# Could someone please point me to an algebraic proof linking the Quadratic Coefficients to the Cramer's Rule Formula given 3 points on the curve?

Could someone please point me to an algebraic proof linking the Quadratic Coefficients to the Cramer's Rule Modified Formula given 3 points on the curve?

The internet is crammed with videos proving the Quadratic Formula for x, solving for the Coefficients A, B & C using equation substitution and also solving for A, B & C using the matrix method, but I couldn't find an algebraic proof to solve for the Coefficient A. Solving for B and C seems trivial since they are dependant on knowing A, but how to arrive at the proof for solving for A eludes me.

Given 3 points $$({x_1},{y_1}) , ({x_2},{y_2})$$ and $$({x_3},{y_3})$$ on a parabola.

Knowing the Quadratic Equation $$A(x^2)+B(x)+C = y$$ applies, we have 3 formulae:

EQ01: $$A({x_1}^2)+B({x_1})+C = {y_1}$$

EQ02: $$A({x_2}^2)+B({x_2})+C = {y_2}$$

EQ03: $$A({x_3}^2)+B({x_3})+C = {y_3}$$

Cramer's Rule shows us 3 equations solving for A, B & C, using the matrix manipulation:

EQ04: $$A = \frac{{y_2}({x_3}-{x_1}) - {y_1}({x_3}-{x_2}) - {{y_3}({x_2}-{x_1})}}{{x_1}^2({x_2}-{x_3})-{x_3}^2({x_2}-{x_1})-{x_2}^2({x_1}-{x_3})}$$

Knowing A from EQ04:

EQ05:
$$B = \frac{{y_2}-{y_1}+ A({x_1}^2-{x_2}^2)}{{x_2}-{x_1}}$$

Knowing A and B:

EQ06: $$C = {y_1}-(A{x_1}^2)-(B{x_1})$$

Could someone please point to (or post) the proof algebraically (not matrix derived) linking equations EQ01, EQ02 and EQ03 with EQ04?

• Yes, I have mistakenly added squared x terms with the B coefficient, thanks, good eye! Apr 19 at 20:36

We have

\begin{align} Ax_1^2 \; + \; Bx_1 + C \; &= \; y_1 & (1) \\ Ax_2^2 \; + \; Bx_2 + C \; &= \; y_2 & (2) \\ Ax_3^2 \; + \; Bx_3 + C \; &= \; y_3 & (3) \\ {\color{white}x} \\ \end{align}

Apply the following operations:

\begin{align} \Big[x_3 \times \{(2)-(1)\}\Big]: \quad \quad \quad \quad \quad \quad Ax_3(x_2^2-x_1^2) \; + \; Bx_3(x_2-x_1) \; = \; x_3(y_2-y_1) \\ \Big[x_1 \times \{(3)-(2)\}\Big]: \quad \quad \quad \quad \quad \quad Ax_1(x_3^2-x_2^2) \; + \; Bx_1(x_3-x_2) \; = \; x_1(y_3-y_2) \\ \Big[x_2 \times \{(1)-(3)\}\Big]: \quad \quad \quad \quad \quad \quad Ax_2(x_1^2-x_3^2) \; + \; Bx_2(x_1-x_3) \; = \; x_2(y_1-y_3) \\ {\color{white}x} \\ \end{align}

Now by adding the last three equations, all the terms with $$B$$ drop out, leaving

$$\small{A\left[ x_1(x_3^2-x_2^2) + x_2(x_1^2-x_3^2) + x_3(x_2^2-x_1^2) \right] = x_1(y_3-y_2) + x_2(y_1-y_3) + x_3(y_2-y_1)} \\$$

Distributing and rearranging the terms on both sides yields

$$\small{A\left[ x_1^2(x_2-x_3) - x_3^2(x_2-x_1) - x_2^2(x_1-x_3) \right] = y_2(x_3-x_1) - y_1(x_3-x_2) - y_3(x_2-x_1)}\\$$

from whence your EQ04 immediately follows.

• Thanks for answering, A.J.! You showed the pieces of the puzzle that I was missing there: I wondered about what the reasoning was for ordering the B terms and signs a specific way and I didn't twig on that it was designed to enable cancellation of the B terms. Thanks very much! Apr 19 at 21:03