how do you find a point equidistant from three other points?
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?)
1007342
y = -1 is a horizontal line that contains the points (n, -1)\n\nhow far is (3, 4) from that line?  Subtrace the y-coordinates: 4 - (-1) = 5
1148398
the shortest distance between two points
66035
Well suppose you want to find antipodal positions on the earth. The earth is only approximately a sphere (it is actually closer to an oblate spheroid), so the mathematical definition of an antipodal point on the earth does not quite hold. \n\nIf one location on earth is at 23 degrees North latitude, and 10 degrees West longitude, the antipodal point will be at 23 degrees South latitude, and 190 degrees West longitude (but 190 degrees West longitude will be more often called 170 degrees East longitude). \n\nThat is, just add 180 degrees to the longitude, and switch the North latitude to South latitude (or vice versa).
392511
Find the distance between the center and the point (2,3).  This will be your radius.\nThen plug everything into the formula:\n(x-h)^2 + (y-k)^2 = r^2, where (h,k) is the center and r is the radius.
365555
I assume you mean how far away is the horizon, which isn't quite what the wording of your question really asks.\n\nAnyway, it is a simple geometry question. Construct a triangle. One point is the center of the Earth, one point is at your eye and one point is at your horizon. The horizon is that place where your line of sight is tangent to the curved surface of the earth, which means that the angle between the line from your eye to the horizon and the line from the center of the Earth to the same point on the horizon is 90 degrees. So you have a right triangle. The hypotenuse is the line from your eye to the center of the Earth and it's length is the radius of the Earth plus your height. That length, squared, is the sum of the squares of the other two sides. One of those sides is the radius of the Earth and the other is the unknown distance to the horizon you are trying to find. Just do the algebra to solve for the third side, which is the distance to the horizon. I think it is about 20 kilometers for the average person.
71998
multiply 1.414 by te distance between te two points
786277
Along x-axis, the distance between -3 and 3 is 6units\n\nAlong y-axis, the distance between 0 and 6 is 6units\n\nBy Pythogaras Theorem, \n(Distance between the points)^2 = 6^2 + 6^2\nDistance between the points= square root(6^2 + 6^2)\n                                               = square root(72)\n                                               = 6 squareroot(2)\n                                               = 8.485
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)
728290
The method is entirely dependent on what you know about the triangle in the first place.  If you know the 3 sides, you can use Steve's law for the altitude; ex: (alt to side c) = sqrt(4a^2b^2-(a^2+b^2-c^2)^2) / 2c\n\nIn general, the altitudes of a triangle intersect at a point called the orthocenter, whose location is described in detail at\nhttp://www.cut-the-knot.org/triangle/altitudes.shtml
117735
Use the Pythagorean theorem (twice).  The ray from the top light to the observer's eye is a straight line that just grazes the surface of the earth.  One right triangle has the following 3 sides: center of earth (COE) to surface (a known quantity), COE to top light (also known) and distance from tower to graze point (unknown).  The right angle is at the graze point.  Set up the other triangle similarly for the observer's eye, solve for the 2 unknowns, then sum them.
582742
Graph the line 3x + 4y = 12. \nIt divides the plane into one section where 3x + 4y < 12 and one where 3x + 4y > 12\n\nCheck to see see which side of the line (0,0) is on. It satisfies the inequality,  so shade the half plane that contains (0, 0) it's in the solution set.\n\nGraph the line x + 3y = 6. Again, check (0, 0) it satisfies that inequality, so shade everything on that side of the second line.\nNow draw the lines x = 0 and y = 0 (y axis and x axis)\nThose are the boundrys of your solution.
137113
Draw the two points (a,b) and (c,d) on a graph so that the segment connecting them is NOT horizontal or vertical. Then make a right triangle with one vertical leg and one horizontal leg where the connecting segment is the hypotenuse.  \n\nThe length of the horizontal leg is c-a\nThe length of the vertical leg is d-b\n\nSo the length of the hypotenuse, which is the distance from (a,b) to (c,d), is\n\nsquare root of (c-a)^2+(d-b)^2
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.
172928
To find the distance between two points:\n\nSubtract the two first coordinates and square this answer (multiply it by itself)\n\nThen subtract the two second coordinates and square this answer\n\nAdd the two squared answers and do the square root of this sum\n\nThat's the distance\n\nSo, for (3,-2) and (-2, 10)...\n\n3 - -2 = 5, 5 squared is 25\n\n-2 - 10 is -12 squared is 144\n\n144+25=169\n\nsqrt (169) = 13\n\nSo the distance is 13
1031286
victoire = 3 points\nnul = 1 point\ndéfaite = 0 point
247689
Use analytical trig (hint: sin function).  That should give you the answer you are looking for.