--- title: "Draw Bezier curve through 4 control points" subject: "Computer Graphics" chapter: 5 key: "semester_5/computer_graphics:bezier-four-points-p0-0-0" source: https://www.bctengineer.com/q/semester_5/computer_graphics/bezier-four-points-p0-0-0 site: BCT Engineering --- # Draw Bezier curve through 4 control points Computer Graphics · Chapter 5: Curve Modeling - Asked in: 2069 Asar, 8 marks - Video: [Bezier Curve Numerical Problem | Computer Graphics](https://www.youtube.com/watch?v=wZbazXBkGv4) — Vinay Mishra - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/computer_graphics/bezier-four-points-p0-0-0 - Whole chapter: https://www.bctengineer.com/md/semester_5/computer_graphics/chapter-5 ## Question Given four 2-dimensional control points P0(0,0), P1(3,3), P2(-2,-2) and P3(2,3), draw a smooth curve using Bezier spline method. ## Short answer Use cubic Bezier blending: P(u)=(1−u)³P0+3u(1−u)²P1+3u²(1−u)P2+u³P3. Evaluate at u=0.5 gives (0.625,0.75). Curve passes through P0(0,0) and P3(2,3), shaped by control polygon without passing through P1,P2. ## Answer Control points P0(0,0), P1(3,3), P2(−2,−2), P3(2,3). Four points give a cubic Bezier curve. P(u) = (1−u)³P0 + 3u(1−u)²P1 + 3u²(1−u)P2 + u³P3 Power form: a = −P0 + 3P1 − 3P2 + P3 = (9 + 6 + 2, 9 + 6 + 3) = (17, 18); b = 3P0 − 6P1 + 3P2 = (−18 − 6, −18 − 6) = (−24, −24); c = −3P0 + 3P1 = (9, 9). x(u) = 17u³ − 24u² + 9u, y(u) = 18u³ − 24u² + 9u Blending values and points: | u | B0 | B1 | B2 | B3 | x | y | |---|---|---|---|---|---|---| | 0 | 1 | 0 | 0 | 0 | 0 | 0 | | 0.10 | 0.729 | 0.243 | 0.027 | 0.001 | 0.677 | 0.678 | | 0.25 | 0.4219 | 0.4219 | 0.1406 | 0.0156 | 1.016 | 1.031 | | 0.50 | 0.125 | 0.375 | 0.375 | 0.125 | 0.625 | 0.750 | | 0.75 | 0.0156 | 0.1406 | 0.4219 | 0.4219 | 0.422 | 0.844 | | 0.90 | 0.001 | 0.027 | 0.243 | 0.729 | 1.053 | 1.782 | | 1 | 0 | 0 | 0 | 1 | 2 | 3 | Example, u = 0.5: x = 0.375·3 + 0.375·(−2) + 0.125·2 = 1.125 − 0.75 + 0.25 = 0.625; y = 0.375·3 + 0.375·(−2) + 0.125·3 = 1.125 − 0.75 + 0.375 = 0.75. **Drawing steps:** plot the four control points and join them to form the control polygon. Plot the points in the table and join them smoothly in order of u. ```text 3 | o P3 o P1 | 2 | e | 1 | d c b | a 0 |----------------o P0----------------------> x | -1 | | -2 |o P2 -2 0 1 2 3 a, b, c, d, e = curve points at u = 0.1, 0.25, 0.5, 0.75, 0.9 curve order: P0 -> a -> b -> c -> d -> e -> P3 ``` The curve leaves P0 along P0P1, goes out to about (1,1), turns back because of the pull of P2, and then moves up towards P3. It passes only through P0 and P3, not through P1 or P2. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Four control points P0..P3 produce a: A. Cubic Bezier curve B. Quadratic Bezier curve C. Hermite curve D. Straight line Answer: A. n = 3 gives a cubic. (Key point: Curve type) 2. For P0(0,0), P1(3,3), P2(-2,-2), P3(2,3), the x-coefficient of u^3 is: A. 9 B. -24 C. 17 D. 13 Answer: C. a = -0 + 9 + 6 + 2 = 17. (Key point: Power form) 3. What is P(0.5)? A. (1.5, 1.5) B. (0.5, 0.5) C. (1.0, 1.0) D. (0.625, 0.75) Answer: D. x = 0.375*3 + 0.375*(-2) + 0.125*2 = 0.625; y = 1.125 - 0.75 + 0.375 = 0.75. (Key point: Midpoint) 4. Which points does the curve actually touch? A. P0, P1, P2 and P3 B. P0 and P3 only C. P1 and P2 only D. P0 and P2 Answer: B. Intermediate control points only pull the curve. 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