 when my self-professor Prithish Chittay working as an assistant professor in mechanical engineering department in Walshchand Institute of Technology, so today we will discuss the parametric representation of a line which is the part of the course CADCAM CE, we will go ahead, so what you are learning. So, the students will be able to apply the CAD fundamentals like the geometrical transformation like suppose I want to create a line, suppose I want to create a line in the geometrical software, but before that I must know how the parametric representation or the equations of the lines that can be created theoretically and we can analyze the application of geometrical modeling, so these are the things that you are learning in these sessions, so what you are learning actually, so first of all you will learn the parametric representation of a line, we will see the derivation and some of the numericals, so the line, line is a analytical curve, for example there is a line or may be there is a, for example there is a circle arc, triangle, rectangle, the different types of polygons, these are the analytical curves, now whenever we are creating the line behind there is a equation, so what are the equations, yes first we will see there is a u is equal to 0 at the starting point, there is a u is equal to 1 at the end point, u is the parameter and p1 and p2 are the position vector, where p1 is at the initial point and p2 at the end point of the line after that, so unit vector that is p2 minus p1 divided by l, where l is the length of a line, so length of a line how to calculate it, yes everybody knows that that is under root x2 minus x1 bracket square plus y2 minus y1 bracket square plus z2 minus z1 bracket square where x, y, z are the coordinates x1, y1, z1 and x2, y2, z2 these are the start point and the end point of the line respectively, Cartesian form how to write in the Cartesian form, so we are having the equation like in the Cartesian form like x of u is equal to x1 plus x2 minus x1 of u and y of u is equal to y1 plus y2 minus y1 of u and z of u is equal to z1 plus z2 minus z1 of u, in the in general form we can write it as u of u is equal to p1 plus p2 minus p1 into u, so this is the in general equation for the line where p1 and p2 are the position vector and u is the parameter, so these equation we are needed it, so that is why I have written it in the as a parametric form, suppose I have provided the limits or the constraints, so that is p of 0 that is p of u is equal to 0, so keep u is equal to 0 here that is p1 plus p2 minus p1 into 0 that will be p of 0 is equal to 1 or that will be also called as p of u is equal to 0 is equal to p1, similarly you can keep u is equal to 1 that is p1 plus p2 minus p1, so p1 will be deleted and p of u is equal to 1 will be p2 these are the we can call as the limits and the constraints regarding u is equal to 0 and u is equal to 1 tangent vector, so just provide the d by du of this particular equation, so d by du of p1 plus p2 minus p1 into u, so that will be d by du of p2 minus p1 into u that will be p2 minus p1 that will be tangent vector, so p dash of u is equal to p2 minus p1 slope or similarly we can have it for example may be the x dash of u is equal to x2 minus x1, so y dash of u is equal to y2 minus y1 and z dash of u is equal to z2 minus z1, the slope of the curve that is y dash divided by x dash that is called as dy by dx or also we can write it as y2 minus y1 divided by x2 minus y1 what is y dash you can see here y dash is equal to y2 minus y1 what is x dash that is called as x2 minus x1, now so before moving to the numerical part, so which are the different analytical curves yes you can think about this question, yes as we can call it as the we have discussed the different analytical curves like we are discussing the parametric equation of a line or maybe after that in the next session we will discuss the parametric equation of a circle, suppose you are having maybe ellipse hyperbola parabola or maybe you are having maybe the triangle rectangle these are the analytical curves after that, so here you are having one some of the numericals based on the equation of the line, we have to write down the parametric equation form of the line passing through 0 80 and 200 so what we are having it just we will write it, so we are having the parametric form, so x of is equal to x1 plus x2 minus x1 of you just we will provide the values here that is x1 is equal to 0, x2 is equal to 200, yes just I will write it here sorry maybe I will just write it here just give me a second, so this is x1 plus x2 minus x1 this is y1, yes x1 is equal to 0, y1 is equal to 80, this is x2, yes this will be the input data and this is y2 yes after that, so by providing the x1 is equal to 0, x2 is equal to 200 and y x1 is equal to 80, so what is the parametric form in with respect to x of you that will be x of you is equal to 200, after that with respect to y of you that is y1 plus y2 minus y1 of you, we have discussed the parametric form in the earlier equation or with the derivation provide the value of y that is y1 is equal to 80, y2 is equal to 80, y1 is equal to 80, so that will be y of you is equal to 80 that is the final parametric equation, after that in general parametric equation with respect to p1 and p2, where p1 and p2 are the parameters or maybe we can also called as position vector and u is the parameter, so p1, so what is p1 that is x1, y1 that is 0, 80, so p2 what is p2 that is x2, y2 that is 280 minus p1, p1 is the position vector at the start point, at the start point there is a x1, y1 that is 0, 80 and into u, so 0, 80 that will be as it is plus 200 minus 0, 80 minus 80 that will be 200 u, so that will be 80 plus 200 u, so that will be x of you plus y of you that is 80 plus 200 u, so that will be in general parametric equation with the coordinates x1 is equal to 0 and y1 is equal to 80 and x2 is equal to 200 and y2 is equal to 80, we will see these are the very simple problems or the numerical based on the parametric equation of the line. The next problem we have to find out the parametric equation of the line which is passing through x1 is equal to 30, y1 is equal to 30, just we will write the given data first of all, so that will be x1 is equal to 30 after that or may be just I will drop this point, just give me a minute, so this will be x1, yes after that this will be y1 that is 30, x2 is equal to 80 here, after that y2 is equal to 60, yes with these are the input points we are having it, also we have to find out the tangent vector and the slope of the curve. Now, the parametric form in the characterization that we have discussed in the last numerical just we will follow that equations that is x of u is equal to x1 plus x2 minus x1 of u, now x1 is equal to 30, x2 is equal to 80 and again x1 is equal to 30, by keeping those values we will have x of u is equal to 30 plus 50 u, similarly y of u is equal to y1 plus y2 minus y1 of u, where y of u is equal to 30 plus 60 minus 30 of u, where y1 is 30 here, y2 is 60, so by keeping those values we will have y of u is equal to 30 plus 30, after that in general parametric equation. So, p of u is equal to p1 plus p2 minus p1 of u, so p1 that is 30 comma 30, p2 is 80 minus 60 on p1 is 30 comma 30 and by this p of u is equal to 30 comma 30 plus 50 comma 30 of u, tangent vector that is p2 minus p1, so that will be 50 comma 30 of u very simple, the slope of the curve that will be y dash of divided by x dash that is 30 comma 50 or 30 by 50 that will be 0.6, so we have to find out the position, the parametric equation for the position vector that is 1 comma 1 and 4 comma 5, so that is x of u that is 1 plus 3 u, y of u that is 1 plus 4 u, similarly we can find out the general parametric equation that is 1 comma 1 plus 3 comma 4 of u, this is the tangent vector and this is the slope of the curve, these are the references, thank you.