When you have an undefined slope, how do you put that in to point-slope form?
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
568116
It has no y intercept cuz its a vertical line.\nIt's slope would be undefined cuz when you try to find the slope the normal way using two points, say (-7,4) and      (-7,8), the slope would be (8-4)/(-7-(-7))=4/0 which is undefined
819957
Start with a graph of:\nf(x) = x\n\nThat's a straight line going through the origin with a slope of 1.  It should look like a 45 degree angle if you draw it correctly.\n\nNow to transform this to:\nf(x) = | x |\n\nYou have to think of each value smaller than 0.  Rather than -1, for example, you'll have 1.  Rather than -2, you'll have 2, etc.  This is like "mirroring" the graph that is below the x-axis so that it is above the x-axis.\n\nThis results in a "V" shape.  A slope of -1 down to the origin, then a slope of 1 from the origin.\n\nFinally:\nf(x) = | x | - 2\n\nFor every value that results, you now subtract 2.  For example, x = 0, gives -2, rather than 0.  This is true for every point x, making every point y smaller by 2.  It would look like you shifted the whole graph down 2 units.\n\nSo it would be "V" like before, but it would go down at a slope of -1, cross the x-axis at -2, and continue to (0, -2).  Then it would slope up (m=1) through the x-axis at 2 and continue upward.\n\nYou can go to the first link and type 'abs(x) - 2' in the box.  The resulting graph is shown as the 2nd link.
771254
First, you use the point slope formula:\ny-y1=m(x-x1)\n\ny-2=3(x-0)\ny-2=3x\ny=3x+2\n\nStandard form:\nAx+By=C\n\nSubtract 3x on both sides:\n-3x+y=2\nThat's your standard form!
986326
when x is 0, find y. plot the point (0, <<whatever you got>>).\nwhen y is 0, find x. plot the point (<<whatever you got>>, 0).\nConnect those two points, and extend the line.
380192
you have a few right answers here slope = -1/4
835994
The slope of any line is dy/dx. The x axis has no dy; i.e., dy=0 so dy/dx=0. The y axis has no dx; dx=0 so dy/dx = undefined.
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
729746
you solved it right, since the line BC never goes up/down(the y-coordinate never changed, graph it and you'll see) the slope is 0. and 0/7=0, so you're right.\nit would be impossible if you got 7/0, because you can't divide by 0.
921452
Sounds like you need homework help.  Why don't you just post the whole problem?\n\nWhen working with slope, remember to put it in fraction form if it isn't already.  For example a slope of 5 would be the same as a slope of 5/1.  Now, learn the phrase "up and right" because that is how you read slope.  If your slope is 5/1, then from any point on the line, you move up 5 units and right 1 unit to draw the line.  \n\nIf your slope is -3/2 you would move up -3 units (which is the same as down 3) and right 2 units.   If your slope was 9/-7 you would move up 9 units and right -7 units (which is the same as left 7).  Get it?\n\nSo now, back to your question...  One way to write an equation for a line is y=mx+b.  You gave us B, so we know it is y=mx+1/3.  But you didn't give enough information.  \n\nIf that was a mistake and m is supposed to equal 1, then the fraction would be 1/1, and the line would move up 1, right1.  That line would look like it goes up and to the right, in the center at a diagonal straight through the upper right quadrant.\n\nHope that helps
701258
Distance on the vertical (y) axis, Time on the horizontal (x) axis.\nIf your sensor is at distance = 0, and the slopes go from left to right\n1) constant shallow slope down towards zero.\n2) constant steep slope up away from zero\n3) horizontal line\n4) shallow slope down at first, becoming steeper and steeper down, curving towards zero
885688
slope m=(y1-y2)/(x1-x2)\n          m=(1--1)/(3-3)\n          m=2/0    which gives us a singularity,\nthat is, a vertical line.
779591
well, you have to set up two questions in point slope form. (ie. equation for the line)\nso for your first question:\n\ny-9 = 2(x-5)\ny-3 = 2(x-r)\n\nso simpliy to be\n\ny = 2x - 10 + 9\ny = 2x - 2r + 3\n\nand simplify again to be\n\ny = 2x - 1\ny = 2x - 2r + 3\n\nand from here.. i'm drawing a blank. but that should help.. maybe?
743347
It all depends on what result you're hoping to get.  For example, if you want to solve a system of equations in two variables, you would try to get the variables on the same side in both equations.  If you want to express the equation in slope intercept form, you would move the y variables to the left side and the x variables and constants to the right side of the equation.  If you're trying to express the equation in standard form, i.e. ax + by + c = 0, you want both variables on the same side.
407712
slope = -0.25\nintercept = 1
151606
if all the lines are equal in every way (duplicate) then you can put a measurment (dimension) from the existing line to a datum or border whichever you choose and then put a callout like 3@12.00' (3 lines at 12.00 feet)in the dimensionwith a arrow pointing at the line.\n\njeff
1090635
1.  memorize the formulas\n\n- standard form\n- point-slope form\n- slope intercept form\n\nThese are 3 ways to manipulate the same equation.\n\nMemorization is the easiest way.  If you have algebra basics down you can manipulate the equations from one form to the next.   You won't need an explanation.