how can i use 3 coordinates to triangulate a point?
396223
lets assume you are talking about a radio signal just for the example like a lost hiker\ntake a direction reading from each of your coordinates\nlike say from the first your signal comes in best from the west, from point two its south-east, and from point three its north\non your map mark your three coordinates and a line representing the direction of best radio reception\nwhere the three lines converge is the aproximate location of the radio the lost hiker was using\nhope this helps you you understand triangulation better
907364
Almost.  You would actually include the endpoints 0 and -3 (so you need to use a bracket instead of parentheses).\nD = [0, infinity)\nR = [-3, infinity)
947280
y = (2/3)x + 3\nThis is in slope intercept form, y = mx + b\nm = 2/3 is your slope and b = 3 is the y-intercept.\nStart by placing a dot at (0,3)\nSince your slope is 2/3, that means you move up 2 and to the right by 3.  Place a new dot there.  You may repeat this step if you wish.  \nConnect the points with a straight line.
96194
Yes you are on the right track...Sort of.  \n\nSay you plug in any number for x.  We'll use 5\n\n-3(5)+y=-15\n-15+y=-15\n\nNow add 15 from the left to cancel the negative 15.  Then add 15 to the right to make the equation:\n\ny=0\n\nSo if x=5 and y=0 then your points will be (5,0)\n\nSay you have your original equation again\n-3x+y=-15\n\nNow plugin for x any number.  We'll use 1.\n\n-3(1)+y=-15\n-3+y=-15\n\nNow add 3 to the left to cancel out the negative 3 and add 3 to the right to make:\n\ny=-12\n\nSo x=1 and y=-12   (1,-12)\n\nHope that helps
443662
A three dimensional coordinate (point in space) reads (x,y,z).\nSo, (1,2,-2) = (x,y,z) clearly implies that x=1, y=2, and \nz= -2, as you required.
634409
Cool! But if the map is on my back, how can i see the coordinates and point them out?
120928
ok, let the triangle be ABC with sides AB, BC, AC.\ndraw a typical line, get too point on in with the distance of BC, name them B and C. then draw an arc with the center of B and the length of AC and another one eith the center of C and the length of AB. if they had a common point, it would be A and your triangle is there, but if not, there is no triangle with those three sides.
98103
Awesome question.\n\nCall the three points p1, p2, p3\n(in vector form, i.e. p1 = (p1x, p1y, p1z) etc.).\n\nCall c the point equidistant from p1, p2, p3.\nc may not always exist, for defective cases:\n- (ignore the trivial cases where p1, p2, p3 are not distinct points i.e. you only have 1 or 2 unique points)\n- if p1, p2, p3 are collinear on any plane, then c does not exist (radius -> infinity)\n\n- in general a result will exist and is going to be c=s, the center of the sphere of some radius on which all points lie\n- it probably makes life a lot easier to first transform the origin to some center (e.g. the centroid (p1+p2+p3)/3, or orthocenter, or some other vector definition of center). Anyway not sure what the transform is, but let's call the resulting new coords p1', p2' , p3'.\nNow you know the distances from s to p1', p2', p3' must all be equal (to radius r). This gives three equations in three unknowns (sx, sy, sz). i.e.\n|p1'-s| = |p2'-s| = |p3'-s| = r\nIt may be easier to solve the square of that eqn:\n(p1x-sx)^2 + (p1y-sy)^2 + (p1z-sz)^2 = \n(p2x-sx)^2 + (p2y-sy)^2 + (p2z-sz)^2 = \n(p3x-sx)^2 + (p3y-sy)^2 + (p3z-sz)^2 = r^2\nHowever the 3 eqns seem to be quadratic, not linear. So not sure how to generalize the solution.\n\nNotes:\n - in the special 2-dimensional case that p1, p2, p3 lie in a plane AND are all equal distance from some c, then c will be the center of circle containing them. Also c necessarily lies in the same plane.\n- in the special 2-dimensional case that p1, p2, p3 lie in a plane BUT not at equal radii of some circle, it will be back to the general sphere solution with c=s and s does NOT lie in the plane. Interesting.\n\n- The final solution is going to be some vector formula, symmetric in p1, p2, p3 containing something symmetric involving cross-products, either:\np1' x p2' x p3'\nor else p1' x p2', p2' x p3', p3' x p1'\n\nAh yes, here it is, determinant equation (28) in\nhttp://mathworld.wolfram.com/Sphere.html\n(I did not cheat - i searched for these links afterwards)\nFour points uniquely define a sphere. So with only three points you should get an infinite family of spheres.\n\nSee Mike Yoder's comment from comp.graphics.algorithms, 2 Dec 1993: \nThree points can never determine a unique sphere. If they are collinear and distinct, they do not lie on any circle (and hence not on any sphere either). If they are not collinear, they determine a unique circle, which is contained by an infinity of spheres: think of a ring with balls of varying sizes balanced on it. \n\n...as quoted on this guy's webpage, which also discusses efficient numerical algorithms for sphere estimation:\nhttp://stevehollasch.com/cgindex/geometry/sphere4pts.html\n\n(A good side question would be: give only three points, you can effectively choose a sphere of any radius you want, greater than some minimum, r>r_min. How do you use this freedom to simply finding ANY solution in the 3 pts case?)
889456
I'm not sure I understand the question, but if you want to translate the vector from its current position to the origin, then just add (4,-5) to its beginning and end points.\n The vector will now be pointing from the origin to (6,1).\nHope this helps.
913865
http://www.webreference.com/3d/ Try that, good luck with the assignment.
535960
The speed has no part in the solution, as it is the same for all three paths. So, you draw a triangly with legs 2.25 and 3 in lenght. the solution is the lenght of the hypotenusa formed. the answer is 3.75 min.
986326
when x is 0, find y. plot the point (0, <<whatever you got>>).\nwhen y is 0, find x. plot the point (<<whatever you got>>, 0).\nConnect those two points, and extend the line.
269336
To solve this problem, you have to substitue the coordinates into point-slope form using the slope (-5) as m.  The y1 refers to the y coordinate, while the x1 refers to the x coordinate.\n \ny-y1=m(x-x1)\ny+12=-5(x+12)\ny+12=-5x-60\ny=-5x-72\n\nslope=-5\ny-int.=-72\n\nThe person above me has everything right except when she had -12=-5(-12)+b.\n-12=-5(-12)+b\n-12=60+b\n-72=b\nb=y-int=-72
978173
You need to start off with a six-pointed star (constructed with two equilateral triangles).\n\ntriangle AEC and triangle FBD\n\nNow connect A to D, then connect B to E, and finally C to F. If you \ndrew everything correctly, those three segments should meet at a point \nin the center of the star.  \n\nThe 3 line segments you just drew plus the six sides of the star make \n9 lines. There are 5 points of intersection on each line, and the \ntotal number of intersections is 19.
188880
well youhave R^3, that means that you are in 3 dimensions, and points are in (x,y,z).  You have two equations for points.  Basically you could pick a value for x and you would get y and z in terms of x....so solve each equation so you have y and z in terms of x\n\nyour new point is \n\n(x, (-3x)/7, (-6x)/2)\nor\n(x, -3x/7, -3x)\n\nand for every x you have the y and z coordinates.\n\nHope this helps
729410
You can  find two different values for x that are easy to find 1/7 of, such as 7 and 14; see what y would be at those points [ -4, -3]; then plot the points in either order and connect the dots.  It is smart to also plot a third point to reduce the chance of error, because if one of the three is wrong, they won't line up.
843247
You have enough information.\n\nIf I understand you correctly, you have the (N,E) coordinates for points A and F.  So given what you know so far, you should be able to compute coordinates for points B, C, and E.  All you need for a complete picture is coordinates for D.  Lets call them (N(D), E(D)).\n\nNow you know the distance DE. Using the distance formula,\n\n(DE)^2 = [N(E) - N(D)]^2 + [E(E) - E(D)]^2 \nthis is one equation with two unknowns.\n\nAlso, you know the azimuth for CD.  Call it θ.\nThen tan(θ) =   [E(D) - E(C)] / [N(D) - N(C)]\nthis is a second equation with the same two unknowns.\n\nTwo equations, two unknowns.  Solve simultaneously to get\nN(D) and E(D).\n\nOnce you have the coordinates you can compute missing azimuth and missing distance.\nIf you would like to email me the numbers I'll take a look.