how do you find the perpendicular bisectorthrough points (1,3) and (-1,9)?
954966
First find the slope of the segment connecting the points:\n(9-3)/(-1 - 1) = 6/-2 = -3\n\nThen do its opposite reciprocal since perp. lines have opp. recip. slopes:  1/3\n\nThen find the midpoint: ( (1 + -1)/2, (3 + 9)/2) = (0, 6)\n\nSo you need the equation of a line through (0,6) with slope 1/3\n\nLines are y = mx + b where m is slope (1/3) and b is y-intercept (6)\n\nSo y = 1/3 x + 6\n\nHope that helps explain and not just gives the answer
759094
the perpendicular slope is the negative inverse of the origional, meaning, once you change your equation into y=mx+b form, you flip your slope, and make it negative.\n\n-3y=1212-2x\ny=-404-(2/3)x\ny=-(2/3)x-404\n\nso the slope is -2/3. the perpendicular is 3/2.\n\nplug it in with the point:\n\n3=(3/2)(4)+b\nb=-3\n\ny=(3/2)x-3
655971
To find if the vectors are perpendicular, you should find the dot product.\n\n4*4 + 5*-2 + -3*2 = 16 - 10 - 6 = 0\n\nSince the dot product is 0, the vectors are perpendicular.\n\nTo find if they are parallel, you check if the vectors are multiples of each other.  Since they are perpendicular, though, this is unnecessary.
850699
y = mx + b  --> slope-intercept form\ny = 3x + b\nsubstituting point(4, 9)\n9 = 3(4) + b\n9 = 12 + b\nb = -3\n\ntherefore the equation is given by\ny = 3x - 3
713860
1.Xc=1/2pifc\nc=1/2pifXc\n\n2.ib=ibq+rt(2is)\nib(1-q)=rt(2is)\nsquaring\n[ib(1-q)]^2=2is\ns=[ib(1-q)]^2/2i\n\n3.Q=Xl/R\nXl=Q/R\n\n4.Lt=L1+L2+2Lm\n2Lm=Lt-L1-L2\nLm=[Lt-L1-L2]/2\n\n5.Es=E1+E2+E3\nE2=[Es-E1-E3]\n\n6.R1=i2R^2/i2\ni2R^2=i2R1/i2\nR=rt[i2R1/i2]\n\n7.R1/Rx=R2/R3\nRx=R1R3/R2\n\n8.y=mx+b\nb=y-mx\n\n9.i2=i1R1/i1
903948
find the slopes of both lines and if there answers are opposite recipricals, then they are parrallel\n\nheres adn ex:\nslope = 2 and -1/2\n\nin other workds, if you multiply the two slopes, u should get -1
81415
The general method is perhaps best illustrated by working through an \nexample.\n\nFind the shortest distance between the given lines, and the points of \nclosest approach on each line.\n\n  x     y-3     z                  x-5     y-8     z-2 \n --- =  ---  = ---   = s   and    ----- = ----- = ------ = t\n  1      1      -1                  3       7       -1\n\nThe common perpendicular is obtained from the vector product\n\n   | i   j    k |    = i(6) -j(2) + k(4)\n   | 1   1   -1 |\n   | 3   7   -1 |   \n\nSo common perpendicular is the vector (3, -1, 2) which can be written \nas a unit vector in the form  1/(sqrt(14)[3, -1, 2]\n\nThe vector connecting the given point on line (1) with the given point \non line (2) is  [(5-0), (8-3), (2-0)] = (5, 5, 2)\n\nThe scalar product of this with the common perpendicular in unit \nvector form is\n\n    5 x 3 + 5 x (-1) + 2 x 2        14 \n    -------------------------  =  -------   =  sqrt(14)\n            sqrt(14)              sqrt(14)\n\n\nSo the shortest distance is sqrt(14).\n\nNow to find the points of closest approach, we have:\n\nline (1) is  r = (0, 3, 0) + s(1, 1, -1)  =  [s, (3+s), -s]\nline (2) is  r = (5, 8, 2) + t(3, 7, -1)  =  [(5+3t), (8+7t), (2-t)]\n\nand we must find s and t. \n\nLine joining a general point on line(1) to a general point on line(2) \nis the vector\n\n  [(5+3t-s), (8+7t-3-s), (2-t+s)]\n\n  [(5+3t-s), (5+7t-s), (2-t+s)]\n\nand if s and t are points of closest approach this must be parallel to \nthe vector (3, -1, 2)\n\n     5+3t-s     5+7t-s     2-t+s\nso   ------  = -------- =  ------\n       3           -1        2\n\nFrom these equations s = -1  and  t = -1\n\nThe points of closest approach are therefore \n\non line(1)  (-1, 2, 1)      and on line(2)   (2, 1, 3)\n\nCheck that shortest distance is sqrt(14)\n\nshortest distance = sqrt((2+1)^2 + (1-2)^2 + (3-1)^2)\n\n                  = sqrt(9 + 1 + 4)  =  sqrt(14) \n\n\nIf you require the mid-point of these points, you get\n\n   -1+2  2+1   1+3\n   ----, ----, ----  =  [1/2, 3/2, 2]\n     2    2     2
1101023
First you must find the slope of the line with the points (6,3) and (7,7).  Slope equals Y1-Y2 / X1-X2.  So that equation with the numbers substituted is Slope equals 3-7 Divided by 6-7 or -4/-1 or 4.  That is a slope of four.  \n\nA perpendicular line's slope is the opposite reciprocal or -1/4.
533058
Point slope form:\n\ny-(-3)=(1/2)(x-5)
397805
This problem is certainly tractable, you just need to\nfind a consistent/methodical way of counting the\nrectangles -- without missing any and without counting\nany of them multiple times.\n\nThe way I chose is to define each rectangle as being\ndefined by the square in the upper left corner and the\nsquare in the lower right corner.  Consider the following\nsimplified 2x3 case (where each square is labelled by\nits X and Y coordinates:\n\n0,0     0,1   0,2\n1,0     1,1   1,2\n\n\nThis 2X3 example has the following rectangles starting\nat 0,0.  Remember, each rectangle is defined by the\nsquare in its upper left and the square in the lower right:\n0,0   0,0\n0,0   0,1\n0,0   0,2\n0,0   1,0\n0,0   1,1\n0,0   1,2\n\nThen it also has the set anchored in the upper left by 0,1:\n0,1   0,1\n0,1   0,2\n0,1   1,1\n0,1   1,2\n\nThen it has the set anchored in the upper left by 0,2:\n0,2   0,2\n0,2   1,2\n\nThen those anchored by 1,0:\n1,0   1,0\n1,0   1,1\n1,0   1,2\n\nThose anchored by 1,1:\n1,1   1,1\n1,1   1,2\n\nAnd finally those anchored by 1,2:\n1,2   1,2\n\n\nRewriting the matrix another way, showing how many\nrectangles are anchored by each location in the matrix:\n\n2x3    2x2   2,1\n1x3    1x2   1x1\n\n\nand then doing the math:\n\n6      4     2\n3      2     1\n\nfor a total of 6+4+2+3+2+1 = 18\n\n\nDoing the same analysys for a 4x5 gives:\n\n4x5   4x4   4x3   4x2  4x1\n3x5   3x4   3x3   3x2  3x1\n2x5   2x4   2x3   2x2  2x1\n1x5   1x4   1x3   1x2  1x1\n\nor rewriting after doing the multiplication:\n\n20    16    12    8    4\n15    12     9    6    3\n10     8     6    4    2\n 5     4     3    2    1\n\nAdding up the columns gives:\n\n50    40    30   20   10\n\nFor a total of 50+40+30+20+10 = 150 total rectangles
757381
to determine whether two lines are parallel or perpendicular or neither, we need to have their slopes.\nthe slope of a line passing through (x1,y1) and (x2,y2) is given by m=(y2-y1)/(x2-x1)\ntwo lines are parallel, if their slopes are equal\ntwo lines are perpendicular if the product of their slopes is -1\nfirst find the slopes of both the lines\nslope of first line is m1=(-3-2)/(9-5)=(-5/4)\nslope of the second line is m2=(1-6)/(8-4)=(-5/4)\nsince m1=m2, the lines are parallel, but not perpendicular
384702
The first 3 are all straight lines, so you just need to get 2 points in each.\n1. y = |x| + 3 for x = 1 then y = 4, for x = 2 y = 5.\n\n2. y = 3/2x - 2 for x = 2 then y = 1, for x = 4 y = 4\n\n3. y = x - 5 for x = 5 y = 0 for x = 6 y = 1\n\nFor #4. This is a curve shaped like and upside down U that passes through 0,0, for all values of x, y will always be negaive.\n\nThe best way is to substitute values of x and get values of y and sketch.\n\nHope I was some help
823237
There is several way to solve the problem, but I think it is easies, to find the parametric representation, and then use, that the scalar-product of the orientations-vectors is zero, if the line are perpendicular, and the orientations-vectors is linear depended if parallel.\n\nL1:  y = -½x-3 => (x,y)= (1,-½)*t+(0,-3)\n4x-2y+5 = 0 <=> y=2x+5/2=>\nL2: (x,y) = (1,2)*t + (0+5/2)\n\nbecause (1,-½)*(1,2)=1-1=0 the lines are perpendicular
490013
Find their slopes: if slopes are equal, they're parallel; if slopes are opposite reciprocals (like 2 and -1/2) they're perpendicular; if neither then neither.\n\n#1: (2- -8)/9-3 = 10/6 =  5/4\n\n#2: (6 - 10)/(2 - 4) = -4/-2 = 2/1\n\nSo neither.
932620
solve the equations of any two and make sure that the solution satisfies the equation of the third perpendicular bisector
1149797
The first one is (-5, 3). I don't remember how to do the others.
1128164
remember 0 degrees is along the +x axis\n90 degrees is along +y axis\n180 degrees is along -x axis\nso you are moving 12 points in the negative x direction if you start at 10 then 10+(-)12 = -2\nno movement in the y direction so 10+ 0 = 10\nPoint q is at  (-2,10)