 Hello, everyone. Welcome to the second session of the 2D Geometrical Transformation. Myself, Professor Prithish Chitte, working as an assistant professor in mechanical engineering department Valchen Institute of Technology, Solapur. In the second session, we will discuss some numericals based on 2D Geometrical Transformation. Learning outcomes? At the end of the session, the student will be able to solve the different numerical based on the different types of 2D Geometrical Transformation, such as translation, scaling and rotation. These are the contents. First, we will see the recap of the first session. After that, we will see the numericals based on the transformations like the translation, scaling and the rotation. So, what we have discussed in the first session? We have discussed like what is meant by the geometrical transformation. Geometrical transformation will help you to observe the geometrical entity very easily with the help of different geometrical transformations like translation that is moving, rotation. We have to provide the angle to the particular geometrical entity with respect to a coordinate. After that, scaling, we have to zoom in or the zoom out for the particular area so that we can see that particular area may be the surface or may be the edge very easily. We can provide the orientation with respect to the current configuration. What is the in general equation for the geometrical transformation? That is, p dash is equal to L of p where p dash is the modified image and p will be our original image. Here, first we will see the translation. So, in the translation, equation is x dash y dash is equal to x y plus t x ty where t x ty is translation matrix. We have discussed like we have to suppose we have to move a particular point or maybe we have to move a particular geometrical entity from one coordinate to the another coordinate where x dash y dash will be our modified coordinates, x y will be our original coordinates and t x ty will be our translation matrix. For the similarly triangle a b c, so this will be our modified triangle that is triangle a dash b dash c dash that is a dash x, b dash x, c dash x y is equal to a dash sorry a x y b x y and c x y plus t x ty where t x ty is our translation matrix. Rotation, here we are rotating the particular original point that is p through the angle theta and we can get the modified point that is p dash x dash y dash. We know the equation that is x is equal to r cos phi and y is equal to r sin phi with the help of this diagram. So, x dash is equal to r cos theta plus phi, so r cos theta cos phi minus r sin theta sin phi. We can put these values of x and y here from the equation one, so x dash is equal to x cos theta minus y sin theta we will this will be our first equation. So, similarly y dash is equal to r sin theta plus phi, so sin theta plus phi is equal to sin theta cos phi cos theta sin phi, so r into sin theta cos phi plus r into cos theta sin phi. So, we can put the values of x and y here, so y dash is equal to x sin theta plus y cos theta that will be our second equation. So, whenever we are combining the equation one and two regarding the x dash and y dash, so we can get x dash y dash is equal to x y modified coordinates original coordinates into the into the rotational matrix that is cos theta sin theta minus sin theta cos theta this is the rotational matrix. The integral equation is the p dash is equal to p into r the modified coordinate equals to original coordinate into rotational matrix scaling. So, here we are zoom in or the zoom out the particular geometrical entity here we are taking the triangle example for the triangle. So, a b c is our original image and a dash b dash c dash is our modified image by the scaling factor x x and s y. So, x dash is equal to x into s x and y dash is equal to y into y s y. Similarly, a dash x a dash y is equal to a x a y into s x 0 0 s y for the similar triangle a b c. So, triangle a b c, so a dash b dash c dash will be our modified triangle. So, a dash x a dash y b dash x b dash y c dash x c dash y is equal to a x a y b x b y and c x c y this will be our original coordinates of the triangle into s x 0 and 0 s y where s x and s y will be our scaling factors. Let us think about this question write the parametric equations of the types of the geometrical transformations like the translation scaling and the rotation. We have discussed this now we will move to the main part that is the numerical based on the translation scaling and the rotation. First we will see the problem regarding the translation. So, this is the statement how the triangle a b c will be translated with the coordinates a 2 5 b 2 9 and c 8 5 by the 10 units in x direction and 20 units in y direction. So, t x is equal to 10 and t y is equal to 20. So, this will be our the equation for the translation by keeping the values in the equation we are knowing the values of a x y and b x y and c x y. So, a x is equal to 2 a y is equal to 5 b x is equal to 2 b y is equal to 9 and c x is equal to 8 and c y is equal to 5 and we are knowing the values of t x and t y that is 10 and 20 respectively. So, we can get the values of a dash x y b dash x y and c dash x y like 12 25 12 29 and 18 25. So, after the translation the points are a dash 12 25 b dash 12 29 and c dash 18 25. So, this will be the diagram after the translation transformation. This will be our original triangle and after the translation transformation we will get the new modified triangle that is a dash b dash and c dash. We will move to the next numerical rotate a point p 10 5 this is the original point that is p 10 5 by counter clockwise this will be our counter clockwise by 30 degree that is theta and find the rotational matrix and the resultant point. First we have to find out the rotational matrix that is r and the resultant point that is p dash. Now, we know the equation p dash x dash y dash is equal to p x y into cos theta sin theta minus sin theta cos theta where cos theta sin theta minus sin theta cos theta is our rotational matrix. So, here rotational matrix so, cos theta sin theta minus sin theta cos theta. So, theta is 30 here so, cos 30 sin 30 minus sin 30 cos 30 so, 0.66 0.5 minus 0.66 0.5. So, this will be our rotational matrix by keeping the values of the original coordinates. So, 10 and 5 that will be our p x y into the rotational matrix. So, we will get the x dash and y dash here. So, x dash and y dash are 6.16 and 9.33 respectively. So, here this is our original coordinate that is p 10 5 after the rotational by angle theta by 30 degree this will be our rotational angle we will get the point p dash that is 6.16 and 9.33. Now, the third problem scale the rectangle a minus 1 minus 1 b 1 minus 1 and c 1 1 and d minus 1 1 these are the coordinates for the rectangle a b c d by 2 units in x direction and 2 units in y direction means s x is equal to 2 and s y is equal to 2 these are our scaling factors. So, this will be our original rectangle that is a b c d we have to find out a dash b dash c dash d dash. So, we are knowing the s x and s y also we are knowing the equation for the scaling in the last session we have discussed that. So, a dash x y b dash x y c dash x y d dash x y is equal to a x y b x y c x y d x y into s x 0 and 0 s y where these matrix will be called as the scaling matrix we are knowing the values of scaling factors that is s x is equal to 2 and s y is equal to 2 by keeping the values. So, these are the values that will be our original coordinates for the a x y b x y c x y and d x y into s x is equal to 2 and s y is equal to 2. So, we will get the modified coordinates for the points that is a dash minus 2 minus 2 b dash 2 minus 2 c dash 2 2 and d dash minus 2 2. So, this will be our resultant new rectangle after the scaling transformation. These are the references. Thank you.