Δx + a(x1)
Step 11: Substitute the values of a, b, and c back into the quadratic equation.
The equation of the parabola that passes through the three points is:
y = ax^2 + bx + c
y = ax^2 + (m - aΔx - ax1)x + [(y1 - mx1)/Δx + a(x1)]
And there you have it! The equation of the parabola that contains the three non-collinear points (x1, y1), (x2, y2), and (x3, y3).