Are the following lines parallel, perpendicular, or neither?
490013
Find their slopes: if slopes are equal, they're parallel; if slopes are opposite reciprocals (like 2 and -1/2) they're perpendicular; if neither then neither.\n\n#1: (2- -8)/9-3 = 10/6 =  5/4\n\n#2: (6 - 10)/(2 - 4) = -4/-2 = 2/1\n\nSo neither.
860061
It depends on if you a working in two dimensions or three and if you are working in Euclidean geometry or non-Euclidean geometry.\n\nIn Euclidean plane geometry in two dimensions if the two lines are perpendicular to the same line through two distinct points on that line, they will be parallel.\n\nIf you are in three dimensions  going through two distinct points, they could be parallel if lying in the same plane or not if not in the same plane.\n\nIn non-Euclidean elliptic geometry on the surface of a sphere they would always intersect like longitude lines as they are perpendicular to the equator or latitude lines at different points and all intersect at the north and south poles. \nor \nIn non-Euclidean hyperbolic geometry on the surface of a hyperboloid they would be parallel using an analogy to latitude and longitude lines.
572729
When two lines are not perpendicular, the types of angles that can be formed are:\n\n(1) Acute - one of the angles formed by the two lines is less than 90 degrees.\n(2) Obtuse - one of the angles formed by the two lines is more than 90 degrees.\n(3) Straight - the two lines are somewhat inclined in the same direction\n(4) Reflex - one of the angles formed by the two lines is either greater than 180 degrees, whilst the other is less than 180 degrees (which the exception that the two lines form a 270 degree, and a 90 degree angle). This applied to angles of a revolution.
798283
Y = 2 is a horizontal line that passes through 2 on the y axis.  So, any line that is perpendicular to that will be a vertical line.  Yes, you are right, UNDEFINED.
767763
What your proof will consist of depends on which statements you are allowed to use as axioms. For instance, Euclid, in his Elements, _defined_ perpendicular lines as those that created congruent, adjacent angles, and thus a proof of this statement in his system would be superflorous.\n\nSome axiomatic systems in high school give you the twin statements "the angle between perpendicular lines is 90°" and "a straight angle is 180°". In which case your proof would be:\n\nLet a and b be perpendicular lines, and P be their point of intersection. Let A be a point other than P on a, and B and C be on b such that P is between B and C. Then APB and APC are adjacent angles (as they share a common side, and niether one is within the other) which therefore sum to angle BPC. Since BPC is a straight angle, it is 180°. Further, since a and b are perpendicular, at least one of APB and APC are 90°, so the other must be 180°-90°, which is 90° also. Since both angles are 90°, they are congruent. Q.E.D.
105143
Draw a force diagram to help you figure this out.\n\nRemember, perpendicular vectors are independent.\n\nF=ma
941588
Identify the slope of each line and compare the values.\nIf the slopes are the same then they are parallel.\nIf the slopes give a value of -1 when multiplied together, then they are perpendicular.\nIf the slopes differ from that about, then they are neither.\n\nBring the y value to a unit value of one.\n(1) Slope 6∙4 and 6∙4, so it's parallel.\n(2) Slope ½ and ? , second equation is unclear.\nThe slopes are probably 3/2 and -3/2, if so, then they are perpendicular.\n(3) Slope -1/3 and 1/3, so the lines are perpendicular.
145572
Your question is quite vague, and so with your previous question. That is why I have to make an assumption. Please see my solution to your other question.\n\nFor this particular question, if:\n1. The line passes thru y = (1/2)x + 4 & (-3,5)...NO PARTICULAR EQUATION.\n2. The line is parallel to y = (1/2)x + 4 & passes thru \n(-3,5)...SEE SOLUTION A.\n3. The line is perpendicular to y = (1/2)x + 4 & passes \nthru (-3,5)...SEE SOLUTION B.\n\nSOLUTION A: PARALLEL LINE\ny = mx + b ==> equation in slope-intercept form.\nm = slope = 1/2 = same slope as the parallel line.\nb = y-intercept\n\nSolving for b, use the point (-3,5):\ny = mx + b\n5 = (1/2)(-3) + b\n5 = -3/2 + b\nb = 5 + 3/2\nb = 10/2 + 3/2\nb = 13/2\n\nTherefore, equation of line is:\n\ny = mx + b\ny = (1/2)x + 13/2  ANSWER\n\n-------------------------------\nSOLUTION B: PERPENDICULAR LINE\ny = mx + b\nm = slope = -2 = negative reciprocal of the \nperpendicular line\nb = y-intercept\n\nSolving for b, use the point (-3,5):\ny = mx + b\n5 = -2(-3) + b\n5 = 6 + b\nb = 5 - 6 \nb = -1\n\nTherefore, equation of the line is:\n\ny = mx + b\ny = -2x + (-1)\n\ny = -2x - 1  ANSWER
961745
the slope of the line perpendicular to that line would be the inverse negative of the slope of x-3y = 8\n\n(x-8)/3 = y\n\nthe slope = x/3, so the perpendicular line has a slope of -3x\n\ny = mx + b\n\nm = -3\n\nx = 5\n\ny = -4\n\n-4 = -15 + b\n\nb = 11 \n\ny = -3x + 11
342727
No, parallel lines never meet
1175454
y=-3x\nand\ny=3x\n\nare perpendicular, but only if you change the scale into 1/3 of the y-scale(meaning, you divide the y-scale by 3). Otherwise y = -3x and y = 3x are not perpendicular.\n\n^_^
73689
The difference is parallel lines are on the same plane while skew lines are not. You can see the pictures here: http://www.math.com/school/subject3/lessons/S3U1L3GL.html#sm2\nClick on each diagram to learn more about pairs of lines.
1171148
the portion  between the two vector that happened to be perpendicular is BSin(alpha)\nalpha is the angle between the two Vectors, Alpha  is   \n ArcTan(B/A)\nWhere B & A are the two vectors.
811970
ok so parallel lines mean they have the same slopes so u know \n-13=(y-15)/(4-3)\n  remember slope is rise/run or the change in y / change in x\n\nso -13 =y-15  y=-13+15 or   y=2
896230
the y coordinates are the same i.e. -1, \nso the line is parallel to the x axis and passes through -1 on the y axis.\nits slope is parallel to the x axis.
893759
Is a line perpendicular to it. The proof is easy.\nDraw a line not perpendicular to the straight line.\nCall this line L1.  Now draw a line perpendicular to the straight\nline. Call it L2. Note that you have a rt triangle with  sides made of L1 and a side from the intersection of L1 and the original line\nto the intersection of L2 and the original line and a hypotenuse which is L2. That makes L2 longer than L1 . So the shortest distance is necessarily perpendicular.
96323
2 lines are parallel if their slopes are equal. 2 lines are perpendicular if the product of their slopes is -1 (or if the slope of one line is the negative reciprocal of the other)\nFrom your given, the slope of the first equation is 3/5\nusing y = mx + b (m is the slope)\nThe slope of the second equation is -5/3. Notice that their product is -1, so the lines are perpendicular.