Question 684155
Find the equation of the line parallel to the line 2x+3y=6 and passing at a distance {{{5sqrt(13)/13}}} from the point (-1,1)? Please I need your solution and graph! thank you so much...
<pre>

Here is the graph of the line 2x + 3y = 6 and the point (-1,1):

{{{drawing(400,400,-5,5,-5,5, graph(400,400,-5,5,-5,5),

circle(-1,1,.1), line(39,-24,-9,8) )}}}

2x + 3y = 6

is equivalent to

2x + 3y - 6 = 0.

Every line parallel to the line 

2x + 3y - 6 = 0 

is of the form

2x + 3y + K = 0.

The distance from the point (p,q) to the line 
Ax + By + C = 0 is given by the formula:

{{{d}}}{{{""=""}}}{{{"" +- ((Ap+Bq+C)/sqrt(A^2+B^2))}}}

So since the distance from (-1,1) to the line 2x + 3y + K = 0

is to be {{{5sqrt(13)/13}}}, we have

{{{5sqrt(13)/13}}}{{{""=""}}}{{{"" +- ((2(-1)+3(1)+K)/sqrt(2^2+3^2)))}}}

{{{5sqrt(13)/13}}}{{{""=""}}}{{{"" +- ((-2+3+K)/sqrt(4+9))}}}

{{{5sqrt(13)/13}}}{{{""=""}}}{{{"" +- ((1+K)/sqrt(13))}}}

Cross multiply and solve for K, 

using the + we get K = 4 and using th - we get -6 

For K = 4 the solution is

2x + 3y + 4 = 0

which has this graph (in green):

{{{drawing(400,400,-5,5,-5,5, graph(400,400,-5,5,-5,5),
green(line(-11,6,13,-10)),
circle(-1,1,.1), line(39,-24,-9,8) )}}}

using K = -6. we have the solution

2x + 3y + (-6) = 0

2x + 3y - 6 = 0

Wow! That's the GIVEN LINE!!!  As it turns out, the given line 
was itself {{{5sqrt(13)/13}}} units from the point (-1,1).  
However we wanted a line PARALLEL to 2x + 3y - 6 = 0, not that 
line itself.  So the only solution is

2x + 3y + 4 = 0

Edwin</pre>