What is the Mathematical equation that gives you the distance between two coordinates (x1,y1 & x2,y2)?
546400
D=square root ((x1-x2)^2 + (y1-y2)^2)
869813
(y--1)/(-2--1)=(x-1)/(2-1)\n=(y+1)/(-2+1)=(x-1)/(1)\n=>y+1=-1(x-1)\ny+1=-x+1\ny=-x is the required equation
1148398
the shortest distance between two points
829160
y'=2 cos(7x)*(-sin 7x)*7 By the Chain Rule
935919
Please calm down. Now do you want the line between (2,0) and (-2,-2) The way I do it is to get the point vector solution of the line and eliminate t. The point is (2,0). The vector is <-2 - 2, -2 -0>. Just subtract the first coordinate from the second. So the vector is <-4, -2> \nNow set up the point vector equation describing the postion of the graph at time t.\nJust point + vector*t in component form.\nThe x-components: x = 2 - 4t \nThe ys: y = 0 - 2t\n\nThere, that is it. But some nasty professors don't like this form, so we'll eliminate t. We plug y = -2t into first equation and get x = 2 + 2y.\n2y = x - 2. y = x/2 -1.
589798
y' = 2x - 2\n2x - 2 = 0\n2x = 2\nx = 1
796769
The point P moves in such a way that its Cartesian coordinates with respect to an origin O at time t is given by x= e^(-t) , y= 2te^(-t). the distance OP is denoted by r and the angle between OP and the x-axis by θ.\n(a)Using an analytical method, find the value of t when r^2 is stationary. Determine the nature of the stationary point.\n\nr² = Δx² + Δy²\n= (e^(-t) - 0)² + (2te^(-t) - 0)²\n= e^(-2t) + 4t²e^(-2t)\n= e^(-2t)(1 + 4t²)\nd(r²)/dt = -2e^(-2t)(1 + 4t²) +8te^(-2t)\n= 2e^(-2t)(1 - 4t + 4t²)\n= 0 for stationary points\nSo 1 - 4t + 4t² = 0 as e^(-2t) > 0 for all t\nie (1 - 2t)² = 0, ie t = ½, ½\nWhen t <½, d(r²)/dt > 0 and when t >½, d(r²)/dt > 0\nThus when t = ½ there is a stationary inflexion\n\n(b) Find, in terms of t, the rate of change of θ with respect to\n\ntanθ = Δy/Δx\n= 2te^(-t)/e^(-t)\n=2t\n\nsec²θ * dθ/dt = 2\nie\n\ndθ/dt = 2/sec²θ\n= 2/(1 + tan²θ)\n= 2/(1 + 4t²)\n\nCan you show the non analytical method for part (a) too!\n\nx = e^(-t) → -t = lnx ie t = -lnx\n\ny = 2te^(-t) \n= -2xlnx\n\nx: .. 0 ...... 0.25 ... 0.5 .... 0.75 . 1.0 .. 1.25 .. 1.5\ny: undef . -0.15 . -0.36 . -0.22 ... 0 .... 0.28 . 0.61\n... (→0)\n\nLim x → 0 (x lnx) = 0 and OP increases with x \n(limiting min value of OP = 0)\n\nMin value of xlnx occurs when lnx + 1 = 0 ie x = -1/e and at this point OP² will show its inflexion as it cannot be a minimum and nor can it be a maximum so all it can do is slow its rate of growth.
887097
y=1 is a straight line parallel to x axis\nx=(-1/4) is a straight line parallel to y axis
418614
This problem can be described by:\n\ndistance from A to pt = distance from B to pt\ndistance from B to pt = distance from C to pt \n\nsqrt( (x-11)^2 + (y-1)^2 ) = sqrt( (x-3)^2 + (y-6)^2 )\nsqrt( (x-3)^2 + (y-6)^2  ) = sqrt( (x-3)^2 + (y-4)^2 )\n\n\n(x-11)^2 + (y-1)^2 = (x-3)^2 + (y-6)^2\n(x-3)^2 + (y-6)^2  = (x-3)^2 + (y-4)^2\nI think x=87/16 , y=1 is a solution
431677
If you're dealing with a straight line (and a few other types of line I think), the equation will be of the form\ny=mx+c\nThe c part will give you the y-intercept i.e. where the line crosses the y-axis.
525749
The numbers you have are two points on a coordinate grid( the one with the x axis and the y axis. If you have some graph paper you could find the points. The first number in each set is the x-number and the second is the y number. You are trying to find the distance between those points. If you drew the dots on graph paper you could make a triangle out of them. One side is the x side and it is 3 units long(13 -10).  One side of the triangle is the y side and its 4 units long(20-16). Connecting those two points makes the third side or hypotenuse of the triangle. \nSo we use the Pythagorean theorem to find the distance.\nSo its 3squared + 4 squared =9 + 16 = 25.We must take the square root of 25 for the final answer, which is 5. Hope this helps.
1010262
If you mean "equal distance" is the total distance travelled.\nUse the equation\ns=ut+1/2at^2\n\nFor P\nu=0\nt=3\ns=distance=1/2(a1)3^2=1/2(a1)9\n\nFor q\nsimilarly\nu=0\nt=1\ns=1/2(a2)1^2=1/2(a2)\n\nNow both distances are equal therefore\n1/2(a2)=1/2(a1)9\n=>9a1=a2\n\nNote however if you are comparing distance travelled by P in between t=2 and t=3 secs\nThen you have to modify it a bit\ndistance travelled by P in between t=2 and t=3\ns3-s2=1/2(a1)3^2-1/2(a1)2^2\n=1/2(a1)(5)\nIn this case we will have\n1/2(a1)(5)=1/2(a2)1^2\n=>5a1=a2\nI think most probably you mean the second case.\nConfusion arises from wording of the problem.\n3rd second actually means distance travelled between t=2 and t=3 seconds
735689
Basically the equation is \n\n[Sum of (mass*Distance X)]/mass gives the X coordinate,\n[Sum of (mass*Distance Y)]/mass gives the Y coordinate\n[Sum of (mass*Distance Z)]/mass gives the Z coordinate\n\nFor arbitrary shapes that cant be broken down into  point masses,you can use integral equations, but most of them simplify into simple equations, this site has a good list\n\nhttp://mathworld.wolfram.com/GeometricCentroid.html
985185
X                      Y1                        Y2\n-2                     -4                          7\n-1                     -1                          6\n0                       2                           5\n2                       8                           3\n5                       17                         0\n\nX is your domain and Y1 is the range for y=3x+2 and Y2 is the range for y=-x+5.\n\nHope this is what you are looking for.
1010902
This is  basic\nslope= m=(y2-y1)/(x2-x1)=(12-7)/(8-6)=5/2\nangle=arctan(5/2)
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.
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