Is y = - 5/3x + 3/5 perpendicular, parallel, or neither to 3y= -3 + 5 ? How did u know? i dont get it.?
104362
neither... the second eq boils down to y = 2/3 which is a horizontal line\n\nto tell:  if parallel the slopes will be equal\nif perpendicular slopes will be negative reciprocal
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
218561
A parallel line must have the same slope.\nSo you just need to find slopes of the lines that  you have.\n1) 5y= -4x - 11\ny = -4/5x -11/5\nthe slope m = -4/5\n\n2) 4y= 5x-7\ny= 5/4x -7/4\nthe slope m = 5/4
143489
A line is typically defined as y=mx+b where the line has a slope of m. \n\nIf b were 0, the line would run through the origin and quadrants 1 and 3. To get a perpendicular line, you need a line that runs through quadrants 2 and 4.\n\nTo find this line, one simply inverts the coefficient of the slope, m, and reverses the sign of m.\n\nIn this case, you need to solve the original equation for y.\n\n1. 3x-4y=0 (original)\n2. 4y=3x\n3. y=3/4x\n\nFor the original line, the slope is 3/4. The slope of a perpendicular line is the negative inverse, or -4/3.\n\nSo, a perpendicular line is y=-4/3x + c where c can be any number--and in the answer choices, it's 12.
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.
127085
Anything with the same slope as the original problem will be parralell so y>1/3x+5 is one of an infinite number of answers. And a perpendicular line is opposite (in sign) in this case it's positive so we will make it negative and reciprocal of the slope in the original problem. so flip the 1/3 over and change the sign from positve to negative and you have -3x (the slope of the perpendicular). So you can write any equation with the slope  of   -3x. So y>-3x+1.
925148
Hi Alyssa...\n\nSlope equation for a line is y = mx + b where m is the slope. The perpendicular slope is (-1/m).  So, first we need to change your equation to this form.\n\n2x-5y=8\n2x-8=5y\n\nnow, divide by 5\n\n(2/5)x-8/5 = y or y = 2/5 x - 8/5\n\nso the slope of the original line is m=2/5\nthe perpendicular slope for a lin is -1/m = -5/2\n\nNow, let's create a new line using the "Point-Slope" formula:\ny-y1 = m(x-x1) where m is the slope and (x1, y1) is a known point.\n\nWe calculated the slope aready above at m = -5/2\nYou have been given the point as (4, -4)\nNow, just plug the numbers into our Point-Slope equation.\n\ny-(-4) = (-5/2) (x-4)\ny + 4 = (-5/2)x + (-5/2) * (-4)\n\nmultiply by 2\n\n2y + 2*4 = 2*(-5/2)x + 2*(-5/2) * (-4)\n2y + 8 = -5x + 20\n\nor\n\n2y + 5x = 12\n\nNow, check that this line does in fact pass through the point (4, -4) by plugging the numbers and verifying that the equation is true!
345972
If you're parallel to the y axis, you have a vertical line. The equation is x = -1\n\n\nIf you're parallel to the x axis, you have a horizontal line The equation is y = 5
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
959758
( -1 -y) / (x + 4y) = -1/3\n=> (1 + y) = (x + 4y)/3\n=> 3 + 3y = x + 4y\n=> x + y = 3 => x = 3 - y = k  =>\nx = k\ny = 3 - k\n\nit means,for infinite number of k,x = k and y = 3 - k
342522
4x - 5y = 2\n5y = 4x - 2\ny = 4x/5 - 2/5\n\nThe slope of this is 4/5, so the slope of the perpendicular is -5/4 (just take the opposite of the reciprocal).\n\nFor the new line, m = -5/4, and you have a point (2, -1).  Using the point-slope formula,\n\ny - y1 = m(x - x1)\ny - (-1) = (-5/4)(x - 2)\ny + 1 = -5x/4 + 5/2\n5x/4 + y = 3/2\n5x + 4y = 6
965470
The answer is neither. If they are parallel they must have the same slope and if they are perpendicular the slopes cannot have the same sign. The slopes of the lines are both negative and unequal.\n\nfind the slopes of both lines.\n\n5x+2y =3\ny = (3-5x)/2 = -5x/2 +3/2\nslope 1 = -5/2\n\n2x+5y =8\ny= (8-2x)/5 = -2x/5 + 8/5\nslope 2 = -2/5
214212
Hi, here's how ya do the problem:\n\n-5x+y=6\n\ny=5x+6 (move -5x over by adding the opposite)\n\nSo then you substitute it into the problem\n3x-2y=3 so that it looks like\n\n3x-2(5x+6)=3\n\nNext distribute the -2\n3x-10x-12=3\n\nand solve\n-7x-12=3\n-7x=15\nx=2.14\n\nThen you substitute 2.14 into either equation to figure out y\n-5(2.14)+y=6\n-10.70+y=6\ny=16.7\n\nHope this helps!
925167
Hi Karen \n\n5(w+3)+3(w+1)=14\n\n● Step 1\n5(w+3) = \n(5 * w ) + ( 5 * 3 ) =\n5w + 15\n\n● Step 2\n3(w+1) =\n( 3 * w ) + (3 * 1) =\n3w +3 \n\n● Step 3\n5w + 15 + 3w +3  = 14\n( 5w + 3w ) = 14 - 15 - 3\n8w = -1 -3 \n8w = -4\nw = - 4 / 8\nw = -1/2  Or  - 0.5 \n\nGood Luck dear ♣
939650
x+y=-3\nx=-y-3\nx -3  -2  -1  0  1  2  3 \ny  0  -1  -2 -3-4-5 -6\n\nx^2+2y^2-12y-18=0\nwhen y=-1\nx^2+2+12-18=0\nx^2-4=0\nx^2=4\nx=+/-2\nso (-2,-1),(2,-1)\nwhen y=3\nx^2+18-36-18=0\nx^2-36=0\nx=+/-6\nso(-6,3),(6,3)\n\npoints of intersection\n(-2,-1) and (-6,3)
1142461
1. x-y=5; y= x - 5\nAdd y to each side x - y + y = 5 + y, becomes x = 5 + y. Subtract 5 from each side, x - 5 = 5 - 5 + y, becomes x - 5 = y\n\nSolve the rest on your own
24594
Here's a quasi-logical answer:\nBy definition, two lines are parallel if they have the same slope (s1=s2) and they are perpendicular if the product of their slopes is -1 (s1*s2=-1)\n\nSo we have s1=s2 and s1*s2=-1\nTherefore s1^2=-1, so s1=s2=i=sqrt(-1)\nIn conclusion, any two lines y=ix+a and y=ix+b are both parallel and perpendicular at the same time.