If the end points are D ( - 2 , 3 ) and K ( 1 , 2 ) what are the coordinates after a rotation of 90 degrees?
80065
Rotated counter-clockwise:\nD (-3,-2) and K (-2,1)\n\nRotated clockwise:\nD (3,2) and K (2,-1)
779746
assume u1 is angular path1, u2 is angular path2\nu1=w1*t, u2=w2*t, while u1+u2=2*pi rad \nthus 2*pi=(w1+w2)*t, hence t=2*pi/(w1+w2).
355378
Let the two sides be formed from three points, (a,b), (c,d), (e,f). So: Side A is (a,b) to (c,d) and Side B is (c,d) to (e,f).\n\nMidpoint of A: ([a+c]/2, [b+d]/2)\nMidpoint of B: ([c+e]/2, [d+f]/2)\nThe vector formed is:\n([c+e]/2 - [a+c]/2, [d+f]/2 - [b+d]/2) =\n([e-a]/2, [f-b]/2)\nThe third side is formed from (e,f) to (a,b), or\n([e-a], [f-b])\nIt's easy to see then that they are parallel.
1078325
Let as suppose that your ellipse is rotated by 45 degrees to right (or by -45 degrees). The new coordinates are X = x cos (-45) - y sin (-45) and Y = x sin (-45) + y cos (-45) and they are satisfying X^2/a^2 + Y^2/b^2=1. There for (x + y)^2/2a^2 + (x - y)^2/2b^2 = 1 (in a few easy steps). The mayor axis over the x=y is given by |a| > |b| .
412752
This looks like it's been rotated 90 degrees counterclockwise.\n\nSo, I guess it would be as follows:\n\nx = -(9/25)(y-3)^2 + 7
411757
since y=+-(x)^(3/2) Range [0,oo[\nnP=(n,n*n^1/2) and x_n/y_n= [(n^1/2^)/n]=n^-1/2
699547
sin[x-(pi/2)]=kcosx\n-sin[(pi/2)-x]=kcosx\n-cosx=kcosx\nk=-1\ncos[x+{pi/2)]=ksinx\n=cos[(pi/2)+x]=ksinx\n=-sinx=ksinx\nk=-1
768593
To find the inverse of this relation, reverse the coordinates of the ordered pairs.\n\nThe answer to this would be\n {(18, -14), (-18, 3), (-1, -2)}
433823
Revolving about the x-axis yields a cone whose tip is at the origin and opens in the positive x direction.  Taking discrete x slices (dx) yields a disk whose volume is given by pi*(kx)^2 dx.  Integrating from 0 to h gives us (where "S" is used as the integration symbol):\n\nV = S pi*(kx)^2 dx\n= pi * k^2 S x^2 dx\n= (pi * k^2) * (x^3)/3\n\nEvaluating from 0 to h gives us\n= (pi * k^2) * (h^3)/3 - (pi * k^2) * (0^3)/3\n= (pi * k^2 * h^3)/3
677039
(1) { -2x-y-z=0\n(2) { -x-2y-z=0\n(3) { -x-y-2z=0\n\nLet's look at (2):\n-x-2y-z=0 => x=-2y-z\nNow at (3):\n-x-y-2z=0 => x=-y-2z\nNow we can see that\n-2y-z=-y-2z => *** y=z ***\n\nNow let's put these values into (1):\n-2x-y-z=0 => -2x-y-y=0 => -2x-2y=0 => 2x=-2y => *** x=-y ***\n\nFor now, we know that x=-y=-z\nLet's put all of what we know into (2), and express everything using 'x':\n-x-2y-z=0 => -x-(-2x)-(-x)=0 => -x+2x+x=0 => 2x=0 => *** x=0 ***\n\nThus, x=0 => y=-0=0 => z=0\nSo the only solution is (0,0,0).\n\nGood luck!
627342
ok, this is found in the 4th quadrant, the order is as follows\n\n(x,y) 1st\n(-x,y) 2nd\n(-x,-y) 3rd\n(x,-y) 4th\n\nhope this helps\n\nmatttlocke
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
1187062
a) the 90 degree angle
1151668
The x-coordinate decreased by 4 - (-2) = 6 and the y-coordinate decreased by 2 - 0 = 2\n-3 - 6 = -9\n5 - 2 = 3\nAnswer: Q'(-9,3)
794182
What nicknamr said, and also, the vertex would be located at the x-value found for the axis of symmetry, and the y-value for that x-value.  So, -b/(2a) = -12/(2 x -1) = 6.  Then, y= -(6)^2 + 12(6) -32 = -36 + 72 -32 = 4.  The vertex comes out to be (6,4).  (I would have answered all the other stuff too, but he already did it, and I found it rather redundant to repeat it all, so I just added on to it =)  )
533058
Point slope form:\n\ny-(-3)=(1/2)(x-5)
995072
The first point is not at exactly 210 degrees.  It is a little over 212 degrees.\n\nThe second point (3,3) is obviously at 45 degrees, not 30 degrees.\n\nThe distance between (-8, -5) and (3, 3) is\nsqrt((-8 - 3)^2 + (-5 - 3)^2) = sqrt((-11)^2 + (-8)^2) = sqrt(121 + 64)\n = sqrt(185)\n\nThe original angle from (3,3) to (-8, -5) is\natan2(-8 - 3, -5 - 3) = atan2(-11, -8) = 180 degrees + arctan(8/11)\ncounterclockwise.\n\nThe clockwise rotation subtracts thirty degrees from that, so in the final coordinate system the original (-8, -5) is sqrt(185) magnitude in the 150 degrees + arctan(8/11) direction from the new origin at the original (3,3).\n\nLuckily, sin( arctan(8/11) ) = 8 / sqrt(8^2 + 11^2) = 8 / sqrt(185)\nand cos( arctan(8/11) ) = 11 / sqrt(185).\n\nWe have the "new" x is\n\nsqrt(185) cos(150deg. + arctan(8/11)) =\n = sqrt(185) ( cos( 150 deg) cos ( arctan(8/11) ) - sin( 150 deg ) sin( arctan(8/11) ) )\n = 11 cos(150 deg) - 8 sin(150 deg)\n = 11 (-sqrt(3)/2) - 8 (1/2)\n = -(1/2)(8 + 11 sqrt(3)) or approx. - 13.526\n\nand the "new" y is:\n\nsqrt(185) sin(150deg. + arctan(8/11)) =\n = sqrt(185) ( sin( 150 deg ) cos ( arctan(8/11) ) + cos( 150 deg ) sin( arctan(8/11) ) )\n = 11 sin( 150 deg ) + 8 cos( 150 deg )\n = 11 (1/2) + 8 (- sqrt(3) / 2)\n = (1/2)(11 - 8 sqrt(3)) or approx. -1.428\n\nwhich makes sense, as the original slightly over 212 degrees after the rotation is slightly over 212 - 30 = 182 degrees, so both quantities are negative with the magnitude of x large and the magnitude of y small.\n\nCheck: "new x"^2 + "new y"^2 =\n = ( -(1/2)(8 + 11 sqrt(3)) )^2 + ( (1/2)(11 - 8sqrt(3)) )^2\n = (1/4)( 8^2 + 2*8*11 sqrt(3) + 3*11^2 + 11^2 - 2*11*8 sqrt(3) + 3*8^2)\n = (1/4)( 4 (8^2 + 11^2) )\n = 185 as it should be.\n\nCheck: atan2("new x", "new y") = 180 degrees + arctan("new y" / "new x")\n\n"new y" / "new x" =\n = (11 - 8sqrt(3)) / (- 8 - 11sqrt(3))\n = (11 - 8sqrt(3)) (- 8 + 11sqrt(3)) / ( (-8)^2 - (11sqrt(3))^2 )\n = (-88 + 121sqrt(3) + 64sqrt(3) - 88*3) / (64 - 121*3)\n = (- 352 + 185sqrt(3)) / -299 or approx. 0.105587\nwith arctan 6.02737\nand expected value arctan(8/11) - 30 or approx. 6.02737\nso this checks out, too.\n\nChanged coordinates:\n\n( -(1/2)(8 + 11 sqrt(3)), (1/2)(11 - 8 sqrt(3)) ) or approx ( - 13.526, -1.428 )