how to find theslpoe equation of the parallel line that runs throw (2,-2) and the equation is y=3x+4?
953271
if it is parallel the slope is 3 so the equation is\ny = 3x + b\nnow you just solve\n-2 = 3*2 + b\n\nfor b and that is your new y intercept and equation.
718048
x + y = 3\n\nSolve for a variable.  I chose X.\nx+y = 3\n -y    -y\nx = 3 - y\n\nSubstitute the value of X in the first equation.\n2(3-y) + y = 4\nDistribute.\n6-2y+y = 4\nAdd like terms.\n6-y = 4\n-6    -6\n-y = -2\nmultiply by -1.\ny=2\n\nSince you know Y = 2, substitute that back into the second equation that you solved for X with.\nx + y = 3\nX + 2 = 3\n  -2    -2\nX = 1\n\nX = 1\nY = 2
835073
okay well\n\n1) combine two to figure out the first variable\nx-2y+z=4\n-x+2y+3z=-4\nso 4z=0; z=0\nthen figure out the next variable with a different set of equations\nlike and multiply so the ys will cancel [take Zs out cause zero]\nx-2y+z=4\n2(2x+y)=6\n\ntherefore\nx-2y=4\n4x+2y=6\n5x=10\nx=2\n\nthen substitute for y\n\n-x+2y=-4\n-2+2y=-4\n2y=-2\ny=-1\n\nso y=-1, x=2, z=0\n\n\ncontinue the process with the rest.
1160744
Just remember that to make the equation parallel, the slope of the equation have to be the same.  so for the first one, the slope will stay the same, substitute the 5 into the y, plug in 0 for the x solve for the b.  the original equation is, y=mx+b.  \nAgain, same thing for the second question, first solve for the y.  \nwhich becomes, y=2x+7/2.  I think.  I am doing it in my head, if I am wrong.. please correct me.  To make it perpendicular, the slope has to be opposite.  so 2x becomes -2x.  Now, do the same that I stated for the first question to figure out b.  \nSame thing for the third question.  Use the equation in the y=2x+7/2 form... y= - mx+b...(it's perpendicular) then plug in the two points solve for b.  \nI believe these steps are right.  I hope this helps.
952487
x+y=8\nx-y=-4\n______\n2x=4 (because +y and -y cancel out)\nx=2\nthen plug 2 in for x in one of the equations\n2+y=8\nsolve for y\ny=6\nordered pair is (2, 6)
128403
general equation of a line with slope m and y-intercept c is \n\ny = mx + c ....... (1)\n\nsay this line is parallel to 2x+3y+4=0 which means the slopes of two lines are equal\n\nso, m = -2/3  [-(x-coefficient/y-coefficient)]\n\nand c = 6 (given) \n\nsubstituting m and c in (1) \n\nrequired equation is y = -(2/3)x + 6\n\n=> 2x+3y-18=0
1002212
first calculate slopes of each line \nslope of 1st line (1+4)/(3+7)\n                          =5/10\n                          =1/2\nslope of second line (4-2)/(10-6)\n                                 =2/4\n                                 =1/2\nas bothe slopes r equal these lines r parallel...\n..\n\nanywaz..two lines r peripendicular wen product of their slopes   is -1(negative one)
725560
The  y coordinate is the same in both sets of coordinates the equation for this line is line y=-2. \n\nThis line is perfectly horizontal => the slope is 0.
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
975114
i am pretty sure the answer is y= -1\n\nyou put it in y=m(x-x^1)+y^1\nand then you put (4,-2) in for x^1 and y^1 and the slpe in for m so it would be\ny=1/4(x-4)-2\nand then distribute the 1/4 so it is\n\ny=1/4x-1-2\nsimplify\n\ny=1/4x-3\n\nnow substitute in (8,y) 8 goes in for x\nso it is\n\ny=1/4(8)-3\n\nso y=2-1\ny=-1\n\ni hope that is right and it helps!
1106036
Rise over run, or (y2-y1)/(x2-x1).  If your first point was (x1,y1) and your second point (x2,y2), then it's easy:\n\n[8 - (-4)] / [14 - 10] = (8+4)/(14-10) = 12/4 = 3\n\nThe slope of the line is 3.
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
901570
The equation of a straight line passing through two points (x1,y1) and (x2.y2) is given by the formula,(y-y1)/(y1-y2) = (x-x1)/(x1-x2)\nTherefore,the required equation is(y-3)/(3-1)=(x-4)/(4+2)\n=>(y-3)/2=(x-4)/6\n=>2(x-4)=6(y-3)    [after cross multiplication] \n=>2x-8=6y-18\n=>2x-6y-8+18=0\n=>2x-6y+10=0\n=>x-3y+5=0 [dividing all trms by 2]\nTherefore the reqd. equation is x-3y+5=0
930136
Use the slope formula to get the slope.  All parallel lines will have the same slope.\n\nm = (y2-y1) / (x2-x1)\n\nSo, for your points you have the following:\n\nm = (2-1) / (4-15) = 1/(-11) = -1/11\n\nTherefore all lines parallel to this line will also have a slope of -1/11.\n\nHope that helped!\n\n~ ♥ ~
917321
x+y=2\n2x-y=0\n\nx=2-y\n2(2-y)-y=0\n\nX=2-y     \n4-3y=0   \n\nx=2-4/3\ny=4/3
698387
y= -(2/3)x^2 + (2/3)x + 4\n\n\n(P): y = ax^2 + bx + c\n\nPoints (0;4)  (-2;0)  (3;0) are element of (P)\n-->  4 = 0a + 0b + c --> c = 4\n0 = 4a - 2b + c\n0 = 9a + 3b + c\n\n--> 5a + 5b = 0 --> b = -a\n--> 4a + 2a + 4 = 0 --> a = -2/3,  b=2/3
932717
y = (5/2)x + 6\nThe slope of this line is (5/2). A parallel line will have the same slope. \n\nUsing a point-slope form equation...\n(y-1)=(5/2)(x-2)\n\nThis is equal to y = (5/2)x -4.\n\nAnother way to do it is solve it directly in slope-intercept form. We know the slope is (5/2), let's call the y-intercept b.\n\ny = (5/2)x + b.\n\nWe know it works for the point (2,1) so...\n\n1 = (5/2)2 + b\n1 = 5 + b\nb = -4.\n\nSo, y = (5/2)x - 4