 Hello everyone, welcome to the lecture series on computer graphics 2D transformation translation. At the end of this session, students will be able to define 2D transformation, describe and represent 2D translation matrix, solve problems based on translation in computer graphic. In this lecture, we will be understanding the various types of 2D transformation in computer graphic. Understanding 2D translation, we will see how to represent a translation matrix and finally conclude with some practice problems on 2D translation. So what is transformation? In computer graphics, transformation is a process of modifying and repositioning the existing graphic. When a transformation takes place on a 2D plane, it is called as 2D transformation. Whereas when it takes place in three-dimensional plane, it is called as 3D transformation. Transformations are helpful in changing the position, size, orientation, shape of the object. So transformation basically modifies the graphic primitive. The graphic primitive can be anything right from a line, a circle, triangle, any polygon. So transformation in computer graphics are broadly categorized as translation, rotation, scaling, reflection and shearing, also called as shear. In this video lecture, we will focus our discussion to 2D translation. So what is translation? And as we are understanding, 2D means the translation that happens in two-dimension. So 2D translation is a process of moving an object from one position to another in a two-dimensional plane. Here in translation, repositioning of an object along a straight line path from one coordinate location to another takes place. As you can see in the diagram given here, x, y is a point, whereas this point is translated to x dash, y dash, where here the repositioning of the object that is the point happens along a straight line path. This is called as the translation distance which is denoted by Tx and Ty. To understand further, consider a point, object O that has to be moved from one position to another in a 2D plane as shown in the diagram. This is the object O that has to be moved in this two-dimensional plane. Let the initial coordinate of the object O be x old and y old. New coordinates of the object O after translation will be x new and y new. The translation vector or also the shift vector that causes the point to be moved from x old to x new is denoted by Tx and Ty as shown in the diagram. Given a translation vector Tx and Ty, this Tx defines the distance the x old coordinate has to be moved whereas Ty defines the distance y old coordinate has to be moved. Further, when we are representing this coordinates in the homogeneous coordinate representation of x y, it comes out to be x y and 1. Through this representation, all the transformation can be performed using matrix vector multiplication. Thus, the above translation matrix may be represented as a 3 by 3 matrix in the given format as shown here. x new, y new, whereas 1 0 Tx, 0 1 Ty, 0 0 1 into x old, y old upon 1. That is 1. So, this is the representation of the translation matrix using homogeneous coordinates. Now, before we move ahead over we once again repeat that when a point is being shifted from that is shifted means translated from one point location to other along a straight line path, the translation vector Tx and Ty is involved which is represented in a homogeneous representation in the given format. Now using this format, we will try to solve a particular practice problem. Now here the problem is based on 2D translation. I hope you have understood what is the concept of translation, how we represent a translation matrix, what are these translation distances in respect to x and y coordinates and how for the sake of convenience we have represented this translation matrix into a homogeneous coordinate representation. Based on the previous understanding we now try to solve a given problem. Given a square with coordinate points A03, B33, C30, D00, apply the translation with distance 1 towards x axis and 1 towards y axis, obtain the coordinates of the square. So, primarily in order to perform 2D translation we are given a square A, B, C, D with the position in x and y. Now we have been told to translate this square with one unit in both x and y axis. So, the objective here is to find the new coordinates of the translated square. So, for now we will be solving and applying the translation individually for every coordinates of the square. You have to solve this problem in the given stepwise approach. Given all coordinates of the square is equal to A03, B33, C3, 0, D00. Obviously the translation vector given in the problem that is Tx and Ty is one unit, both for x and y. Now let the new coordinates of corner A be x new and y new. Applying the translation equations we have x new is equal to x old plus Tx that is 0 plus 1 because A is 0 and the translation distance in x is 1, y new is equal to y old plus Ty that is 3 plus 1, we get it 4. Thus the new coordinates for corner A of the square is 1 comma 4. We continue this for the next coordinate that is B which is 33. Let the new coordinate of corner B is equal to x new, y new. Applying translation equation we have x new is equal to x old plus Tx again that is 3 plus 1, y is equal to 4, y new is equal to y old plus Ty is equal to 3 plus 1 that is 4. Thus the new coordinates of corner B is equal to 4, 4. Now moving ahead for coordinate 3 of the square ABCD. Let the new coordinates of corner C be x new, y new. Applying the translation equation we have x new is equal to x old plus Tx that is 3 plus 1 because 3 that is x is 3 and translation distance is 1. So we get 4, y new that is y old plus Ty is equal to 0 plus 1 that is 1. Thus the new coordinates of corner C is equal to 4, 1 and for the last coordinate that is the new coordinates of corner D is equal to x new, y new. Applying the translation equations we have x old plus Tx is equal to 0 plus 1 that is 1 and y new is equal to y old plus Ty is equal to 0, 1 that is 1. Thus the new coordinates of the square ABCD is D is equal to 1, 1. Now as we can see here the new coordinates of the square are with respect to previous 1, 1, 4, 4, C, 1, 4, 1 and D, 1, 1. See this is the diagram before translation and this is the one which is a representation after applying the translation with one unit in x and y direction. I hope you have understood the problem. Now I request the students to pause the video for some time and solve the given problem. Here the problem is given a circle with radius 10 and center coordinates 1, 4. Apply the translation with distance y towards x axis and 1 towards y axis. Obtain the new coordinates of C without changing its radius. Think for a while apply the translation homogeneous coordinate matrix and get the answer. Pause the video for some time. Moving ahead as you can see this is the new center for C after translation and the new coordinates are 6 and 5. The diagram shows the circle after translation. I hope you have understood the concept of 2D translation. These are the references. Thank you for your patient listening.