Perpendicular distance of a point from linear regression line?

[Winny]

Hi all,

First post for me on here.


I have some Excel charts, with least squares fit lines through the
data. On each chart I have a single extra data point (it's own series)
and I want to know the perpendicular distance from the line to this
point.

This is a physical distance in the real world, as my axes are both in
milimeters but to different scales.

Thanks for any help.

--Winny

[Jon Peltier]

You know slope and intercept of fitted line, and coordinates of point,
so you can calculate dx and dy, the horizontal and vertical distance
from point to line. By similar triangles, the shortest distance h from
line to point is:

h = dx dy / SQRT[(dx)^2 + (dy)^2]

- Jon
-------
Jon Peltier, Microsoft Excel MVP
Peltier Technical Services
Tutorials and Custom Solutions
http://PeltierTech.com/
_______


Winny wrote:

> Hi all,
>
> First post for me on here.
>
>
> I have some Excel charts, with least squares fit lines through the
> data. On each chart I have a single extra data point (it's own series)
> and I want to know the perpendicular distance from the line to this
> point.
>
> This is a physical distance in the real world, as my axes are both in
> milimeters but to different scales.
>
> Thanks for any help.
>
> --Winny
>

[Winny]

Hi Jon,

Thanks very much.

I should have been able to work that out for myself really, but the
trig part of my brain has been unused for a couple of decades.

--Winny

[Tushar Mehta]

Hi Jon,

I cannot derive the formula you got but it cannot work for any vertical
or horizontal line.

For an approach from MathWorld see
http://mathworld.wolfram.com/Point-L...mensional.html

It gives the distance of (x0,y0) from ax+by+c=0 as
abs(a*x0+b*y0+c)/sqrt(a^2+b^2)

--
Regards,

Tushar Mehta
www.tushar-mehta.com
Excel, PowerPoint, and VBA add-ins, tutorials
Custom MS Office productivity solutions

In article <#[email protected]>,
[email protected] says...
> You know slope and intercept of fitted line, and coordinates of point,
> so you can calculate dx and dy, the horizontal and vertical distance
> from point to line. By similar triangles, the shortest distance h from
> line to point is:
>
> h = dx dy / SQRT[(dx)^2 + (dy)^2]
>
> - Jon
> -------
> Jon Peltier, Microsoft Excel MVP
> Peltier Technical Services
> Tutorials and Custom Solutions
> http://PeltierTech.com/
> _______
>
>
> Winny wrote:
>
> > Hi all,
> >
> > First post for me on here.
> >
> >
> > I have some Excel charts, with least squares fit lines through the
> > data. On each chart I have a single extra data point (it's own series)
> > and I want to know the perpendicular distance from the line to this
> > point.
> >
> > This is a physical distance in the real world, as my axes are both in
> > milimeters but to different scales.
> >
> > Thanks for any help.
> >
> > --Winny
> >
>

[Jon Peltier]

He he, I used a much less advanced technique than you did.

- Jon
-------
Jon Peltier, Microsoft Excel MVP
Peltier Technical Services
Tutorials and Custom Solutions
http://PeltierTech.com/
_______


Tushar Mehta wrote:

> Hi Jon,
>
> I cannot derive the formula you got but it cannot work for any vertical
> or horizontal line.
>
> For an approach from MathWorld see
> http://mathworld.wolfram.com/Point-L...mensional.html
>
> It gives the distance of (x0,y0) from ax+by+c=0 as
> abs(a*x0+b*y0+c)/sqrt(a^2+b^2)
>