How do I construct a parallel line through a given point using a straight edge and compass?
156603
Draw a line.  Set the compass to the distance you want the point to be from the line so the circle passes through the point at the top of its arc.  Draw the circle.  Then draw the same size circle again at another point on the line and connect the tops of both circles.
832822
Before graphing, treat the inequality as an equal sign, and put the equation in a form you know how to graph.  The easiest is probably the slope intercept form, y = mx + b, where m is the slope and b is the y intercept.\nGraph the line that you see, but because it is an inequality, graph it as a dashed line. If the equation is greater than or equal to, or less than or equal to then graph it as a solid line.\nPick a point that is not on the line and plug it into your equation. If the line doesn't cross the origin, that would probably be a good point to choose.  Plug in the point you have chosen, and if the inequality holds true for that point, shade in the direction of the point from the line.
585436
For an arbitrary angle, no.  There is an interesting algebraic proof to show that it is not possible to trisect an angle with a straightedge (no markings) and compass alone through the construction of a field.  Now it is obviously possible for certain angles such as the right angle to be trisected.  However, the subset of angles which can be trisected has measure 0 in the set of all possible angles (countable in an uncountable set).\n\n(See source below for more.)
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.
974393
If you mean parallel to y=2x + 5, then the slope will be 2\n\nSo you need to use:\ny = 2x + b\n\nPlug in your point (-1, 4) and solve for b:\n4 = 2(-1) + b\n4 = -2 + b\nb = 6\n\nSo your line is:\ny = 2x + 6
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.
481919
It is impossible to trisect an arbitrary angle using only compass and straightedge (but bisection is possible). No there is no mathematical technique (i.e a theoretical construction technique to do so, even if impractical). It cannot be done exactly. The best that can be done is approximation. One can imagine, I suppose, a radial arm (our radius rotating about a fixed  point) whose rotation speed was controlled very precisely by a computer and the speed could be extremely slow and a laser could be used with a super accurate quartz timer to 'print' a dot at fixed intervals thus dividing an arbitrary angle as desired. Certainly, technologically feasible, but I haven't heard anything about it, and it would be a technological technique, not a mathematical "breakthrough".
684619
Draw a horizontal staight line to be the base of the trapezoid.\nMark a point to be a corner of the trapezoid. Label it A.\nSet your compass to the length of the base.\nMeasure that distance from the first mark and make a second mark on the line. Lable it B\n\nWithout changing the width of the compass, put the point at A and swing an arc above the line. Move the point of the compass to B and swing and arc. Lable the point where the arcs cross C.\n\nDraw the lines AC and BC. You now have an equilateral triangle. The angles are all 60 degress.\n\nSet your compass to the distance of the leg length.\nPut the point at A and swing an arc across AC. Label the intersection D.\nMove the point to B and swing an arc across BC. Label the intersection E.\nDraw the line DE.\nDone.
241573
You can't just pick a point on one plane and compare it to an arbitrary point on the other plane as the previous user suggested. You could get different answers depending on what points you pick.\n\nThe equation of a plane can be written as ax + by + cz + d = 0. The equations of two parallel planes will have the same a, b, and c but different d values (let's call it d1 for the first plane and d2 for the second). Pick an arbitrary point on the first plane (call it (x1,y1,z1)) and then compute the distance from that point to the second plane using the following formula:\n\n(a*x1 + b*y1 + c*z1 + d2)/sqrt(a^2 + b^2 + c^2)\n\nIn general you will not get the same answer if you pick an arbitrary point on the second plane and use the distance between those two points (unless you happen to pick a point on the second plane such that the line between the two points you picked is perpendicular to both of the planes).
739738
Statements\n1) PT tangent to circle O, PAB is a secant.\nReason: Given\n2)Draw line TB\nreason: two points determine a line\n3) angle mtp is congruent to angle mtb\nreason: an angle bisector(line MT) bisects an angle into two congruent angles\n4) draw tangent XY that meets at point T and is parallel to line segment PB\nreason: a tangent may be drawn to one point of a circle\n5) angle mtb is congruent to angle pmt \nreason: when a transversal crosses two parallel lines the alternate interior angle formed are congruent\n6) angle mtp is congruent to angle pmt\nreason: i forgot the name of this theorom look it up(something like if x=y and y=z then x=z)\n7) triangle PMT is isoceles \nreason= converse of the base angles theorm\n\ni hope i was able to help somewhat, i tried to draw a diagram to the best of my ability with the given information. by the way i know there was another way to do the proff with circle properties but i think it was much easier to draw a tangent parallel to PB saved a lot of steps also. good luck!
544028
You have to have a right triangle.\nThen you figure out which is side is opposite or adjacent from the angle you need.
570630
well you can make a pretend display on a real compass, IE when its pointing north make the paper display say whatever direction you want, compasses work by pointing north  http://science.howstuffworks.com/compass1.htm
661921
On your number line, go one third of the way between 0 and 1 and put your point on the 0 side of one third.  If you are using graph paper, make three lines equal one unit so that each line it one third unit apart.  This way will make one third be the first line past the number 0.
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.
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
882048
Here's the answer you want:  With a compass set to the radius of the circle, start marking off radii around the circle, dividing it into 6 equal parts.  Connect any 3 alternate points and you'll have your triangle.\n\nNO MATH REQUIRED!
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