RE: back to the well again... parallel trend lines

[john hamon <jhamon@xxxxxxxxxxx>]

gary,

thanks for the response.

problem is... how do you specify the angle of the trendline?  i see no way
documented in my "exhaustive" easy language reference.

thanks for the prompt reply.

jh

-----Original Message-----
From: Gary Fritz [mailto:fritz@xxxxxxxx]
Sent: Thursday, January 04, 2001 9:57 AM
To: jhamon@xxxxxxxxxxx
Cc: omega-list@xxxxxxxxxx
Subject: Re: back to the well again... parallel trend lines


> please note that i don't want to connect two known points.  i want
> to draw a parallel line through a known trend line or pair of
> points.

If you know two points, it should be pretty easy.  If your points are
X1,Y1 and X2,Y2, then why not draw a line from X1+Z,Y1 to X2+Z,Y2?

That draws the trendline Z units vertically above the existing line.
If you want the parallel line to be a specific distance away from the
trendline, say H (for Hypotenuse :-), then you just need to do a
little trig.

The angle of the trendline is
  theta = arctangent((Y2-Y1)/(X2-X1)).

Theta is also the angle between a vertical line and a line
perpendicular to your trendline.  So

  Z = H * cosine(theta)

This is off the top of my head but I think it's right...

(View this in a fixed-pitch font like Courier:)

                               . X2+Z,Y2
                            .
                         .
                      .
                   .
                .
             .  |              . X2,Y2
          . \   |           .
       .     \  | Z      .
X1+Z,Y1     H \ |     .
               \|  .
                .
             .
          .     theta
X1,Y1  .   ----------------

Gary