Given P1(1,-4,7) and P2(3,-3,2).   Find the Distance between P1 and P2?
150854
i bet your thinking, my god, it has 3 values (x,y,z), how the heck does one attack this\n\nbelieve it or not, its simple\n\njust like you would in R^2\n\nsqrt[ (x-x1)^2 + (y-y1)^2 + (z-z1)^2]\n\nso \n\nsqrt [ (3-1)^2 + (-3- -4)^2 + (2-7)^2]\n\nso\n\nsqrt[ (2)^2 + 1^2 + (-5)^2]\n\nso\n\nsqrt [ 4 +1+25]\n\nso\n\nsqrt [30]\n\nhope that helps
508385
3-4(2)+2(2)+1=0\n2(-3)+1-[3-4(-3)]=-20\n3-4[2(1)+1]=-9\n\ng(x)=(x-4)/-5\n3-4[(x-4)/-5]=x\n{[3-4(x)]-4}/-5=x
768807
distance formula\ndistance between two points =Square Root of ( (x1-x2)^2 + (y1-y2)^2)\nSquare root of ((4-1)^2 + (-8 -3)^2)\nSquare root of (9 + 121) \nSquare root of (130)\nuse calculator\n11.4
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
265237
3s = -6\ndivide each side by 3\ns = -2
35596
The answer above didn't go far enough down the page. That's just the equation of the line.\n\nIf the point is (x_0, y_0), then the distance is:\n\n(|ax_0+by_0+c|)/(sqrt(a^2+b^2)).
238182
p1    p2\n--- = ---\nq1    q2\n\n 5     2\n--- = ---\n 2    q2\n\nCross multiply,\n\n4/5
805734
Point A: (1,4)\nPoint B: (2,1)\n\nTake the coordinates one at a time.  From A to B, the X coordinate goes from 1 to 2, a difference of 1.  Twice the existing difference of 1 is 2, which we add to Point B's X coordinate, 2, to get 4.\n\nDo the same thing with the Y coordinates.  From A to B, the Y coordinate goes from 4 to 1, a difference of -3.  Twice the existing difference is -6, which we add to Point B's Y coordinate, 1, to get -5.\n\nCombining the X and Y coordinates, we get the final answer, (4,-5).
705901
Here is how you do it:\n(2,1) and (3,4) = [(2)(3) + (1)(4)]/(2)(1) - (3)(4)\n                        = (6 + 4)/(2-12)\n                        = 10/-10\n                        = -1\n\nTada
66380
You have to use the formula -\nm1s1(t3-t1)=m2s2(t2-t3)\n=>191(4.184)(30.5-25)=38s2(192-30.5)\n=>s2=191(4.184)(5.5)/((38)(161.5)\n=>s2=0.716 J/g.k
735022
There will be a term proportional to distance added to a constant delay term, so the form will be t=kD + c. Solving for k and c:\nt1=.152=100k + c\nt2=.229=170k + c\nt2-t1=.077=70k\nThen k=.0011\nt1=.152=.11 + c --> c=.042\nt2=.229=.187 + c --> c=.042\nResult is\nt=.0011*D + .042
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.
1073694
If you're given that:\n\nP1 is at (x1,y1) \nP2 is at (x2,y2)\n\nand you're looking for: \n\nP3 at (x3,y3)\n\nIf the length of the line from P1 to P3 is L1\nand the length of the line from P2 to P3 is L2\n\nThen you can solve for the single point which\nsatisfies both of:\n\nL1 = SQRT( (X3-X1)**2 + (Y3-Y1)**2 )\nL2 = SQRT( (X3-X2)**2 + (Y3-Y2)**2 )\n\nTo simplify the math, square both sides of each eq:\n\nL1**2 = (X3-X1)**2 + (Y3-Y1)**2 \nL2**2 = (X3-X2)**2 + (Y3-Y2)**2\n\nSince you're given the first two points and\nthe lengths to the last point, the terms\nX1, Y1, X2, Y2, and L1, L2 are all constants. \nOnce you plug in the constant values, you're\nleft with two equations and just two unknowns \n(X3 and Y3).\n\nSolve for X3 and Y3 using a little algebra.
848382
To figure this out, you need to know that an ordered pair is in the form: (x, y). Just take each ordered pair and stick it into the equation:\na) 3(1) = 2(-1) -1 ... 3 = -3 (Not correct.)\nb) 3(-1) = 2(-1) - 1 ... -3 = -3 (Correct.)\nc) 3(-2) = 2(-5/2) - 1 ... -6 = -6 (Correct.)\nd) 3(0) = 2(-1/2) - 1 ... 0 = -2 (Not correct.)\n\nSo, the correct answers are (b) and (c).
819448
2p^3 = 2p(p + 2)\n2p^3-2p(p+2)=0\n2p(p^2-p-2)=0\n2p(p-2)(p+1)\np=0\np-2=0\np=2\np+1=0\np=-1\n\nroots are \np=0, 2, -1
407918
(3,-7)(6,15)\nm = (y2 - y1)/(x2 - x1)\n\nm = (15 - (-7))/(6 - 3)\nm = (15 + 7)/3\nm = (22/3)\n\nANS : (22/3)
1136505
(7,0) \nXdiff = X1 - X2\nYdiff = Y1 - Y2\nThe first (X1 or Y1) should be the larger number of the 2\nso Xdiff = 4 - -3 = 7\nand Ydiff = 0 - 0 = 0