Write the equation of the line through point (-5,0) and parallel to the y-axis.?
436432
X = (0*Y) -5\nY = 0 * X\n\nbut basically just write X = -5 and thats it
247531
The general purpose straight line has the equation Y = mX.  If you know the slope (m) to get the equation of a line thru a given point, say (h,k) just substitute the slope for the m, (X-h) for the X, and (Y-k) for the Y.  E.g. a line with slope 2 thru (-5,7) has the equation (Y-7)=2(X+5).  Then do a little algebra to get the slope-intercept form;\nY=2X+17.\n\nIf slope is not given, but instead you have coordinates of two points, find the slope by dividing the difference between the Y coordinates of the points by that between the X coordinates.  It doesn't matter which point you subtract, as long as you do it the same way for both the X and Y coordinates.   Then use that slope with the coordinates of either one of the points as above to get the equation.\n\nIn the example given, you're apparently supposed to find a line thru (5,1) that intercepts the line Y=4X-7.  Any two lines will intercept somewhere, so two quick answers are X=5 or Y=-7!  You can substitute X=5 into the equation to get the point of intersection of the first: (5,13).  OR\n\nA line thru (5,1) parallel to the line y=4X-7 will have the same slope.  So its equation will be (Y-1)=4(X-5) or \nY=4X-19.\n\nA line thru (5,1) and perpendicular to the line y=4X-7 will have a slope that is the negative reciprocal of 4 = -1/4 or -0.25.  So its equation will be (Y-1)=(-0.25)(X-5) which simplifies to Y = -0.25X + 1.25 + 1 or Y = -0.25X + 2.25.\n\nThink about it.  Thinking always helps.  Yahoo won't be there for your tests.
197692
y-5-y-5=7y\n-10=7y\ny=-10/7
293210
the graph of |x - 2| + |y - 5| = 5 consists of four lines which are graphs of the following functions:\nx-2+y-5=5 which is  line A: x+y=12\n-x+2-y+5=5 which is line B: x+y=2\n-x+2+y-5=5 which is line C: x-y=-8\nx-2-y+5=5 which is line D: x-y=2\nnext thing to do is graph these four lines. the easiest way is to find the 0s. \nline A: x+y=12\n        x=0 y=12 point A1(0,12)\n        y=12 x=0 point A2(12,0)\nline B:  x+y=2\n         x=0 y=2 point B1(0,2)\n         x=2 y=0  point B2(2,0)\nline C: x-y=-8\n         x=0 y=8 point C1(0,8)\n         x=-8 y=0 point C2(-8,0)\nline D: x-y=2\n        x=0 y=-2  point D1(0,-2)\n        x=2 y=0   point D2(2,0)\nthese four lines enclose a figure that is a rectangle. lets find the length of each side. after you plot the four lines you will notice that line x+y=12 intersects with line x-y=2 lets call the intersection point I1 to find I1 coordinates we rearrange y=12-x and y=x-2 and from y=y we get x-2=12-x or 2x=14 or x=7 if x=7 y=12-7=7-2=5 the point I1 is I1(7,5).\nline x+y=12 also intersects with line x-y=-8. Lets call this intersection point I2. to find coordinates we have y=12-x and y=x+8 y=y or 12-x=x+8 or 2x=4 or x=2 for x=2 y=12-2=2+8=10 so the point is I2(2,10) \nside I1I2 of the rectangle length is side a. to find length of a we do\na^2=(7-2)^2+(10-5)^2=25+25=50\na=sqrt(50)\nwe need the second side, lets call it b.\nto find length of side b we need at least one more point. lets find intersection of x-y=2 and x+y=2 and call it point I3. y=x-2 and y=2-x or x-2=2-x or 2x=4 or x=2. for x=2 y=2-2=2-2=0 point I3(2,0)\nside b:\nb^2=(7-2)^2+(5-0)^2=50\nb=sqrt(50)\nit means a=b or our rectangle is a square!\nso the area of the square is Area=side^2 since our square&#92;s side is a=b=sqrt(50), the area is 50.\nI hope this helps, I really don't have time to explain it in more details.
96194
Yes you are on the right track...Sort of.  \n\nSay you plug in any number for x.  We'll use 5\n\n-3(5)+y=-15\n-15+y=-15\n\nNow add 15 from the left to cancel the negative 15.  Then add 15 to the right to make the equation:\n\ny=0\n\nSo if x=5 and y=0 then your points will be (5,0)\n\nSay you have your original equation again\n-3x+y=-15\n\nNow plugin for x any number.  We'll use 1.\n\n-3(1)+y=-15\n-3+y=-15\n\nNow add 3 to the left to cancel out the negative 3 and add 3 to the right to make:\n\ny=-12\n\nSo x=1 and y=-12   (1,-12)\n\nHope that helps
711572
You are graphing y < -4/5  x + 9/5\n\nOn the y-axis, above zero, graph 1  4/5 (a little less than 2).  From that point, move down 4 squares and right 5 squares and put another point.  Connect this point with the 1 4/5 on the y-axis and draw a dotted line between them.   Shade BELOW  this line.
549653
Subtract 2x from each side\n-y = 0 - 2x\nMultiply each side by -1\ny = 2x
818056
3x - 5y = 7\n5y = 3x - 7\ny = (3/5)x -7/5\nA line parallel to it is y = (3/5)x + b\n(6 , -7) on it, so -7 = (3/5)*6 + b\n-35 = 18 + 5b\n5b = -53 and b = -53/5\nThe wished line is y = (3/5)x - 53/5 or\n5y = 3x -53\n\nTh
97021
Using the points you can work out the slope. That is 0.\nThen use one set of points and the slope to work out the equation of a line formula which is:   \n         y-y1=m(x-x1)\nm is the slope \n       y+4=0(x-11)\n       y+4=0\n       y=-4
645941
4x + y >= 4\ny >= -4x +4\n\n-------\ny = 0\n0 >= -4x + 4\nx = 1\n-------\nx = 0 \ny >= 0 + 4\ny = 4\n-------\n\n\nTherefore, shade the region above the line.\n\nTwo points on the line are:\nx-intercept: (0, 4)\ny-intercept: (1, 0)
920988
if the slope is undefined,the line will be parallel to the y axis and of the form x=k\ny-2=1/0(x+3)\n0(y-2)=x=3\nx+3=0\nx=-3
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
194847
Im not gonna give u the answer, coz u wont learn otherwise, but i can tell you how to solve it...\n\nWell, remember that the standard equation for a line is \n\ny = mx + c\n\nwhere m is the slope of the line... Now, there is a relation between the slopes of two lines if they are parallel, look it up yourself...\n\nSo u have the slope of the line whose equation u need to find... U also have a point thru which it passes... Another representation of the equation of a line is...\n\ny - y1 = m(x - x1)\n\nwhere x,y are the variables for the eqn, m is the slope of the line, and x1, y1 are the (x,y) co-ordinates of a point thru which the line passes...\n\nYouve got all the stuff, throw it in, and wolla...
375327
Where x = 0, the line intercepts the y-axis:\n\n1. 5x-2y =10 \n2y = 5x - 10\ny = (5/2)x - 5 \ny-int = (0, -5)\n\n2. 2y+7x =14\n2y = -7x + 14\ny = (-7/2)x + 7\ny-int = (0, 7)\n\n3. x=3\nHas no y-intercept\n\n4. 1/2x-2y=8\n2y = (1/2)x - 8\ny = (1/4)x - 4\ny-int = (0, -4)
951551
A line that runs parallel to the Y axis.
789536
If you are supposed to do it by graph, then it's easy.  Just place the points on the graph (scaled, of course) and see when x=0, namely, when the line crosses the y-axis.  That will be your answer.  If you are looking for the formula way, here you go:\n\nm=(4-0)/(10-(-10)) = 4/20 = 1/5\n\n0 = 1/5(-10) + b\nb= 2\n\ny=x/5+2\n\nYour y-intercept, or f(0), is 2
674651
put it in standard form :\n\ny < -1/2x + 3/2 \nand \ny > 2/3x + 2\n\nthen you know the first line passes thru the y axis at 3/2 and the second at 2\n\nthe slopes are -1/2 and 2/3 .... then shade below the line on the first one cuz y is less than .\n\nthen shade above the second line because y is greater than