How do you find a point at a given distance?
541010
1.  Find the line that goes through d that is perpendicular to line l.  The slope of the perpendicular line is -1/m, where m is the slope from line l.  Since you have the slope for that line, then just plug in d to get the intercept, which I will call c.\n\n2.  Find the point (x,y) where the two lines meet by solving the system of equations.\n\ny = mx + b\ny = (-1/m)x + c\n\nI will call this point e.\n\n3.  Find the distance between d and e.  If d = (x1,y1) and e = (x2,y2) then the distance between d and e is\n\nsqrt((x1-x2)^2 + (y1-y2)^2)
789577
The distance between two points given by positions:\n{x1,y1,z1} and {x2,y2,z2} \nis equal to:\nsqrt[(x1-x2)^2+(y1-y2)^2+(z1 - z2)^2 ]
729745
No, it does not matter which point you layble what... just make sure you specify which point is which in your answer with the distances. If you need help computing the distances, create another question, or email me... glad to help
424777
I'd think you could find two points with the same single variable like temperature, or something easier to visualize like height, but two variables is questionable. Suppose you had 2 altimeters that you could position remotely (and they automatically stayed at opposite points). Idealize height as being a smooth function of location. \nInitially the altimeters indicate unequal heights. When you move them one way the difference increases. Then move the opposite way to decrease the difference. Move a little bit, then reassess the relative changes, find a new direction if needed, and make another small step toward decreasing difference. Now mentally superpose the two surfaces you are surveying. If there are no height maxima or minima and the surfaces intersect, repeating this procedure should find a pair of zero-difference points. Now you can follow the contour of intersection of the surfaces and the difference remains zero. If the surfaces diverge in every direction there must be an altitude maximum or minimum in one or both survey areas; move until the extreme is off the survey area and there is overlap, and repeat the zero-finding procedure. There is an overwhelming probability that such a pair of areas, and therefore of points, can be found. Earth does not have a situation where one hemisphere's altitude range is completely separate from the other hemisphere's. The same goes for temperature and pressure. \nWhat about a second variable, say pressure? Suppose it too can be mapped to two intersecting surfaces. Project the zero-height-difference and zero-pressure-difference contours onto the same plane. If the two contours intersect anywhere, the two-variables-equal point pair has been found. If they don't, you can't find one here; move to a whole new area.\nBy this logic you're extremely likely to find a pair of one-variable equality points, and pretty likely to find a two-variable pair of points. I know this is not a good mathematical answer, which should be a proof or disproof, but then it may not be provable or disprovable.
931645
first answer was close to right.\n\nthere is one qualification that seems need.\n(do this all in meters for easyness of calculation)\n\nR = radius of earth = 6400 km = 6.4 * 10^6 meters\nd = distance to horizon in meters\nh = height in meters\n\nSimilar triangles:\n\nR / d  = d / 2h\n\nd^2 = 2Rh\n\nd = sqrt (2Rh)\n\n....  works pretty well for small "h"\n\neg 2 meters height =>\n\nd = sqrt (2 * 6.4 * 10^6 * 2 ) = sqrt (25.6 * 10^6) \n=5.06 km\n\n[[ online calculator gives 5.5 km ]\n\nmine's a LOT easier to use ...\n\nI use "2h" instead of "h" for similar triangles\nbecause of discrepancy of right angle at the\nradius from eearth-center to observer and\nr-a from earth-center to distant object.\n[intersection happens "h" below earth surface]\n\nthat sounds confusing but I can't draw a picture\n\nmy eqn is easiest possible tho ..
958876
A simple algorithm is to first take all n(n-1)/2 pairs of points. Each pair defines a line, count how many of the remaining points lie on the line. This will be a cubic algorithm in n, the number of points.\n\nA clever algorithm using sorting takes time O( n^2 log n). See the Sources.
10096
Writing this from scratch would involve getting a complete zipcode database with lattitude/longitude. These will approximate zipcode areas.\n\nOnce you have 2 points, you apply Pythagorean Theorem:\n\na^2 + b^2 = c^2\n\nthus, for two points (a,b) and (c,d) in an x/y coordinate system, the straight-line distance between those points is:\n\n $distance = sqrt(pow((a-c),2) + pow((b-d),2)); \n\nSince there are 100,000 possible zipcodes, you don't want to perform the calculation above for all combinations. That would be far too costly. We have to find a way to reduce the number of possible zipcodes before we find the ones to check.\n\nIf your desired maximum radius is $r, and your starting lat/long is $x_start, $y_start, then you know the coordinates of all target zipcodes must be within the box defined by\n\n($x_start - $r/$k, $y_start - $r/$k) and ($x_start + $r/$k, $y_start + $r/$k)\n\nwhere $k is the number of miles (or km) per degree of lat/long. That is, the circumference of the earth divided by 360. (I know, it's approximate, and not the same for lat as long, but you get the idea) We'll call this $r/$k = $R\n\nThus you can first select only those zipcodes from your database:\n\nselect zip, x, y from ziptable where x between $x_start - $R and $x_start + $R and y between $y_start - $R and $y_start + $R\n\nYou could go one step further and have MySQL do your work for you.\n\nselect zip, sqrt(pow(x - $x_start, 2) + pow(y - $y_start, 2))*$k as distance from ziptable where x between $x_start - $R and $x_start + $R and y between $y_start - $R and $y_start + $R order by distance\n\n\nThat will give you the zipcodes you're looking for within the radius you specify.\n\nObviously you must modify to suit your needs.
105416
go to maps.yahoo.com and you can put in exact addresses from point A to point B and it'll tell you the distance in miles, time and driving directions
267671
First find its slope using the formula slope=(y2 - y1)/(x2 - x1) where x2 = the second x-coordinate, etc.\n\nslope = (-5 - 5)/(4 - -2) = -10/6 = -5/3\n\nthen choose either point and use the form for an equation of a line:  y - y1 = slope(x - x1)\n\nso either y - 5 = -5/3(x - -2)  [which becomes (x+2)]  or\n\ny - -5 which is y + 5 = -5/3(x-4)
735493
The distance between two points is defined by the following equation derived from the pythagorean theorum:\n\nd = sqrt((x2 - x1)^2 + (y2-y1)^2)\n\nWhere (x1, y1) is the coordinates of the first point and (x2, y2) is the coordinates of the second point.\n\nFor your problem,\n\n(x1, y1) = (4, -3)\n(x2, y2) = (-4, 3)\n\nd = sqrt((-4 - 4)^2 + (3 - -3)^2)\nd = sqrt((-8)^2 + (6)^2)\nd = sqrt(64 + 36)\nd = sqrt(100)\nd = 10\n\nThe distance between the two points is 10.\n\nI hope this helps!
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.
952205
You are missing a constraint in your problem.  If your only constraints are \n- theta must be less than or equal to pi/2 \n- you want to maximize theta\nthen theta = pi/2.\n\nSo how far should you be?  0 meters.
977003
Draw a straight line .now plot three points S,T and R on it ,taking ST=3.8 cm and TR=1.4 cm.\nWrite the distances 38 in betweenS and T and 14 between Tand R.your sketch is complete.\nNow ,rite the followin unde the sketch:-\nSR=ST+TR=38+14=52.
762369
Find the coordinates of the points where the straight line meets 'y=1' meets the circle 'x^2+y^2-14x+2y+45=0'.\n\n'x^2+1^2-14x+2(1)+45=0\nx^2 -14x +48 = 0\nthe solutions are:\nx=(14 +- sqrt( 14^2 -4(48) ) /2\nx=14+- sqrt(4) )/2\nx= (14 +2)/2 or x= (14-2)/2\nx=8 or x = 6\n.
895746
distance from a point in the circle to (1,1) is\nsqrt( (cost -1)^2 + (sint -1)^2 )\n\nas you can see this is a function of t,\nbut let us work with the square.\nf(t) = (cost -1)^2 + (sint -1)^2 \n= cos^2 t -2cost +1 + sin^2 t -2sint +1\n= 3 - 2cost -2 sint\n\nnow to find the minimum, or any other critical point,\nwe get the derivative of f(t)\n\nf'(t) = 2 sint -2 cost =0\nif sint = cost\nt= pi/4, 7pi/4, etc\n\nf''(t)= 2cost + 2sint\nf''(pi/4) = 2 cos(pi/4) + 2sin(pi/4) >0, so pi/4 is a loc minimum\nand\nf''(7pi/4) =2 cos(7pi/4) + 2sin(7pi/4) =0\n'
1088640
edit: (sorry hehe)\nthey'd be 500 miles apart...\n\nuse the pythagorean theorem...\n\nat 3 pm, one would be 300 miles west and the other would be 400 miles north...\n\nto get the distance between them (x):\n\nx=square root of 300*300 + 400*400\nx=500
949810
Use the distance formula:\n\nD = √ [ (y2-y1) ² + (x2-x1) ² ]\n\n1)  D = √ [ (-2+4) ² + (-2+4) ² ] = √(4+4) = √8 = 2√2\n\n2)  D = √ [ (3+1) ² + (-2+7) ² ] = √(16+25) = √41\n\n3)  D = √ [ (4-5) ² + (6+3) ² ] = √(1+81) = √82\n\nNRT