What is the midpoint Formula?
95382
Taken from about.com:\nTo find the midpoint of a line section the formula is: \n\nx = (x1 + x2)/2 \nand \ny = (y1 + y2)/2 \n\nwhere x and y are the coordinates of the midpoint, and (x1,y1) are the coordinates of the first point and (x2,y2) are the coordinates of the second point. Let's plug it in to the equation. \n\nx = (7+10)/2 = 17/2 = 8.5 \ny = (3+6)/2 = 9/2 = 4.5 \n\nSo the midpoint is (8.5,4.5)
411613
Kick point is listed at "mid-hi"\n\n\n65 - S* ANV65S .335/.625 67g 3.5 46" mid-hi
877827
Is that a single diameter?  Your statement is ambiguous.\n\nAssuming that it is, find the distance between those two points.  Halfway between the two points is the center
960360
To do this accurately, you need to find the midpoint of each side of the triangle. Connecting each vertex to the midpoint of the opposite side will give you the Centroid(the place where all the medians meet).  I graphed it out and it appears to be (-7,-3).
482401
Assume starts at Zero speed.\nMax speed is after 20 sec during which added 1.5m/s of speed each of these seconds.\nDistance during this phase d = Half a * (t squared)\nDistance during middle phase is  Max speed (as above) times 120 (NOT 2) seconds.\nDeceleration in third phase is Max speed (as above) divided by each of the 15 seconds it took to lose that speed. (If it does not work out as 4/3rds the acceleration, you have a problem!)\nDistance in the third phase is again d = Half a t^2  (Strictly speaking ut + 0.5at^2 with negative value of a, but no difference in this case as comes to a stop)\nDraw graph of speed v time (NB seconds - don't mix with minutes) find area by geometry.\nEdit - give you a clue - the distance is not 3900 and the deceleration is not 0.5 (nor even -0.5) as above.  Have you ever been tempted to give someone the wrong answer?
829618
1.find the midpoint of BC usingthe midpoint formula\n(x1+x2/2,y1+y2/2) say (x,y)\n2.find the slope os BC using theformula \ny2/y1/x2-x1 say m\n3.the slope of the perpendicular bisector will be the negative inverse of this -1/m\n4. now use the point (x,y) and slope (-1/m) and find the equation of the perpendicular bisector
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)
434240
Generally, midpoints divide the lengths into half. So then, the whole perimeter of LMN is 1/2 of ABC. So the ratio is 1/2.\n\nYou can pick two CORRESPONDING sides, and their ratio is still 1/2. To find the area, there's a formula:\n\n(Side 1/Side 2)^2\n\n(One length of LMN/One length of ABC)^2\n= (1/2)^2\n= (1^2)/(2^2)\n= (1)/(2*2)\n= 1/4\n\nNote: one length of LMN doesn't have to be 1 and another length of ABC doesn't have to be 2. They can share common multiples, but you end up simplifying to ratio 1/2.\n\nratio of perimeter of triangles LMN to ABC is 1/2\nratio of areas of triangles LMN to ABC is 1/4
533058
Point slope form:\n\ny-(-3)=(1/2)(x-5)
73340
It is by definition: your two points are "satisfying" the same equation.  They must be on the same line.\n\nKeep it simple.
817425
(a) By drawing the points on a graph sheet, Midpoint is determined as (11,16)\n\n(b) Again by drawing on a graph sheet, by using Hypotenuse theorem; \nRadius = Longest side of right angled triangle \n            = square root of (4^2+5^2) \n            = square root of (16+25) \n            = square root of (41) \n            = 41^0.5 \n            = 6.4
640114
Ellipse first:\nThe center is at the origin, because you have x^2 and y^2. If there is a translation, you would see something like (x+5)^2, that would be a translation of 5 units LEFT. The "a" value is always the bigger value. If it is under the x, then the ellipse is long the horizontal way, if it is under the y, then long the vertical way. From the center, count "a" units the long way, and "b" units the short way to find the shape of the ellipse. There is a formula for finding the focus points also. It is c^2=a^2-b^2, solve for c, and go that distance from the center along the long axis (or the major axis) of the ellipse.\n\nHyperbola:\nNote that the same translation holds for ellipse, but for the formula you put down, the center is at the origin. If the x^2 term comes first, the hyperbola opens left/right. If the y^2 term comes first, the hyperbola opens up/down. For a hyperbola, the "a" term is ALWAYS first, and "b" is next. Count "a" units left and right (if x^2 term is first) to find the vertices. Also, it is helpful to count "b" units up and down, and label those. Then, I have my students draw a rectangle between the a and the b markings, these are helper lines, and the diagonals of the rectangle are the asymptotes. Use basic slope-intercept to find the equations of the asymptotes. The focus has a formula, it is c^2=a^2+b^2, you can find "c" and label it in the bowl of the hyperbola.\n\nHope all that helped!
466006
Pick a few easy points like pi/2, pi, 3pi/2, 2pi.  You know the period.  Then just graph it.
316654
The midpoint of that segment is (3, -2) - all you do is take the average of the x coordinates and the average of the y coordinates.\n\nThe slope of that segment is found by\n\nm = (y2 - y1)/(x2 - x1)\n= (1 - (-5))/(4 - 2)\n= 6/2\n= 3\n\nSince you're looking for a line perpendicular to this one, take the slope, invert it, and change the sign.  So the slope of the new line will be -1/3.\n\nNow you have a point and a slope for the new point: m = -1/3, and point (3, -2).  Use the point-slope formula to find the equation of the line:\n\ny - y1 = m(x - x1)\ny - (-2) = (-1/3)(x - 3)\ny + 2 = (-1/3)x + 1\ny = (-1/3)x - 1
325102
sphere points: A, B, C as vectors from sphere center O.\n\ncompute:\n\nX = A x B\nY = A x C\n\n\nU = X x (A + B)\nV = Y x (A + C)\n\ncenter points:\n\nP = ± U x V\n\n(if necessary normalize P to sphere radius)\n\n--------------\n\ncross product: A x B\n\nsketch:\nhttp://static.flickr.com/48/129027255_d097e01a2d_o.jpg\n\nfurther reading: spherical Voronoi Diagrams, projective geometry\n\nthanks Neodiogenes. my solution was incomplete. i corrected it.
844246
Okay, to begin with, rewrite the equation as so:\n\n(x/5)^2 + ((y+5)/6)^2 = 1\n\nso, if y = 0, the x = +/-sqrt(275/36)\n\nif x = 0, y = +1 or -11\n\nSo, the ellipse is symmetrical about the y axis, with the major axis endpoints at (0,1) and (0,-11)\n\nThe minor axis will be halfway between the major axis endpoints, so it will be when y = -5.\n\nWhen y = -5, x = +/-5\n\nSo, the minor axis endpoints are at (-5,-5) and (5,-5)\n\nNow, for the second part:\n\nFor an ellipse, let's say the distance between the foci is d. So, the distance from the furthest focus to the opposite major axis endpoint is d+L, where L is the distance from the nearest focus to the major axis endpoint. At the end of the minor axis endpoint, the distance from each focus is p. Now, 2p + d is equal to 2d+2L. We know that the entire length of the long axis is 11, and that is d+2L.\n\nSo, if 2d+2L = 2p+d, then d+2L = 2p\n\nd+2L = 11, so 2p = 11, or p=5.5\n\nNow, the right angle triangle formed by the endpoint of the minor axis, the intersection of the minor axis and the y axis, and one focus has sides 5 on one side, 5.5 on the hypotenuse and d/2 on the other side. Using Pythagoras' theorem to solve for d, we get:\n\n25+d^2/4 = 30.25\n\nso, d^2 = 21, or d=sqrt(21)\n\nSo, the focal points are at (0,-2.71) and (0, -7.29)\n\nqed
1043340
What is n?  what is m?  What is q?  How can we come up with an answer without this data?\n\nI'm going to assume that q is the midpoint the segment or segments, and that m and n are also equidistant from q.\nThen you set up a right triangle, you're looking for the hypoteneuse length and the legs measure 10 and 16.\n\nIf that's so, then lm = √(10²+16²)\nlm = 2√(5²+8²)\nlm = 2√89