if u look at two linear equations (only equations) how do u tell that the corresponding lines are parallel?
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Answer:\nIf the slopes are the same but the y-intercepts are different then the lines are different. (Also, if the y-intercepts are the same but the slopes are different then the lines are different).\n\nExplanation:\nIf you have the two equations in the slope-intercept form (y=mx+b):\n\nLine 1: y'=m'x'+b'\nLine 2: y"=m"x"+b"\n\nThen,\na) If m'=m", then the lines are parallel.\nb) If b'=b", then the lines intercept the y-axis at the same point.\nc) If m'=m" AND b'=b" then the lines are parallel AND they intercept the y-axis at the same point. In this case, if you look at the two equations, you will see that the two equations are identical - this means that the two lines are identical. Line1 = Line2.\n\nHere's an illustration of the concept:\nhttp://i36.photobucket.com/albums/e49/UKild/SameAndDifferentSlopes.jpg
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Any line parallel to x - 2y = 5 will have the same slope as x - 2y = 5. Therefore you must first find the slope of x - 2y = 5. To easily find the slope, you re-write the equation in the form y = mx + c where m is the slope.\n\nx - 2y = 5\n- 2y = -x + 5\ny = 1/2 x - 5/2\n\nOR \n\ny = 0.5x - 2.5\nThe slope is therefore 0.5\n\nGiven the point (2, -4) and slope 0.5\nthe equation is found as follows:\n\ny - (-4) = 0.5(x - 2)\ny + 4 = 0.5x - 1\ny = 0.5x - 1 - 4\ny = 0.5x - 5\n\nYou can re-write that equation in any form you like.
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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
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I'll get you started by answering the first question.\n\nSay that you have two parallel planes in the form\n\nax+by+cz = d1\nax+by+cz = d2\n\nThe vector normal to these planes is N = (a,b,c) i.e., if you draw pictures of these planes then the vector N defines a line that goes perpendicularly through them both.  This line intersects the planes at the points\n\nP1 = d1/(a^2 + b^2 + c^2)*N\nP2 = d2/(a^2 + b^2 + c^2)*N\n\nThe distance between the two planes is simply the distance between these points:\n\ndist = | P1-P2 | = |d1-d2|/sqrt{a^2 + b^2 + c^2}\n\nFor your equations, I'm going to multiply the second equation by 3:\n\n15x+6y-6z=7\n15x+6y-6z=18\n\nThe distance between these planes is\n\ndist = |18-7|/sqrt{15^2 + 6^2 + 6^2} = √33/9
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If it has no solutions, then the two lines are parallel but different.  If it has infinitely many solutions, the two lines are actually the same line (think of them as one right on top of another).\n\nWhen you try to solve them algebraically, in both cases the x and y (or whatever two variables are used ) will both cancel out.  If there is no solution, you will end up with two unequal numbers, such as 0 = 3.  If there are infinitely many, you will end up with two equal numbers, such as 6 = 6.
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alternative angles you mean? It's kinda like the corners of the letter Z. The angles are equal. The bottom and top line is parallel. And both angles are at the opposite sides of the line.
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The answer to the question depends on exactly what kind of geometry you are dealing with and what "points" and "lines" mean. \nIf you are talking about ordinary lines and ordinary geometry, then parallel lines do not meet. For example, the line x=1 and the line x=2 do not meet at any point, since the x coordinate of a point cannot be both 1 and 2 at the same time. \n\nIn this context, there is no such thing as "infinity" and parallel lines do not meet. \n\nHowever, you can construct other forms of geometry, so-called non-Euclidean geometries. For example, you can take the usual points of the plane and attach to them an additional point called "infinity" and consider all lines to also include this additional point. In this context, there is a single "infinity" location where all lines meet. In a geometry like this, all lines intersect at infinity, in addition to any finite point where they might happen to meet. \n\nOr, you could attach not just one additional point, but a whole collection of additional points, one for each direction. Then you can consider two parallel lines to meet at the extra point corresponding to their common direction, whereas two non-parellel lines do not intersect at infinity but intersect only at the usual finite intersection point. This is called projective geometry
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They are the values that make the equation true.  If the equation is true then the equation represents a straight line.
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because the solution of the equation is where the two lines of the system intersect, or cross. parallel lines never meet, so they never intersect, so there is no solution.
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No. The diagonals of a parallelogram bisect each other, but they don't bisect the angles.
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Both equations are in slope-intercept form (y = mx + b).\n\nIn your case, both lines have the same slope (-2/3) but different y-intercepts (+1 for the first, -4 for the second).  Lines with the same slope but different y-intercepts are PARALLEL.\n\n(Note: if you got slopes that were the same and the intercepts were the same, the lines would be coincident -- the same line.  If the slopes were negative inverses, e.g. 2/3 and -3/2, then they would be perpendicular.  And if the slopes were completely different the lines would be skew.)
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To the asker: is -2x+5=0 the correct equation? Did you mean for -2x+5=y? If you did mean -2x+5=0, then x=2.5.\n\nA line parallel to x = 2.5 through (8,7) would just be x=8
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Since you already know what vectors are, take two vectors from this plane as a basis for all vectors in the plane. The easiest basis is just to take two sides from the triangle formed by the 3 given points. \nFrom your study of linear algebra, you may remember that one way of checking whether a set of n vectors was linearly independent was to take the n-volume of their parallelotope (their determinant). If the n-volume was 0, then obviously the set of vectors makes less than an n-dimensional parallelotope, so they can't span an n-dimensional space. We also define a vector v to be orthogonal to a vector w if their inner product is 0.\nThus, we are looking for the vector w = (w1, w2, w3) such that the determinant (u, v, w) is non-zero and u*w=0 and v*w=0. If we expand the determinant above along the unknown vector w, choosing the n-volume to be the same as the area of the parallelegram spanned by u and v and expand the dot products, we find we have system of 3 linear equations in 3 variables. The solution vector of this system is called cross product of u and v, the formula for which you can find in the source below. It is a vector perpendicular to u and v that has as magnitude the area of the parallelegram spanned by u and v.
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Properties of angles associated with parallel lines\nIf a transversal meets two parallel lines, then:\ncorresponding angles are equal.\nalternate angles are equal.\nco-interior angles are supplementary.
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I need to do a drawing to answer this. Anyway, here are some usefull sites:\n\nhttp://www.bymath.com/studyguide/geo/sec/geo14a.htm\n\nThe parallel postulate. If parallel lines are cut by a transversal, then corresponding angles are congruent.
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to find if a line is paralel to each other, you had to find out it's gradient or slope. If the gradient is the same, then the lines are paralel.\n\nTo find gradient is rather easy.\n\nYou just plug in the points into (y2 - y1)/(x2 - x1)\n\nSo the first line is:\n(3 - - 7) / (1 - - 4)\n= (3 + 7) / (1 + 4)\n= 10/5\n= 2\n\nand the second line is:\n(10 - 6) / (4 - 2)\n= 4 / 2\n= 2\n\nyou see here that 2 = 2, so the two lines are paralel
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In addition to what “Spoonish1” told you I suggest you to write a program calculating minimal distance between 2 straight lines in 3D. Be this distance =0, then and only then you may state that 2 lines intersect. A classic line equation in 3D looks like:\n(y-y1)/(y2-y1)=(x-x1)/(x2-x1)=(z-z1)/(z2-z1) – just as a matter of fact. Has your desire faded? Or more help?