ANSWERS: 4
  • As far as I can tell, the best way would be to graph it, find two points (one on each line) that line up vertically on the y axis, and then count out the horizontal distance between them on the x axis.
  • from: http://www.answerbag.com/q_view/606046 vector calc based solution: first we define the lines L1: p1 + t*v1 L2: p2 + u*v1 where pi is a point, and v1 is a vector and t & u are time notice I only use v1, NOT v1 & v2. That's because it's the same vector (parallel lines) not this problem turns into finding the distance between p1 and L2 the smallest distance would be the length of a perpendicular from p1 to L2, intersecting at a point, g draw a (right) triangle with (p2,p1) being the hypotenuse, and (p1,g), (p2,g) beings legs we know (by pythagorous) (p2,p1)#(p2,p1) = (p1,g)#(p1,g) + (p2,g)#(p2,g), where '#' is being used as the dot operator (p2,p1) is known, since we defined the lines with that also, (p2,g) is the 'projection' of (p2,p1) onto v1- so to find it's distance take (p2,p1)#v@ where v@ is the 'unit vector' of v1- or more simply: v@ = v1/||v1|| now you have enough to solve for (p1,g)#(p1,g)!
  • "High school oriented solution" a point on the first line: (0,4) draw a line *perpendicular* to both lines y = -(1/2)x + 4 (0,4) is also on that line that line also intersects the second line: 2x+1 = -(1/2)x + 4 2x = -(1/2)x + 3 2.5 x = 3 x= 1.2 and y = -(1/2)*1.2+4 = 3.4 so, starting from (0,4) and going perpendicularly towards line two, we would hit (1.2, 3.4); why is it important to go perpendicular? so now, we need the distance between (0,4) and (1.2, 3.4) distance formula sqrt(1.2 ^2 + (-.6)^2) = 1.34164079 and THAT is the distance between the two lines!
  • Try a symultaneous equation and then solve it

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