name the geometry of given a line and a point not on the line....only one line can be drawn through the point?
476914
In plain geometry :\n*Only one line can be drawn through that point perpendicular to the given line\n*Only one line can be drawn through that poin parallel to the given line\nIn space geometry :\nOnly one line can be drawn through that poin parallel to the given line
787261
Let the 2 points be (x1,y1)  and  (x2,y2)\n\neqn of line s given by\n\n(y-y1)/(y2-y1) = (x-x1)/(x2-x1)\n\nwhere  (y2-y1)/(x2-x1) is the slope of line
278253
5 lines?? must they be straight lines or can they be curved or another circle inside??  If you put the four lines thru like 2 x's then draw a small circle it is 16 pieces.
510419
Mathematics can be defined as the logically rigorous study of topics such as quantity, structure, space, and change. Another view, held by many mathematicians, is that mathematics is the body of knowledge justified by deductive reasoning, starting from axioms and definitions.\n\nA spatial point is an entity with a location in space but no extent (volume, area or length). In geometry, a point therefore captures the notion of location; no further information is captured.\n\nA line, or straight line, can be described as an (infinitely) thin, (infinitely) long, perfectly straight curve (the term curve in mathematics includes "straight curves"). In Euclidean geometry, exactly one line can be found that passes through any two points. The line provides the shortest connection between the points.\n\nThere u are! :)
666292
Take two points A and B. There are uncountably many lines through A and only countably many of these have points of X. Similarly for lines through B. Now one line out of each category and make sure the two are not parallel. The path that goes from A to the intersection point, then to B is a path in R^2/X.
1113346
Get a compass (not the navigation kind...the adjustable point-and-pencil kind), and set it at some length greater than 1/2 the segment's length.  Draw an arc through the segment, above and below the segment, from both endpoints of the segment.  There are two points of intersection for these arcs that you just drew: one above and one below the original segment.  When you connect these intersection points, you have bisected the segment with your new line, which is perpendicular to the original.
151877
Theorytical if the line are formed on a flat gemetrical plane they will not intersect.Or if two tangent in a in the same circle are parallel they will never intersect.\nThe fith postulate of Euclid tackled this question.Later mathematicians, like gauss proved that they could intersect. But this proof was based on a different idea=paralell line having curvatures! Of course they would intersect. So the geometry which based on curvatures was started and was called Non Euclid Geometry.\nActually this could have been proven alot fast with some common sense! In reality it is imposible to construct a perfect circle or a perfect straight line(this is my own postulate).   This is because we are not able to have a perfect resolution of measurement.\nNot matter how you try to measure two parallel line  the results is ether convergence or divergence of the two lines if you make the line long enough!So in practice the 5th postulate of Euclid did have a practical application.
927347
Draw lines from the top of the vertical lines in the H to the opposite lower corners. There you go.\n\nIn response to the comment below:\nBTW\nI guess the middle line of the H would have to be not directly centered between the lines.  It has to be a little highter or lower so the 2 lines do not intersect on the center line.  Then you come out with 5 individual triangles and not triangles within triangles.
829001
find the equation of the line that passes through the point of intersection of the lines x-3y=4 and 3x+y=2 and is perpendicular to the line 3x+4y=0. Give your final answer in the form ax+by+c=0 where a,b and c are integers to be determined.\n \nfirst we find the point of intersection\n\nx-3y=4,3x+y=2\n\n>>>>x=1,y= -1\n\ntherefore, point of intersection of two lines\nis(1,-1)\n\n3x+4y=0\ny=(-3/4)x\n\nthe gradient of line perpendicular to this line\n= -1/gradient\n\nperpendicular line passes through (1,-1) and \nhas a gradient of 4/3\n\nequation of perpendicular is\n\ny=4/3x+c  say\n\nsubstitute values at point(1,-1)\n\n-1= 4/3+c\n\n>>>>>c= -7/3\n\nequation of line is y=4/3x-7/3\n\n>>>>>3y-4x+7=0 \n\ntherefore, the equation of the line that passes through the point of intersection of the lines x-3y=4 and 3x+y=2 and is perpendicular \nto the line 3x+4y=0 is\n\n3y-4x+7=0\n\nwhen doing these problems,it helps to draw \na rough graph of the situation\n\nanother answer could be\n\n4x-3y-7=0 or 300y-400x+700=0\n\ni hope that this helps
245713
More completely, given two lines ax1 + bx2 = c and dx1 + ex2 = f, they can either intersect completely, in which case the solution set is a line, or they can intersect in one point, or they can not intersect at all (parallel lines). This is due to Euclidean geometry.\nIf you're more of an algebraist, suppose they intersected at more than one point (x1, x2) and (y1, y2). Then ax1 + bx2 = ay1 + by2 and dx1 + ex2 = dy1 + ey2. But then (a-d)x1 + (b-e)x2 = (a-d)y1 + (b-e)y2, so (d-a)/(b-e) = (y2-x2)/(y1-x1). Suppose we chose another point (z1, z2) on the line formed by (y1, y2) and (x1, x2). But note that (z1, z2) will also satisfy the equations, hence the entire line satisfies the equations whenever more than one point satisfies the equation. This is just the first case given above (the lines completely overlap). \nNote the degenerate condition (x1,x2)=(y1,y2) of a single solution gives an undefined quotient 0/0.
48873
well i would like to pretend its a regular hexagon, in this case you would have to draw a line from one vertex to an opposite one, the line should meet at the center of gravity (the point where all 3 lines intersect)
921939
Good Evening\n\nI think we've been avoiding this question because we can't graph here... however to get you started, create an x|y chart for each line, plug in a couple of values, and connect the dots  (for a line you only need two points, but some books recommend finding three, so you can verify you didn't make a mistake somewhere)\n\nx+y = 8\nx|y\n----\n4|4\n3|5\n2|6\n\nx - y = -4\nx|y\n0|4\n1|5\n2|6\n\nGraph the two lines, and they should intersect at (2,6) since it appears in both of the charts...\n\n~i~
818092
When Graphed\n\nEquations can be any line\nLinear Equations form only straight lines\nFunction is only a function if you don't have 2 identical x values.
639268
a line with no slope..a straight line ----------
506764
well, id need a pic to explain it...but, you can draw two lines from the midpoint of one of the sides, and then you have 4 triangles...the big one, and three little ones inside...
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.
188576
I think I get it.  You're given the two circles (which I assume do not intersect and one is not inside the other) and then you have to come up with the equations for two tangential lines to both of these circles.  \n\nOnce you have the two lines this becomes a linear inequality system.  What you're asking is if the point is in the area bounded by one line on top and the other line on the bottom, so write these two lines as two inequalities.   Example: \n\ny1 < mx1 + b\n\ny2 > mx2 + b\n\nI think this is the format, it's been a while since I did stuff like this.  You can graph it by shading the area between the lines.\n\nAnyway, once you have this it's pretty easy: if a point satisfies both inequalities then it's inside the area.  If not, then it's outside.\n\nOkay, how to draw the tangent lines?  Well, this is a little trick to explain, it would be easier if I could graph it.  But the thing to remember is that by definition a tangent line is perpendicular to the circle at that point.  Thus if you draw a line connecting the centers of the two circles, and then draw one tangent line to the circles, then draw a line connecting the center of each circle to the tangent point on each circle, you will see that you have two similar right triangles that share the same angle.\n\nAt this point I don't think you need to use trignometry to find the equations of the tangent line ... but my mind is slipping a bit at this point and I need a moment to think it out.\n\nupdate:  I found the solution to finding the tangent to 2 circles knowing only the center points and the radii, but I'd have to draw you a picture to show you how.  It's complicated, as it involves a bunch of similar right triangles.