parametric form of a line?
963964
P=(1,2,1) Q=(1,5,5)\n\nThe parameterisation X(t) with parameter t generates all points on the line (or line segment, for 0<=t<=1)\nwith X(0)=P, X(1)=Q :\n\nX(t) = t*P + (1-t)*Q\n       =t*(1,2,1) +(1-t)*(1,5,5)\n(multiplying in, which is generally not needed)\n       = (1, 5-3t, 5-4t)
1161836
One way to write the equation of a line is\ny - y1 = m(x - x1)\nwhere m is the slope and (x1, y1) is a point on the line.\n\nGiven your information,\ny - 4 = -2(x - 1)\nThat is the point-slope form of the line.  This can be changed to slope-intercept form:\ny - 4 = -2x + 2\ny = 2x + 6
915931
There can be several parameterization\n\n1. The specification of a curve, surface, etc., by means of one or more variables which are allowed to take on values in a given specified range. \n\n2. A parameterization of a surface  X(u,v) in uand v is regular if the tangent vectors \ndX/du and dX/dv are always linearly independent. \n[THE DERIVATIVE ARE PARTIAL HERE.]\n\n3. A surface in 3-space can be parameterized by two variables (or coordinates) u and u such that \n\n   (1)  x= x(u,v)\n   (2)  y= y(u,v)\n   (3)  z= z(u,v)\nIf a surface is parameterized as above, then the tangent vectors \n\n (4) Tu =   (dX/du)x^ + (dY/du)x^ + (dZ/du)x^ \n\n (5)  Tv =  (dX/dv)x^ + (dY/dv)x^ + (dZ/dv)x^ \nare useful in computing the surface area and surface integral. \n\n[Tv means T with suffix v, (dX/dv) means partial derivative of X with v and x^ means x with a cap].\n\nSorry, but I can not write euqations so well without equation editor.
150933
What do you call the line that intersects a circle in only 1 point?
151606
if all the lines are equal in every way (duplicate) then you can put a measurment (dimension) from the existing line to a datum or border whichever you choose and then put a callout like 3@12.00' (3 lines at 12.00 feet)in the dimensionwith a arrow pointing at the line.\n\njeff
985391
Use point-slope form:\nFirst, calculate the slope: (-2 - 0) / (5 - 0) = -2 / 5.\nHere is the general point-slope form:\n(y-y1) = m*(x-x1)\n(y+2) = (-2/5)*(x-5)\ny + 2 = -2/5*x + 2\ny = -2/5*x\n\n______\n\n\nFor parametric form, there's not that much to it:\nx = 0 + 5t\ny = 0 + -2t\n\nFor clarity, here's an example with another set of points:\n(2,-3) and (4,8)\nx = 2 + 2t\ny = -3 + 11t
838376
Does it have to be in any particular form?\n\nThe really easy form is point-slope form, which is\ny - y1 = m(x - x1)  where (x1, y1) is a point on the line and m is the slope.\n\nSo your first one would be y - 0 = 2/3 (x - 6)\n\nwhich simplifies to y = 2/3 x - 4\n\nDo the second just like that.
998290
mathematically speaking   if you had two points on a line\n\nPoint Dexter is some unknown point right of  the center\nand Point Sinister is some unknown point left of the center.
538540
format>paragraph>look for line spacing and select double
902740
In addition to what “Spoonish1” told you I suggest you to write a program calculating minimal distance between 2 straight lines in 3D. Be this distance =0, then and only then you may state that 2 lines intersect. A classic line equation in 3D looks like:\n(y-y1)/(y2-y1)=(x-x1)/(x2-x1)=(z-z1)/(z2-z1) – just as a matter of fact. Has your desire faded? Or more help?
919933
I assume this is equal to y?\n\nIf it is, it's definitely a maximum because it's a parabola that opens downward\n\nNow I'll work on the vertex
1112728
to draw lines u hav to select 2 points at a time (as no 3 r collinear)\n\nthus no of selections of such points is nC2= n!/ [ 2! (n-2)!]
932841
This might help --> http://www.uwm.edu/~ericskey/TANOTES/Geometry/node15.html
712712
[on] [y=] [(] [x,t,o,n] [+] [2] [)] [/] [x,t,o,n] [enter] [5] [-] [2] [x,t,o,n] [enter] [(-)] [2] [x,t,o,n] [-] [3] [enter] [window] [(-)] [5] [enter] [5] [enter] [1] [enter] [(-)] [5] [enter] [5] [enter] [graph]\n\nYou should have on screen a vertical asymptote at x=0, horizontal asymptote at y=1, a curve going to negative infinity towards 0 on the left, a curve going to positive infinity towards 0 on the right, on the left tangent line y=-2x-3 touches point(-1,-1), and on the right tangent line y=5-2x touches point(1,3).
966909
I am assuming that all you want is that simple pattern drawn.\n\nFor a predefined size of the triangle LINELIMIT:\n\nint main() {\n  int head = 1. prev; line = 0, col = 0;\n\n  for (line = 0; line < LINELIMIT; line++) {\n    prev = head;\n    for (col = 0; col < line; col++) {\n      printf("%d ", prev);\n      prev *= 2;\n    }\n\n    printf("&#92;n");\n    head *= 2;\n  }\n\n  return 0;\n};
512252
Standard form of a parabola that opens left or right is\n\nx = c(y - h)^2 + k\n\nFor your equation, complete the square to get it in standard form:\n\nx = 3y^2 + 6y + 4\n= 3(y^2 + 2y) + 4\n= 3(y^2 + 2y + 1) + 4 - 3\n= 3(y + 1)^2 + 1
1079894
The basic definition of a parabola states that it is the locus of points that are equidistant from a given point (the focus), and a line (the directrix). Therefore, we have to find two distances:\n\nPoint-to-Point: d1 = sqrt( (x-11)^2 + (y+7)^2 )\nPoint-to-Line: d2 = abs( x-5 ) / sqrt (1^2) = (x - 5)\n\nThen, make the two distances equal, d1 = d2:\n(x-11)^2 + (y+7)^2 = (x-5)^2\n........simplifying........\ny^2 + 14y + 49 - 12x + 96 = 0\n\n(y + 7)^2 = 12*(x - 8)\n\nThis is the equation for our parabola.