2D Transformation in Computer Graphics
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2D transformations are important operations in computer graphics that allow modifying the position, size, and orientation of objects in a 2D plane. There are several types of 2D transformations including translation, rotation, scaling, and more. Transformations are represented using matrix math for efficient application of sequential transformations. Key techniques include homogeneous coordinates to allow different types of transformations to be combined into a single matrix operation.
![• Ynew = Yold x Sy = 0 x 3 = 0
Thus, New coordinates of corner C after scaling = (6, 0).
For Coordinates D(0, 0)
Let the new coordinates of corner D after scaling = (Xnew, Ynew).
Applying the scaling equations, we have-
• Xnew = Xold x Sx = 0 x 2 = 0
• Ynew = Yold x Sy = 0 x 3 = 0
Thus, New coordinates of corner D after scaling = (0, 0).
Thus, New coordinates of the square after scaling = A (0, 9), B(6, 9), C(6, 0), D(0, 0).
Derivation of scaling matrix based on arbitrary point
Step-01: Translate
tx = - Px, ty = - Py
[T1] = |
1 0 𝑡𝑥
0 1 𝑡𝑦
0 0 1
| = |
1 0 −𝑃𝑥
0 1 −𝑃𝑦
0 0 1
|
Step-02: Scaling about the origin
[S0] = |
𝑆𝑥 0 0
0 𝑆𝑦 0
0 0 1
|
Step-03: Back Translate
tx = Px, ty = Py
[T2] = |
1 0 𝑡𝑥
0 1 𝑡𝑦
0 0 1
| = |
1 0 𝑃𝑥
0 1 𝑃𝑦
0 0 1
|](https://image.slidesharecdn.com/2dtransformatiincomputergraphics-211026180432/85/2D-Transformation-in-Computer-Graphics-8-320.jpg)
![Step-04:
Scaling matrix about arbitrary point
[Sp] = [T2] [S0] [T1]
= |
1 0 𝑃𝑥
0 1 𝑃𝑦
0 0 1
| |
𝑆𝑥 0 0
0 𝑆𝑦 0
0 0 1
| |
1 0 −𝑃𝑥
0 1 −𝑃𝑦
0 0 1
|
= |
1 0 𝑃𝑥
0 1 𝑃𝑦
0 0 1
| |
𝑆𝑥 0 −𝑆𝑥. 𝑃𝑥
0 𝑆𝑦 −𝑆𝑦. 𝑃𝑦
0 0 1
|
= |
𝑆𝑥 0 −𝑆𝑥. 𝑃𝑥 + 𝑃𝑥
0 𝑆𝑦 −𝑆𝑦. 𝑃𝑦 + 𝑃𝑦
0 0 1
|
=|
𝑆𝑥 0 𝑃𝑥(1 − 𝑆𝑥)
0 𝑆𝑦 𝑃𝑦(1 − 𝑆𝑦)
0 0 1
|
Problem:
Magnify the triangle with vertices A(0, 0), B(1, -1) and C(4, 2) to twice its size by keeping the
vertex C(4, 2) fixed.
Solution:
Here, scaling factor along X-axis, Sx= 2 and scaling factor along Y-axis, Sy=2 and the arbitrary
point Px=4 and Py=2
We know that,
[
𝑥′
𝑦′
1
] = [
𝑆𝑥 0 𝑃𝑥(1 − 𝑆𝑥)
0 𝑆𝑦 𝑃𝑦(1 − 𝑆𝑦)
0 0 1
] [
𝑥
𝑦
1
] (1)
For vertex A(0, 0) the equation (1) becomes
[
𝑥′
𝑦′
1
] = [
2 0 4(1 − 2)
0 2 2(1 − 2)
0 0 1
] [
0
0
1
]
= [
2 0 −4
0 2 −2
0 0 1
] [
0
0
1
]](https://image.slidesharecdn.com/2dtransformatiincomputergraphics-211026180432/85/2D-Transformation-in-Computer-Graphics-9-320.jpg)
![= [
0 + 0 − 4
0 + 0 − 2
0 + 0 + 1
]
= [
−4
−2
1
]
Therefore A’ = (x’, y’) = (-4, -2)
For vertex B(1, -1) the equation (1) becomes
[
𝑥′
𝑦′
1
] = [
2 0 4(1 − 2)
0 2 2(1 − 2)
0 0 1
] [
1
−1
1
]
= [
2 0 −4
0 2 −2
0 0 1
] [
1
−1
1
]
= [
2 − 0 − 4
0 − 2 − 2
0 + 0 + 1
]
= [
−2
−4
1
]
Therefore B’ = (x’, y’) = (-2, -4)
2D Rotation in Computer Graphics
In Computer graphics,
2D Rotation is a process of rotating an object with respect to an angle in a two dimensional plane
Consider a point object O has to be rotated from one angle to another in a 2D plane.
Let-
• Initial coordinates of the object O = (Xold, Yold)
• Initial angle of the object O with respect to origin = Φ
• Rotation angle = θ
• New coordinates of the object O after rotation = (Xnew, Ynew)](https://image.slidesharecdn.com/2dtransformatiincomputergraphics-211026180432/85/2D-Transformation-in-Computer-Graphics-10-320.jpg)
![Derivation of rotation matrix based on arbitrary point
Step-01: Translate
tx = - Px, ty = - Py
[T1] = |
1 0 𝑡𝑥
0 1 𝑡𝑦
0 0 1
| = |
1 0 −𝑃𝑥
0 1 −𝑃𝑦
0 0 1
|
Step-02: Rotate about origin
[R0] = |
𝑐𝑜𝑠𝜃 −𝑠𝑖𝑛𝜃 0
𝑠𝑖𝑛𝜃 𝑐𝑜𝑠𝜃 0
0 0 1
|
Step-03: Back Translate
tx = Px, ty = Py
[T2] = |
1 0 𝑡𝑥
0 1 𝑡𝑦
0 0 1
| = |
1 0 𝑃𝑥
0 1 𝑃𝑦
0 0 1
|
Step-04: Rotation matrix about arbitrary point
[Tp] = [T2] [R0] [T1]
= |
1 0 𝑃𝑥
0 1 𝑃𝑦
0 0 1
| |
𝑐𝑜𝑠𝜃 −𝑠𝑖𝑛𝜃 0
𝑠𝑖𝑛𝜃 𝑐𝑜𝑠𝜃 0
0 0 1
| |
1 0 −𝑃𝑥
0 1 −𝑃𝑦
0 0 1
|
= |
1 0 𝑃𝑥
0 1 𝑃𝑦
0 0 1
| |
𝑐𝑜𝑠𝜃 −𝑠𝑖𝑛𝜃 −𝑃𝑥𝑐𝑜𝑠𝜃 + 𝑃𝑦𝑠𝑖𝑛𝜃
𝑠𝑖𝑛𝜃 𝑐𝑜𝑠𝜃 −𝑃𝑥𝑠𝑖𝑛𝜃 − 𝑃𝑦𝑐𝑜𝑠𝜃
0 0 1
|
= |
𝑐𝑜𝑠𝜃 −𝑠𝑖𝑛𝜃 −𝑃𝑥𝑐𝑜𝑠𝜃 + 𝑃𝑦𝑠𝑖𝑛𝜃 + 𝑃𝑥
𝑠𝑖𝑛𝜃 𝑐𝑜𝑠𝜃 −𝑃𝑥𝑠𝑖𝑛𝜃 − 𝑃𝑦𝑐𝑜𝑠𝜃 + 𝑃𝑦
0 0 1
|
= |
𝑐𝑜𝑠𝜃 −𝑠𝑖𝑛𝜃 𝑃𝑥(1 − 𝑐𝑜𝑠𝜃) + 𝑃𝑦𝑠𝑖𝑛𝜃
𝑠𝑖𝑛𝜃 𝑐𝑜𝑠𝜃 𝑃𝑦(1 − 𝑐𝑜𝑠𝜃 − 𝑃𝑥𝑠𝑖𝑛𝜃
0 0 1
|](https://image.slidesharecdn.com/2dtransformatiincomputergraphics-211026180432/85/2D-Transformation-in-Computer-Graphics-14-320.jpg)
![Problem
Perform a 45 degree rotation of triangle A(0, 0), B (1, 1) and C(5, 2)
i. About the origin
ii. About P(-1, -1)
Solution
i. We know that,
[
𝑥′
𝑦′
1
] = [
𝑐𝑜𝑠𝜃 −𝑠𝑖𝑛𝜃 0
𝑠𝑖𝑛𝜃 𝑐𝑜𝑠𝜃 0
0 0 1
] [
𝑥
𝑦
1
] (2)
For vertex A(0, 0) the equation (2) becomes
[
𝑥′
𝑦′
1
] = [
𝑐𝑜𝑠450
−𝑠𝑖𝑛450
0
𝑠𝑖𝑛450
𝑐𝑜𝑠450
0
0 0 1
] [
0
0
1
]
= [
1
√2
−
1
√2
0
1
√2
1
√2
0
0 0 1
] [
0
0
1
]
= [
0 − 0 + 0
0 + 0 + 0
0 + 0 + 1
]
= [
0
0
1
]
Therefore A’ = (x’, y’) = (0, 0)
For vertex B(1, 1) the equation (2) becomes
[
𝑥′
𝑦′
1
] = [
𝑐𝑜𝑠450
−𝑠𝑖𝑛450
0
𝑠𝑖𝑛450
𝑐𝑜𝑠450
0
0 0 1
] [
1
1
1
]
= [
1
√2
−
1
√2
0
1
√2
1
√2
0
0 0 1
] [
1
1
1
]
= [
1
√2
−
1
√2
+ 0
1
√2
+
1
√2
+ 0
0 + 0 + 1
]](https://image.slidesharecdn.com/2dtransformatiincomputergraphics-211026180432/85/2D-Transformation-in-Computer-Graphics-15-320.jpg)
![= [
0
2
√2
1
]
=[
0
√2
1
]
Therefore B’ = (x’, y’) = (0, √2)
For vertex C(5, 2) the equation (2) becomes
[
𝑥′
𝑦′
1
] = [
𝑐𝑜𝑠450
−𝑠𝑖𝑛450
0
𝑠𝑖𝑛450
𝑐𝑜𝑠450
0
0 0 1
] [
5
2
1
]
= [
1
√2
−
1
√2
0
1
√2
1
√2
0
0 0 1
] [
5
2
1
]
= [
5
√2
−
2
√2
+ 0
5
√2
+
2
√2
+ 0
0 + 0 + 1
]
= [
3
√2
7
√2
1
]
=
[
3√2
2
7√2
2
1 ]
Therefore C’ = (x’, y’) = (
3√2
2
,
7√2
2
)
ii. We know that,
[
𝑥′
𝑦′
1
] = [
𝑐𝑜𝑠𝜃 −𝑠𝑖𝑛𝜃 𝑃𝑥(1 − 𝑐𝑜𝑠𝜃) + 𝑃𝑦𝑠𝑖𝑛𝜃
𝑠𝑖𝑛𝜃 𝑐𝑜𝑠𝜃 𝑃𝑦(1 − 𝑐𝑜𝑠𝜃) − 𝑃𝑥𝑠𝑖𝑛𝜃
0 0 1
] [
𝑥
𝑦
1
] (3)
For vertex A(0, 0) the equation (3) becomes
[
𝑥′
𝑦′
1
] = [
𝑐𝑜𝑠450
−𝑠𝑖𝑛450
−1(1 − 𝑐𝑜𝑠450) + (−1)𝑠𝑖𝑛450
𝑠𝑖𝑛450
𝑐𝑜𝑠450
−1(1 − 𝑐𝑜𝑠450) − (−1)𝑠𝑖𝑛450
0 0 1
] [
0
0
1
]](https://image.slidesharecdn.com/2dtransformatiincomputergraphics-211026180432/85/2D-Transformation-in-Computer-Graphics-16-320.jpg)
![=[
1
√2
−
1
√2
−1 (1 −
1
√2
) + (−1)
1
√2
1
√2
1
√2
−1 (1 −
1
√2
) − (−1)
1
√2
0 0 1
] [
0
0
1
]
=[
1
√2
−
1
√2
−1 (1 −
1
√2
) −
1
√2
1
√2
1
√2
−1 (1 −
1
√2
) +
1
√2
0 0 1
] [
0
0
1
]
=
[
1
√2
−
1
√2
− (
√2−1
√2
) −
1
√2
1
√2
1
√2
− (
√2−1
√2
) +
1
√2
0 0 1 ]
[
0
0
1
]
=
[
1
√2
−
1
√2
−√2+1−1
√2
1
√2
1
√2
−√2+1+1
√2
0 0 1 ]
[
0
0
1
]
=
[
1
√2
−
1
√2
−√2
√2
1
√2
1
√2
−√2+2
√2
0 0 1 ]
[
0
0
1
]
=[
1
√2
−
1
√2
−√2
√2
1
√2
1
√2
√2 − 1
0 0 1
] [
0
0
1
]
=[
0 − 0 +
−√2
√2
0 + 0 + (√2 − 1)
0 + 0 + 1
]
=[
−1
√2 − 1
1
]
Therefore A’ = (x’, y’) = (-1, √2 − 1)
For vertex B(1, 1) the equation (3) becomes
[
𝑥′
𝑦′
1
] = [
𝑐𝑜𝑠450
−𝑠𝑖𝑛450
−1(1 − 𝑐𝑜𝑠450) + (−1)𝑠𝑖𝑛450
𝑠𝑖𝑛450
𝑐𝑜𝑠450
−1(1 − 𝑐𝑜𝑠450) − (−1)𝑠𝑖𝑛450
0 0 1
] [
1
1
1
]](https://image.slidesharecdn.com/2dtransformatiincomputergraphics-211026180432/85/2D-Transformation-in-Computer-Graphics-17-320.jpg)
![=[
1
√2
−
1
√2
−1 (1 −
1
√2
) + (−1)
1
√2
1
√2
1
√2
−1 (1 −
1
√2
) − (−1)
1
√2
0 0 1
] [
1
1
1
]
=[
1
√2
−
1
√2
−1 (1 −
1
√2
) −
1
√2
1
√2
1
√2
−1 (1 −
1
√2
) +
1
√2
0 0 1
] [
1
1
1
]
=
[
1
√2
−
1
√2
− (
√2−1
√2
) −
1
√2
1
√2
1
√2
− (
√2−1
√2
) +
1
√2
0 0 1 ]
[
1
1
1
]
=
[
1
√2
−
1
√2
−√2+1−1
√2
1
√2
1
√2
−√2+1+1
√2
0 0 1 ]
[
1
1
1
]
=
[
1
√2
−
1
√2
−√2
√2
1
√2
1
√2
−√2+2
√2
0 0 1 ]
[
1
1
1
]
=[
1
√2
−
1
√2
−1
1
√2
1
√2
√2 − 1
0 0 1
] [
1
1
1
]
=[
1
√2
−
1
√2
− 1
1
√2
+
1
√2
+ (√2 − 1)
0 + 0 + 1
]
=
[
1−1−√2
√2
1+1+√2(√2−1)
√2
1 ]
=
[
−√2
√2
2+(√2)2−√2
√2
1 ]](https://image.slidesharecdn.com/2dtransformatiincomputergraphics-211026180432/85/2D-Transformation-in-Computer-Graphics-18-320.jpg)
![=[
−1
2+2−√2
√2
1
]
=[
−1
4−√2
√2
1
]
=[
−1
2√2√2−√2
√2
1
]
=[
−1
√2(2√2−1)
√2
1
]
=[
−1
2√2 − 1
1
]
Therefore B’ = (x’, y’) = (-1, 2√2 − 1)
For vertex C(5, 2) the equation (3) becomes
[
𝑥′
𝑦′
1
] = [
𝑐𝑜𝑠450
−𝑠𝑖𝑛450
−1(1 − 𝑐𝑜𝑠450) + (−1)𝑠𝑖𝑛450
𝑠𝑖𝑛450
𝑐𝑜𝑠450
−1(1 − 𝑐𝑜𝑠450) − (−1)𝑠𝑖𝑛450
0 0 1
] [
5
2
1
]
=[
1
√2
−
1
√2
−1 (1 −
1
√2
) + (−1)
1
√2
1
√2
1
√2
−1 (1 −
1
√2
) − (−1)
1
√2
0 0 1
] [
5
2
1
]
=[
1
√2
−
1
√2
−1 (1 −
1
√2
) −
1
√2
1
√2
1
√2
−1 (1 −
1
√2
) +
1
√2
0 0 1
] [
5
2
1
]
=
[
1
√2
−
1
√2
− (
√2−1
√2
) −
1
√2
1
√2
1
√2
− (
√2−1
√2
) +
1
√2
0 0 1 ]
[
5
2
1
]
=
[
1
√2
−
1
√2
−√2+1−1
√2
1
√2
1
√2
−√2+1+1
√2
0 0 1 ]
[
5
2
1
]](https://image.slidesharecdn.com/2dtransformatiincomputergraphics-211026180432/85/2D-Transformation-in-Computer-Graphics-19-320.jpg)
![=
[
1
√2
−
1
√2
−√2
√2
1
√2
1
√2
−√2+2
√2
0 0 1 ]
[
5
2
1
]
=[
1
√2
−
1
√2
−1
1
√2
1
√2
√2 − 1
0 0 1
] [
5
2
1
]
=[
5
√2
−
2
√2
− 1
5
√2
+
2
√2
+ (√2 − 1)
1
]
=
[
5−2−√2
√2
5+2+√2(√2−1)
√2
1 ]
=
[
3−√2
√2
7+(√2)2−√2
√2
1 ]
=
[
3−√2
√2
9−√2
√2
1 ]
=
[
3√2−2
2
9√2−2
2
1 ]
=
[
3√2−2
2
9√2−2
2
1 ]
=
[
3√2
2
− 1
9√2
2
− 1
1 ]
Therefore C’ = (x’, y’) = (
3√2
2
− 1,
9√2
2
− 1)](https://image.slidesharecdn.com/2dtransformatiincomputergraphics-211026180432/85/2D-Transformation-in-Computer-Graphics-20-320.jpg)
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2D Transformation in Computer Graphics
- 1. 2D Transformation in Computer Graphics In Computer graphics, Transformation is a process of modifying and re-positioning the existing graphics. • 2D Transformations take place in a two dimensional plane. • Transformations are helpful in changing the position, size, orientation, shape etc of the object. Coordinate Transformation: The object is held stationary while the coordinate system is transformed relative to the object. Geometric Transformation: The object itself is transformed relative to the coordinate system or background. Transformation Techniques- In computer graphics, various transformation techniques are- 1. Translation 2. Rotation 3. Scaling 4. Reflection 5. Shear Homogenous Coordinates To perform a sequence of transformation such as translation followed by rotation and scaling, we need to follow a sequential process − • Translate the coordinates, • Rotate the translated coordinates, and then • Scale the rotated coordinates to complete the composite transformation. To shorten this process, we have to use 3×3 transformation matrix instead of 2×2 transformation matrix. To convert a 2×2 matrix to 3×3 matrix, we have to add an extra dummy coordinate W. In this way, we can represent the point by 3 numbers instead of 2 numbers, which is called Homogenous Coordinate system. In this system, we can represent all the transformation equations in matrix multiplication.
- 2. What is the significance of 2D Transformation? • Transformations are the building blocks of computer graphics. • Positioning, shaping, viewing positions are done by transformations. • Transformation is also used for determining the perspective views. 2D Translation in Computer Graphics In Computer graphics, 2D Translation is a process of moving an object from one position to another in a two dimensional plane. Consider a point object O has to be moved from one position to another in a 2D plane. Let- • Initial coordinates of the object O = (Xold, Yold) • New coordinates of the object O after translation = (Xnew, Ynew) • Translation vector or Shift vector = (Tx, Ty) Given a Translation vector (Tx, Ty)- • Tx defines the distance the Xold coordinate has to be moved. • Ty defines the distance the Yold coordinate has to be moved.
- 3. This translation is achieved by adding the translation coordinates to the old coordinates of the object as- • Xnew = Xold + Tx (This denotes translation towards X axis) • Ynew = Yold + Ty (This denotes translation towards Y axis) In Matrix form, the above translation equations may be represented as- • The homogeneous coordinates representation of (X, Y) is (X, Y, 1). • Through this representation, all the transformations can be performed using matrix / vector multiplications. The above translation matrix may be represented as a 3 x 3 matrix as- Problem Given a circle C with radius 10 and center coordinates (1, 4). Apply the translation with distance 5 towards X axis and 1 towards Y axis. Obtain the new coordinates of C without changing its radius.
- 4. Solution Given- • Old center coordinates of C = (Xold, Yold) = (1, 4) • Translation vector = (Tx, Ty) = (5, 1) Let the new center coordinates of C = (Xnew, Ynew). Applying the translation equations, we have- • Xnew = Xold + Tx = 1 + 5 = 6 • Ynew = Yold + Ty = 4 + 1 = 5 Thus, New center coordinates of C = (6, 5). Problem Given a square with coordinate points A(0, 3), B(3, 3), C(3, 0), D(0, 0). Apply the translation with distance 1 towards X axis and 1 towards Y axis. Obtain the new coordinates of the square. Solution Given- • Old coordinates of the square = A (0, 3), B(3, 3), C(3, 0), D(0, 0) • Translation vector = (Tx, Ty) = (1, 1) For Coordinates A(0, 3) Let the new coordinates of corner A = (Xnew, Ynew). Applying the translation equations, we have- • Xnew = Xold + Tx = 0 + 1 = 1 • Ynew = Yold + Ty = 3 + 1 = 4 Thus, New coordinates of corner A = (1, 4).
- 5. For Coordinates B(3, 3) Let the new coordinates of corner B = (Xnew, Ynew). Applying the translation equations, we have- • Xnew = Xold + Tx = 3 + 1 = 4 • Ynew = Yold + Ty = 3 + 1 = 4 Thus, New coordinates of corner B = (4, 4). For Coordinates C(3, 0) Let the new coordinates of corner C = (Xnew, Ynew). Applying the translation equations, we have- • Xnew = Xold + Tx = 3 + 1 = 4 • Ynew = Yold + Ty = 0 + 1 = 1 Thus, New coordinates of corner C = (4, 1). For Coordinates D(0, 0) Let the new coordinates of corner D = (Xnew, Ynew). Applying the translation equations, we have- • Xnew = Xold + Tx = 0 + 1 = 1 • Ynew = Yold + Ty = 0 + 1 = 1 Thus, New coordinates of corner D = (1, 1). Thus, New coordinates of the square = A (1, 4), B(4, 4), C(4, 1), D(1, 1).
- 6. 2D Scaling in Computer Graphics In computer graphics, scaling is a process of modifying or altering the size of objects. • Scaling may be used to increase or reduce the size of object. • Scaling subjects the coordinate points of the original object to change. • Scaling factor determines whether the object size is to be increased or reduced. • If scaling factor > 1, then the object size is increased. • If scaling factor < 1, then the object size is reduced. Consider a point object O has to be scaled in a 2D plane. Let- • Initial coordinates of the object O = (Xold, Yold) • Scaling factor for X-axis = Sx • Scaling factor for Y-axis = Sy • New coordinates of the object O after scaling = (Xnew, Ynew) This scaling is achieved by using the following scaling equations- • Xnew = Xold x Sx • Ynew = Yold x Sy In Matrix form, the above scaling equations may be represented as- For homogeneous coordinates, the above scaling matrix may be represented as a 3 x 3 matrix as-
- 7. Problem Given a square object with coordinate points A(0, 3), B(3, 3), C(3, 0), D(0, 0). Apply the scaling parameter 2 towards X axis and 3 towards Y axis and obtain the new coordinates of the object. Solution Given- • Old corner coordinates of the square = A (0, 3), B(3, 3), C(3, 0), D(0, 0) • Scaling factor along X axis = 2 • Scaling factor along Y axis = 3 For Coordinates A(0, 3) Let the new coordinates of corner A after scaling = (Xnew, Ynew). Applying the scaling equations, we have- • Xnew = Xold x Sx = 0 x 2 = 0 • Ynew = Yold x Sy = 3 x 3 = 9 Thus, New coordinates of corner A after scaling = (0, 9). For Coordinates B(3, 3) Let the new coordinates of corner B after scaling = (Xnew, Ynew). Applying the scaling equations, we have- • Xnew = Xold x Sx = 3 x 2 = 6 • Ynew = Yold x Sy = 3 x 3 = 9 Thus, New coordinates of corner B after scaling = (6, 9). For Coordinates C(3, 0) Let the new coordinates of corner C after scaling = (Xnew, Ynew). Applying the scaling equations, we have- • Xnew = Xold x Sx = 3 x 2 = 6
- 8. • Ynew = Yold x Sy = 0 x 3 = 0 Thus, New coordinates of corner C after scaling = (6, 0). For Coordinates D(0, 0) Let the new coordinates of corner D after scaling = (Xnew, Ynew). Applying the scaling equations, we have- • Xnew = Xold x Sx = 0 x 2 = 0 • Ynew = Yold x Sy = 0 x 3 = 0 Thus, New coordinates of corner D after scaling = (0, 0). Thus, New coordinates of the square after scaling = A (0, 9), B(6, 9), C(6, 0), D(0, 0). Derivation of scaling matrix based on arbitrary point Step-01: Translate tx = - Px, ty = - Py [T1] = | 1 0 𝑡𝑥 0 1 𝑡𝑦 0 0 1 | = | 1 0 −𝑃𝑥 0 1 −𝑃𝑦 0 0 1 | Step-02: Scaling about the origin [S0] = | 𝑆𝑥 0 0 0 𝑆𝑦 0 0 0 1 | Step-03: Back Translate tx = Px, ty = Py [T2] = | 1 0 𝑡𝑥 0 1 𝑡𝑦 0 0 1 | = | 1 0 𝑃𝑥 0 1 𝑃𝑦 0 0 1 |
- 9. Step-04: Scaling matrix about arbitrary point [Sp] = [T2] [S0] [T1] = | 1 0 𝑃𝑥 0 1 𝑃𝑦 0 0 1 | | 𝑆𝑥 0 0 0 𝑆𝑦 0 0 0 1 | | 1 0 −𝑃𝑥 0 1 −𝑃𝑦 0 0 1 | = | 1 0 𝑃𝑥 0 1 𝑃𝑦 0 0 1 | | 𝑆𝑥 0 −𝑆𝑥. 𝑃𝑥 0 𝑆𝑦 −𝑆𝑦. 𝑃𝑦 0 0 1 | = | 𝑆𝑥 0 −𝑆𝑥. 𝑃𝑥 + 𝑃𝑥 0 𝑆𝑦 −𝑆𝑦. 𝑃𝑦 + 𝑃𝑦 0 0 1 | =| 𝑆𝑥 0 𝑃𝑥(1 − 𝑆𝑥) 0 𝑆𝑦 𝑃𝑦(1 − 𝑆𝑦) 0 0 1 | Problem: Magnify the triangle with vertices A(0, 0), B(1, -1) and C(4, 2) to twice its size by keeping the vertex C(4, 2) fixed. Solution: Here, scaling factor along X-axis, Sx= 2 and scaling factor along Y-axis, Sy=2 and the arbitrary point Px=4 and Py=2 We know that, [ 𝑥′ 𝑦′ 1 ] = [ 𝑆𝑥 0 𝑃𝑥(1 − 𝑆𝑥) 0 𝑆𝑦 𝑃𝑦(1 − 𝑆𝑦) 0 0 1 ] [ 𝑥 𝑦 1 ] (1) For vertex A(0, 0) the equation (1) becomes [ 𝑥′ 𝑦′ 1 ] = [ 2 0 4(1 − 2) 0 2 2(1 − 2) 0 0 1 ] [ 0 0 1 ] = [ 2 0 −4 0 2 −2 0 0 1 ] [ 0 0 1 ]
- 10. = [ 0 + 0 − 4 0 + 0 − 2 0 + 0 + 1 ] = [ −4 −2 1 ] Therefore A’ = (x’, y’) = (-4, -2) For vertex B(1, -1) the equation (1) becomes [ 𝑥′ 𝑦′ 1 ] = [ 2 0 4(1 − 2) 0 2 2(1 − 2) 0 0 1 ] [ 1 −1 1 ] = [ 2 0 −4 0 2 −2 0 0 1 ] [ 1 −1 1 ] = [ 2 − 0 − 4 0 − 2 − 2 0 + 0 + 1 ] = [ −2 −4 1 ] Therefore B’ = (x’, y’) = (-2, -4) 2D Rotation in Computer Graphics In Computer graphics, 2D Rotation is a process of rotating an object with respect to an angle in a two dimensional plane Consider a point object O has to be rotated from one angle to another in a 2D plane. Let- • Initial coordinates of the object O = (Xold, Yold) • Initial angle of the object O with respect to origin = Φ • Rotation angle = θ • New coordinates of the object O after rotation = (Xnew, Ynew)
- 11. This rotation is achieved by using the following rotation equations- • Xnew = Xold x cosθ – Yold x sinθ • Ynew = Xold x sinθ + Yold x cosθ In Matrix form, the above rotation equations may be represented as- For homogeneous coordinates, the above rotation matrix may be represented as a 3 x 3 matrix as-
- 12. Problem Given a triangle with corner coordinates (0, 0), (1, 0) and (1, 1). Rotate the triangle by 90-degree anticlockwise direction and find out the new coordinates. Solution We rotate a polygon by rotating each vertex of it with the same rotation angle. Given- • Old corner coordinates of the triangle = A (0, 0), B(1, 0), C(1, 1) • Rotation angle = θ = 90º For Coordinates A(0, 0) Let the new coordinates of corner A after rotation = (Xnew, Ynew). Applying the rotation equations, we have- Xnew = Xold x cosθ – Yold x sinθ = 0 x cos90º – 0 x sin90º = 0 Ynew = Xold x sinθ + Yold x cosθ = 0 x sin90º + 0 x cos90º = 0 Thus, New coordinates of corner A after rotation = (0, 0). For Coordinates B(1, 0) Let the new coordinates of corner B after rotation = (Xnew, Ynew). Xnew = Xold x cosθ – Yold x sinθ = 1 x cos90º – 0 x sin90º = 0 Ynew = Xold x sinθ + Yold x cosθ = 1 x sin90º + 0 x cos90º = 1 + 0
- 13. = 1 Thus, New coordinates of corner B after rotation = (0, 1). For Coordinates C(1, 1) Let the new coordinates of corner C after rotation = (Xnew, Ynew). Xnew = Xold x cosθ – Yold x sinθ = 1 x cos90º – 1 x sin90º = 0 – 1 = -1 Ynew = Xold x sinθ + Yold x cosθ = 1 x sin90º + 1 x cos90º = 1 + 0 = 1 Thus, New coordinates of corner C after rotation = (-1, 1). Thus, New coordinates of the triangle after rotation = A (0, 0), B(0, 1), C(-1, 1).
- 14. Derivation of rotation matrix based on arbitrary point Step-01: Translate tx = - Px, ty = - Py [T1] = | 1 0 𝑡𝑥 0 1 𝑡𝑦 0 0 1 | = | 1 0 −𝑃𝑥 0 1 −𝑃𝑦 0 0 1 | Step-02: Rotate about origin [R0] = | 𝑐𝑜𝑠𝜃 −𝑠𝑖𝑛𝜃 0 𝑠𝑖𝑛𝜃 𝑐𝑜𝑠𝜃 0 0 0 1 | Step-03: Back Translate tx = Px, ty = Py [T2] = | 1 0 𝑡𝑥 0 1 𝑡𝑦 0 0 1 | = | 1 0 𝑃𝑥 0 1 𝑃𝑦 0 0 1 | Step-04: Rotation matrix about arbitrary point [Tp] = [T2] [R0] [T1] = | 1 0 𝑃𝑥 0 1 𝑃𝑦 0 0 1 | | 𝑐𝑜𝑠𝜃 −𝑠𝑖𝑛𝜃 0 𝑠𝑖𝑛𝜃 𝑐𝑜𝑠𝜃 0 0 0 1 | | 1 0 −𝑃𝑥 0 1 −𝑃𝑦 0 0 1 | = | 1 0 𝑃𝑥 0 1 𝑃𝑦 0 0 1 | | 𝑐𝑜𝑠𝜃 −𝑠𝑖𝑛𝜃 −𝑃𝑥𝑐𝑜𝑠𝜃 + 𝑃𝑦𝑠𝑖𝑛𝜃 𝑠𝑖𝑛𝜃 𝑐𝑜𝑠𝜃 −𝑃𝑥𝑠𝑖𝑛𝜃 − 𝑃𝑦𝑐𝑜𝑠𝜃 0 0 1 | = | 𝑐𝑜𝑠𝜃 −𝑠𝑖𝑛𝜃 −𝑃𝑥𝑐𝑜𝑠𝜃 + 𝑃𝑦𝑠𝑖𝑛𝜃 + 𝑃𝑥 𝑠𝑖𝑛𝜃 𝑐𝑜𝑠𝜃 −𝑃𝑥𝑠𝑖𝑛𝜃 − 𝑃𝑦𝑐𝑜𝑠𝜃 + 𝑃𝑦 0 0 1 | = | 𝑐𝑜𝑠𝜃 −𝑠𝑖𝑛𝜃 𝑃𝑥(1 − 𝑐𝑜𝑠𝜃) + 𝑃𝑦𝑠𝑖𝑛𝜃 𝑠𝑖𝑛𝜃 𝑐𝑜𝑠𝜃 𝑃𝑦(1 − 𝑐𝑜𝑠𝜃 − 𝑃𝑥𝑠𝑖𝑛𝜃 0 0 1 |
- 15. Problem Perform a 45 degree rotation of triangle A(0, 0), B (1, 1) and C(5, 2) i. About the origin ii. About P(-1, -1) Solution i. We know that, [ 𝑥′ 𝑦′ 1 ] = [ 𝑐𝑜𝑠𝜃 −𝑠𝑖𝑛𝜃 0 𝑠𝑖𝑛𝜃 𝑐𝑜𝑠𝜃 0 0 0 1 ] [ 𝑥 𝑦 1 ] (2) For vertex A(0, 0) the equation (2) becomes [ 𝑥′ 𝑦′ 1 ] = [ 𝑐𝑜𝑠450 −𝑠𝑖𝑛450 0 𝑠𝑖𝑛450 𝑐𝑜𝑠450 0 0 0 1 ] [ 0 0 1 ] = [ 1 √2 − 1 √2 0 1 √2 1 √2 0 0 0 1 ] [ 0 0 1 ] = [ 0 − 0 + 0 0 + 0 + 0 0 + 0 + 1 ] = [ 0 0 1 ] Therefore A’ = (x’, y’) = (0, 0) For vertex B(1, 1) the equation (2) becomes [ 𝑥′ 𝑦′ 1 ] = [ 𝑐𝑜𝑠450 −𝑠𝑖𝑛450 0 𝑠𝑖𝑛450 𝑐𝑜𝑠450 0 0 0 1 ] [ 1 1 1 ] = [ 1 √2 − 1 √2 0 1 √2 1 √2 0 0 0 1 ] [ 1 1 1 ] = [ 1 √2 − 1 √2 + 0 1 √2 + 1 √2 + 0 0 + 0 + 1 ]
- 16. = [ 0 2 √2 1 ] =[ 0 √2 1 ] Therefore B’ = (x’, y’) = (0, √2) For vertex C(5, 2) the equation (2) becomes [ 𝑥′ 𝑦′ 1 ] = [ 𝑐𝑜𝑠450 −𝑠𝑖𝑛450 0 𝑠𝑖𝑛450 𝑐𝑜𝑠450 0 0 0 1 ] [ 5 2 1 ] = [ 1 √2 − 1 √2 0 1 √2 1 √2 0 0 0 1 ] [ 5 2 1 ] = [ 5 √2 − 2 √2 + 0 5 √2 + 2 √2 + 0 0 + 0 + 1 ] = [ 3 √2 7 √2 1 ] = [ 3√2 2 7√2 2 1 ] Therefore C’ = (x’, y’) = ( 3√2 2 , 7√2 2 ) ii. We know that, [ 𝑥′ 𝑦′ 1 ] = [ 𝑐𝑜𝑠𝜃 −𝑠𝑖𝑛𝜃 𝑃𝑥(1 − 𝑐𝑜𝑠𝜃) + 𝑃𝑦𝑠𝑖𝑛𝜃 𝑠𝑖𝑛𝜃 𝑐𝑜𝑠𝜃 𝑃𝑦(1 − 𝑐𝑜𝑠𝜃) − 𝑃𝑥𝑠𝑖𝑛𝜃 0 0 1 ] [ 𝑥 𝑦 1 ] (3) For vertex A(0, 0) the equation (3) becomes [ 𝑥′ 𝑦′ 1 ] = [ 𝑐𝑜𝑠450 −𝑠𝑖𝑛450 −1(1 − 𝑐𝑜𝑠450) + (−1)𝑠𝑖𝑛450 𝑠𝑖𝑛450 𝑐𝑜𝑠450 −1(1 − 𝑐𝑜𝑠450) − (−1)𝑠𝑖𝑛450 0 0 1 ] [ 0 0 1 ]
- 17. =[ 1 √2 − 1 √2 −1 (1 − 1 √2 ) + (−1) 1 √2 1 √2 1 √2 −1 (1 − 1 √2 ) − (−1) 1 √2 0 0 1 ] [ 0 0 1 ] =[ 1 √2 − 1 √2 −1 (1 − 1 √2 ) − 1 √2 1 √2 1 √2 −1 (1 − 1 √2 ) + 1 √2 0 0 1 ] [ 0 0 1 ] = [ 1 √2 − 1 √2 − ( √2−1 √2 ) − 1 √2 1 √2 1 √2 − ( √2−1 √2 ) + 1 √2 0 0 1 ] [ 0 0 1 ] = [ 1 √2 − 1 √2 −√2+1−1 √2 1 √2 1 √2 −√2+1+1 √2 0 0 1 ] [ 0 0 1 ] = [ 1 √2 − 1 √2 −√2 √2 1 √2 1 √2 −√2+2 √2 0 0 1 ] [ 0 0 1 ] =[ 1 √2 − 1 √2 −√2 √2 1 √2 1 √2 √2 − 1 0 0 1 ] [ 0 0 1 ] =[ 0 − 0 + −√2 √2 0 + 0 + (√2 − 1) 0 + 0 + 1 ] =[ −1 √2 − 1 1 ] Therefore A’ = (x’, y’) = (-1, √2 − 1) For vertex B(1, 1) the equation (3) becomes [ 𝑥′ 𝑦′ 1 ] = [ 𝑐𝑜𝑠450 −𝑠𝑖𝑛450 −1(1 − 𝑐𝑜𝑠450) + (−1)𝑠𝑖𝑛450 𝑠𝑖𝑛450 𝑐𝑜𝑠450 −1(1 − 𝑐𝑜𝑠450) − (−1)𝑠𝑖𝑛450 0 0 1 ] [ 1 1 1 ]
- 18. =[ 1 √2 − 1 √2 −1 (1 − 1 √2 ) + (−1) 1 √2 1 √2 1 √2 −1 (1 − 1 √2 ) − (−1) 1 √2 0 0 1 ] [ 1 1 1 ] =[ 1 √2 − 1 √2 −1 (1 − 1 √2 ) − 1 √2 1 √2 1 √2 −1 (1 − 1 √2 ) + 1 √2 0 0 1 ] [ 1 1 1 ] = [ 1 √2 − 1 √2 − ( √2−1 √2 ) − 1 √2 1 √2 1 √2 − ( √2−1 √2 ) + 1 √2 0 0 1 ] [ 1 1 1 ] = [ 1 √2 − 1 √2 −√2+1−1 √2 1 √2 1 √2 −√2+1+1 √2 0 0 1 ] [ 1 1 1 ] = [ 1 √2 − 1 √2 −√2 √2 1 √2 1 √2 −√2+2 √2 0 0 1 ] [ 1 1 1 ] =[ 1 √2 − 1 √2 −1 1 √2 1 √2 √2 − 1 0 0 1 ] [ 1 1 1 ] =[ 1 √2 − 1 √2 − 1 1 √2 + 1 √2 + (√2 − 1) 0 + 0 + 1 ] = [ 1−1−√2 √2 1+1+√2(√2−1) √2 1 ] = [ −√2 √2 2+(√2)2−√2 √2 1 ]
- 19. =[ −1 2+2−√2 √2 1 ] =[ −1 4−√2 √2 1 ] =[ −1 2√2√2−√2 √2 1 ] =[ −1 √2(2√2−1) √2 1 ] =[ −1 2√2 − 1 1 ] Therefore B’ = (x’, y’) = (-1, 2√2 − 1) For vertex C(5, 2) the equation (3) becomes [ 𝑥′ 𝑦′ 1 ] = [ 𝑐𝑜𝑠450 −𝑠𝑖𝑛450 −1(1 − 𝑐𝑜𝑠450) + (−1)𝑠𝑖𝑛450 𝑠𝑖𝑛450 𝑐𝑜𝑠450 −1(1 − 𝑐𝑜𝑠450) − (−1)𝑠𝑖𝑛450 0 0 1 ] [ 5 2 1 ] =[ 1 √2 − 1 √2 −1 (1 − 1 √2 ) + (−1) 1 √2 1 √2 1 √2 −1 (1 − 1 √2 ) − (−1) 1 √2 0 0 1 ] [ 5 2 1 ] =[ 1 √2 − 1 √2 −1 (1 − 1 √2 ) − 1 √2 1 √2 1 √2 −1 (1 − 1 √2 ) + 1 √2 0 0 1 ] [ 5 2 1 ] = [ 1 √2 − 1 √2 − ( √2−1 √2 ) − 1 √2 1 √2 1 √2 − ( √2−1 √2 ) + 1 √2 0 0 1 ] [ 5 2 1 ] = [ 1 √2 − 1 √2 −√2+1−1 √2 1 √2 1 √2 −√2+1+1 √2 0 0 1 ] [ 5 2 1 ]
- 20. = [ 1 √2 − 1 √2 −√2 √2 1 √2 1 √2 −√2+2 √2 0 0 1 ] [ 5 2 1 ] =[ 1 √2 − 1 √2 −1 1 √2 1 √2 √2 − 1 0 0 1 ] [ 5 2 1 ] =[ 5 √2 − 2 √2 − 1 5 √2 + 2 √2 + (√2 − 1) 1 ] = [ 5−2−√2 √2 5+2+√2(√2−1) √2 1 ] = [ 3−√2 √2 7+(√2)2−√2 √2 1 ] = [ 3−√2 √2 9−√2 √2 1 ] = [ 3√2−2 2 9√2−2 2 1 ] = [ 3√2−2 2 9√2−2 2 1 ] = [ 3√2 2 − 1 9√2 2 − 1 1 ] Therefore C’ = (x’, y’) = ( 3√2 2 − 1, 9√2 2 − 1)
- 21. 2D Reflection in Computer Graphics • Reflection is a kind of rotation where the angle of rotation is 180 degree. • The reflected object is always formed on the other side of mirror. • The size of reflected object is same as the size of original object Consider a point object O has to be reflected in a 2D plane. Let- • Initial coordinates of the object O = (Xold, Yold) • New coordinates of the reflected object O after reflection = (Xnew, Ynew) Reflection on X-axis This reflection is achieved by using the following reflection equations- • Xnew = Xold • Ynew = -Yold In Matrix form, the above reflection equations may be represented as- For homogeneous coordinates, the above reflection matrix may be represented as a 3 x 3 matrix as-
- 22. Reflection on Y-axis This reflection is achieved by using the following reflection equations- • Xnew = -Xold • Ynew = Yold In Matrix form, the above reflection equations may be represented as- For homogeneous coordinates, the above reflection matrix may be represented as a 3 x 3 matrix as- Problem Given a triangle with coordinate points A(3, 4), B(6, 4), C(5, 6). Apply the reflection on the X axis and obtain the new coordinates of the object. Solution Given- • Old corner coordinates of the triangle = A (3, 4), B(6, 4), C(5, 6) • Reflection has to be taken on the X axis For Coordinates A(3, 4) Let the new coordinates of corner A after reflection = (Xnew, Ynew).
- 23. Applying the reflection equations, we have- • Xnew = Xold = 3 • Ynew = -Yold = -4 Thus, New coordinates of corner A after reflection = (3, -4). For Coordinates B(6, 4) Let the new coordinates of corner B after reflection = (Xnew, Ynew). Applying the reflection equations, we have- • Xnew = Xold = 6 • Ynew = -Yold = -4 Thus, New coordinates of corner B after reflection = (6, -4). For Coordinates C(5, 6) Let the new coordinates of corner C after reflection = (Xnew, Ynew). Applying the reflection equations, we have- • Xnew = Xold = 5 • Ynew = -Yold = -6 Thus, New coordinates of corner C after reflection = (5, -6). Thus, New coordinates of the triangle after reflection = A (3, -4), B(6, -4), C(5, -6).
- 24. Problem Given a triangle with coordinate points A(3, 4), B(6, 4), C(5, 6). Apply the reflection on the Y axis and obtain the new coordinates of the object. Solution Given- • Old corner coordinates of the triangle = A (3, 4), B(6, 4), C(5, 6) • Reflection has to be taken on the Y axis For Coordinates A(3, 4) Let the new coordinates of corner A after reflection = (Xnew, Ynew). Applying the reflection equations, we have- • Xnew = -Xold = -3 • Ynew = Yold = 4 Thus, New coordinates of corner A after reflection = (-3, 4). For Coordinates B(6, 4) Let the new coordinates of corner B after reflection = (Xnew, Ynew). Applying the reflection equations, we have- • Xnew = -Xold = -6 • Ynew = Yold = 4 Thus, New coordinates of corner B after reflection = (-6, 4). For Coordinates C(5, 6) Let the new coordinates of corner C after reflection = (Xnew, Ynew). Applying the reflection equations, we have- • Xnew = -Xold = -5 • Ynew = Yold = 6 Thus, New coordinates of corner C after reflection = (-5, 6). Thus, New coordinates of the triangle after reflection = A (-3, 4), B(-6, 4), C(-5, 6).
- 25. 2D Shearing in Computer Graphics In Computer graphics, 2D Shearing is an ideal technique to change the shape of an existing object in a two dimensional plane. In a two dimensional plane, the object size can be changed along X direction as well as Y direction. So, there are two versions of shearing- 1. Shearing in X direction 2. Shearing in Y direction Consider a point object O has to be sheared in a 2D plane. Let- • Initial coordinates of the object O = (Xold, Yold) • Shearing parameter towards X direction = Shx • Shearing parameter towards Y direction = Shy • New coordinates of the object O after shearing = (Xnew, Ynew)
- 26. Shearing in X-axis Shearing in X axis is achieved by using the following shearing equations- • Xnew = Xold + Shx x Yold • Ynew = Yold In Matrix form, the above shearing equations may be represented as- For homogeneous coordinates, the above shearing matrix may be represented as a 3 x 3 matrix as- Shearing in Y-axis Shearing in Y axis is achieved by using the following shearing equations- • Xnew = Xold • Ynew = Yold + Shy x Xold In Matrix form, the above shearing equations may be represented as-
- 27. For homogeneous coordinates, the above shearing matrix may be represented as a 3 x 3 matrix as- Problem Given a triangle with points (1, 1), (0, 0) and (1, 0). Apply shear parameter 2 on X axis and 2 on Y axis and find out the new coordinates of the object. Solution Given- • Old corner coordinates of the triangle = A (1, 1), B(0, 0), C(1, 0) • Shearing parameter towards X direction (Shx) = 2 • Shearing parameter towards Y direction (Shy) = 2
- 28. Shearing in X-axis For Coordinates A(1, 1) Let the new coordinates of corner A after shearing = (Xnew, Ynew). Applying the shearing equations, we have- • Xnew = Xold + Shx x Yold = 1 + 2 x 1 = 3 • Ynew = Yold = 1 Thus, New coordinates of corner A after shearing = (3, 1). For Coordinates B(0, 0) Let the new coordinates of corner B after shearing = (Xnew, Ynew). Applying the shearing equations, we have- • Xnew = Xold + Shx x Yold = 0 + 2 x 0 = 0 • Ynew = Yold = 0 Thus, New coordinates of corner B after shearing = (0, 0). For Coordinates C(1, 0) Let the new coordinates of corner C after shearing = (Xnew, Ynew). Applying the shearing equations, we have- • Xnew = Xold + Shx x Yold = 1 + 2 x 0 = 1 • Ynew = Yold = 0 Thus, New coordinates of corner C after shearing = (1, 0). Thus, New coordinates of the triangle after shearing in X axis = A (3, 1), B(0, 0), C(1, 0). Shearing in Y-axis For Coordinates A(1, 1) Let the new coordinates of corner A after shearing = (Xnew, Ynew). Applying the shearing equations, we have- • Xnew = Xold = 1 • Ynew = Yold + Shy x Xold = 1 + 2 x 1 = 3 Thus, New coordinates of corner A after shearing = (1, 3). For Coordinates B(0, 0) Let the new coordinates of corner B after shearing = (Xnew, Ynew). Applying the shearing equations, we have- • Xnew = Xold = 0 • Ynew = Yold + Shy x Xold = 0 + 2 x 0 = 0
- 29. Thus, New coordinates of corner B after shearing = (0, 0). For Coordinates C(1, 0) Let the new coordinates of corner C after shearing = (Xnew, Ynew). Applying the shearing equations, we have- • Xnew = Xold = 1 • Ynew = Yold + Shy x Xold = 0 + 2 x 1 = 2 Thus, New coordinates of corner C after shearing = (1, 2). Thus, New coordinates of the triangle after shearing in Y axis = A (1, 3), B(0, 0), C(1, 2).