# Proving $x_1 = (x_0+x_2)/2$ on a quadratic Bezier curve

Can someone please prove that $x_1 = (x_0 + x_2)/2$ on a quadratic Bezier curve.

Where: $x_0$ is the first control points, $x_2$ is the second control point, $x_1$ is the middle control point located where the tangents at $x_0$ and $x_2$ meet

For the function $f(x) = ax^2 + bx + c$ at points $P_0 = (x_0, y_0)$, $P_1 = (x_1, y_1)$, $P_2 = (x_2, y_2)$.

Thanks

• Maybe you mean a quadratic Bezier ? – Yves Daoust Oct 23 '16 at 14:44
• I changed $x_3$ to $x_1$. Is that what you meant ?? – bubba Oct 23 '16 at 14:49

The tangent line at $P_0$ is $y-y_0 = (2ax_0 + b)(x-x_0)$.
The tangent line at $P_2$ is $y-y_2 = (2ax_2 + b)(x-x_2)$.
Confirm that the $x$-coordinate of the intersection point is $x_1$.