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Question

to find the distance between parallel lines with equations x + 3y = 6 and x + 3y = -14, you first have to find a point on one line, find the equation through that point perpendicular to the first line and then find the point where that perpendicular intersects the second line. The equation of a line perpendicular to x + 3y = 6 is 3x – y = 0. Find the point where 3x – y = 0 intersects x + 3y = -14.

Answer 1

x+ 3y = 6

x+3y = -14

vertical distnce = (6--14)/3 = 20/3 = 6 2/3= 6.666...

but you want the perpendical distance which should be a little less than 6 2/3

slope of the two parallel lines is = -1/3

construct a right triangle, vertical side = 20 = hypotenuse of a right triangle

horizontal side = base = (20/3)cos(tan^-1(3)) = about 2.108

vertical side = altittude = (20/3)sin(tan^-1(3)) = about 6.32455532

distance between the two lines = about 6.3

Answer 2

problem seems incorrectly written, so far three answers all same for distance between parallel lines described, However problem wants intersection of x+3y=-14 line and perpendicular line through point (0,0). certainly (0,0) is not on line x+3y=6. Perhaps problem also wants intersection of perpendicular line through origin and x+3y=6 line ----and then to use distance formula with the two points found. Result=2 times square root 10.