On the x-coordinate, there is straight line AB, point A is fixed on the x-coordinate, point B is located at any point x-coordinates, to describe this function.
Perhaps he has a typo. Near the end meant to say: "... point B is located at any point y-coordinates,... Answer to question he asked is: y=0 for all x such that A<x<B plus point A & B if you want them included on the Line segment.
no , point B is located at any x-coordinate solutions : point A=a , point B=x , y=AB a) y=|a-x| b) y=-|a-x| c) y=a-x d) y=x-a Mathematics official says geometric objects measures must be positive numbers, say my function to geometric objects measures may be negative numbers question - how to make it look proceedings graphics of my functions in the plane (Cartesian coordinate system)
This is the function point (A - is a constant, B - is a variable) length of straight line (AB) is the solution functions
I doubted anyone asking, especially so poorly that James could not guess what the question was, would know either that vertical bars indicate absolute value or that is always a non-negative difference, and this form is some what ambiguous about the status of his points A & B (part of the line segment or not?) But worse is you have defined a single point at (|a- b|, 0, 0) in standard Cartesian coordinates, not a line segment, and that point may not even be on the line segment he wanted defined! I think you really mean: |a-b|> x for a < x < b , which will at least let x have various values, but that is not as clear or simple as the form I gave.
Yeah, a-b is the length of the line segment it is not x as I stated. I was trying (incorrectly) to have him understand that he is describing a line on the x axis so there is no y or z component.
Yes. that is exactly what I said in post 3: "Answer to question he asked is: y=0 for all x such that A<x<B" It really does not get more simply and clearly stated than that bold part.
Please Register or Log in to view the hidden image! Seemed to be a family of line segments that has one of its endpoints located on the line x=a Since there is nothing said about the coordinates of point A and point B, assume they can take any y values in reals Let A=(a,y1), B(x,y2) with x, a, y1, y2 in reals and a fixed. Let (x',y') be a point on the function The required function is Domain: x' in [a,x] Codomain y' in [y1,y2] (y2-y1)/(x-a)=(x'-a)/(y'-y1) (y2-y1)(y'-y1)=(x'-a)(x-a) y2y'-y1y'-y2y1+y1^2=x'x-ax-x'a+a^2 y'(y2-y1)+x'(a-x)+(y1^2-a^2+ax-y2y1)=0 ------(*) which is a straight line segment with endpoints A and B for any y1,y2,a,x The length of each line segments in the family is then l=sqrt((y2-y1)^2+(x-a)^2) Interesting cases of (*) 1. When y1=y2=c (c in reals) (*) is the horizontal line segment [a,x] at y=c When c=0, (*) is the interval [a,x] 2. When y2<y1 and a<x OR y2>y1 and a>x, the slope is negative, otherwise it is nonnegative In particular slope is zero when y1=y2=c as mentioned above and infinite when a=x 3. When y1=y2 and x=a, you get a contradiction. I have not attempt to analyse the limit of (*) as y2 approaches y1 and x approaches a yet I am not sure if there are other interesting cases for (*), but it seems its length is never negative valued
there is only one y ( no y1,y2 ) The mapping function from the x-coordinates of the plane (Cartesian coordinate system) y = x-a, x and a remain on the x-coordinate, y goes to the y-coordinate. view photo https://pkxnqg.bn1302.livefilestore...WSh5idkVdC-swrTkqYaXV8fmts9x7Ks/ii.png?psid=1 the lines of x and a parallel to the y-coordinates line of y parallel to the x-coordinate formed at the intersection of real points A and B points A and B are combined and gets straight line AB is given by x = 4, a = 2, y = 2 Repeat for x = 3.5, a = 2, y = 1.5, view photo formed at the intersection of real points C and D points C and D are combined and received straight line CD https://befwwg.bn1302.livefilestore...DKraCcJKIy-UHkR4VeCHL_PmPvJTSMeM/i.png?psid=1 connect the dots AC (BD) straight lines AB and CD ABDC points form the surface of 4≥x≥3.5 Draw a graph of the function at the current proceedings for a) y=|a-x| b) y=-|a-x| c) y=a-x d) y=x-a e) y={|a-x|}\cup{-|a-x|}
What you described is either a undirected line segment or an ideal point. A "function" would require at least two ordered pairs of coordinates to describe, even if such a description was "piece-wise" or discontinuous.
a) y=|2-x| graph, the red surface https://cfxpzq.bn1302.livefilestore...duf35TDz7kJFlvpinPGfiGmOhMAVbDw/01.png?psid=1 b) y=-|2-x| graph, the red surface https://nq6hfq.bn1302.livefilestore...xhUjRvcF6i6WXHS-qNoA1O50MwSx-Kw/02.png?psid=1 c) y=2-x graph, the red surface https://0nivia.bn1302.livefilestore...qqqkUyEXwbMEkel843ZQ20Rq__x6V-A/03.png?psid=1 d) y=x-2 graph, the red surface https://d6pekg.bn1302.livefilestore...0AJlHabqjfVBAgiGORjpymT7vzmMKCA/04.png?psid=1 e) y={|2-x|}[LATEX]\cup[/LATEX]{-|2-x|} or y={2-x}[LATEX]\cup[/LATEX]{x-2} graph, the red surface https://qhdsnq.bn1302.livefilestore...HeXxWALY-BwcLW4G5ObhnQSYfKmEaVQ/05.png?psid=1 which are geometric objects obtained for valuesx and y , shape a≥x≥b ( a≥y≥b ) ? , you have a graph
