---
title: "Rotate triangle 45° about (3,3)"
subject: "Computer Graphics"
chapter: 3
key: "semester_5/computer_graphics:rotate-triangle-45-pivot-3-3"
source: https://www.bctengineer.com/q/semester_5/computer_graphics/rotate-triangle-45-pivot-3-3
site: BCT Engineering
---

# Rotate triangle 45° about (3,3)

Computer Graphics · Chapter 3: Two –Dimensional Transformations

- Asked in: 2076 Asoj, 6 marks
- Video: [Rotation of a Triangle about  a Point in Computer Graphics in Hindi  Lec-40](https://www.youtube.com/watch?v=GRJDZ1nY14U) — Zeenat Hasan Academy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/computer_graphics/rotate-triangle-45-pivot-3-3
- Whole chapter: https://www.bctengineer.com/md/semester_5/computer_graphics/chapter-3

## Question

Rotate a triangle A(5,6), B(6,2) and C(4,1) by 45 degree about an arbitrary pivot point (3,3).

## Short answer

Translate pivot (3,3) to origin, rotate by 45°, translate back. Using x' = 3 + (x-3)cos45° - (y-3)sin45°, y' = 3 + (x-3)sin45° + (y-3)cos45°. Vertices: A(5,6)→(2.29,6.54), B(6,2)→(5.83,4.41), C(4,1)→(5.12,2.29).

## Answer

Rotation about a pivot (xr, yr) = (3, 3) by θ = 45°: translate the pivot to the origin, rotate, translate back.

**Composite matrix** M = T(3,3) · R(45°) · T(−3,−3)

cos45° = sin45° = 0.7071

```text
        | 1 0 3 | | 0.7071 -0.7071 0 | | 1 0 -3 |
M     = | 0 1 3 | | 0.7071  0.7071 0 | | 0 1 -3 |
        | 0 0 1 | | 0       0      1 | | 0 0  1 |

        | 0.7071 -0.7071  3.0000 |
M     = | 0.7071  0.7071 -1.2426 |
        | 0       0       1      |
```

Using x' = xr + (x − xr)cosθ − (y − yr)sinθ and y' = yr + (x − xr)sinθ + (y − yr)cosθ:

| Point | (x−3, y−3) | Rotated about origin | + (3, 3) = new point |
|---|---|---|---|
| A(5,6) | (2, 3) | (2·0.7071 − 3·0.7071, 2·0.7071 + 3·0.7071) = (−0.707, 3.536) | A'(2.29, 6.54) |
| B(6,2) | (3, −1) | (2.121+0.707, 2.121−0.707) = (2.828, 1.414) | B'(5.83, 4.41) |
| C(4,1) | (1, −2) | (0.707+1.414, 0.707−1.414) = (2.121, −0.707) | C'(5.12, 2.29) |

**Answer:** A'(2.29, 6.54), B'(5.83, 4.41), C'(5.12, 2.29).

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What is the matrix for rotation by 45° about (3,3)?
   A. R(45°)·T(3,3)
   B. T(-3,-3)·R(45°)·T(3,3)
   C. R(45°) only
   D. T(3,3)·R(45°)·T(-3,-3)
   Answer: D. Translate, rotate, translate back. (Key point: Composite)

2. What is A(5,6) after translating the pivot to origin?
   A. (2, 3)
   B. (8, 9)
   C. (5, 6)
   D. (-2, -3)
   Answer: A. Subtract (3,3). (Key point: First step)

3. Which is B'?
   A. (2.29, 6.54)
   B. (5.83, 4.41)
   C. (5.12, 2.29)
   D. (6, 2)
   Answer: B. (3,-1) rotates to (2.828, 1.414), plus (3,3). (Key point: Result)

4. What is the top-right entry of the composite matrix?
   A. -1.2426
   B. 0.7071
   C. 3.0000
   D. 0
   Answer: C. 3 - 3cos45° + 3sin45° = 3. (Key point: Translation entries)

## Related in Computer Graphics

- [Rotation about a pivot point](https://www.bctengineer.com/question.md?key=semester_5%2Fcomputer_graphics%3Arotation-about-pivot-point-derive) — Computer Graphics, chapter 3, asked 2×
- [Scaling about arbitrary fixed point](https://www.bctengineer.com/question.md?key=semester_5%2Fcomputer_graphics%3Ascaling-about-fixed-point) — Computer Graphics, chapter 3
- [3D scaling about arbitrary point](https://www.bctengineer.com/question.md?key=semester_5%2Fcomputer_graphics%3Ascale-3d-arbitrary-point) — Computer Graphics, chapter 4
- [3D rotation about arbitrary axis](https://www.bctengineer.com/question.md?key=semester_5%2Fcomputer_graphics%3Arotate-3d-arbitrary-axis-steps) — Computer Graphics, chapter 4, asked 2×
- [Intensity at centroid of a triangle](https://www.bctengineer.com/question.md?key=semester_5%2Fcomputer_graphics%3Atotal-intensity-triangle-centroid) — Computer Graphics, chapter 8
- [Rotate line about Y-axis by 90°](https://www.bctengineer.com/question.md?key=semester_5%2Fcomputer_graphics%3Arotate-line-y-axis-90) — Computer Graphics, chapter 4

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