Question 1014674
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reduce -3x - 5y + 11 = 0 to normal form and give the distance of the line from the origin.
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Normal form equation of this straight line is

3x + 5y = 11.

The distance of the line from the origin (0,0) is 


d = {{{abs(3*0 + 5*0 -11)/sqrt(3^2 + 5^2)}}} = {{{11/sqrt(34)}}}.


For details, see the lesson <A HREF=http://www.algebra.com/algebra/homework/Vectors/The-distance-from-a-point-to-a-straight-line-in-a-coordinate-plane.lesson>The distance from a point to a straight line in a coordinate plane</A> in this site.
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