45 degree line in a graph

[feaner]

How do i put a 45 degree line in a graph? Then how would I go about finding the intersection of that line and a trendline? Any help would be greatly appreciated

 

[BAlGaInTl]

I assume that by 45 degree line, you mean a line with a slope of 1 that passes through 0, 0?

If so, you can make another data series and plot that.

1, 1
2, 2
3, 3

That will give you a picture of that line.

The other part of the problem is not as direct. Basically, you will have to set equation of one line, equal to the other to find the X intersect.

So, if you have the following

f(x) = x (this is your reference line)

and

g(x) = mx + b

Then your x intersect would be

x = b/(1-m)

Translated to an Excel Formula, that would be something like:

=INTERCEPT(F19:F23,E19:E23)/(1-SLOPE(F19:F23,E19:E23))

Where Column F has your Y values, and column E has your X values.

So to get your Y value for the intersect, you would use something like

=SLOPE(F19:F23,E19:E23)*INTERCEPT(F19:F23,E19:E23)/(1-SLOPE(F19:F23,E19:E23))+INTERCEPT(F19:F23,E19:E23)

.....

I think....

 

[BAlGaInTl]

If you get an error... that means that the lines don't intersect.

 

[BAlGaInTl]

I actually just realized it is a "bit" simpler.

Since your first function is f(x) = x, the x point and y point will always be equal.

So you really only need to solve for the x intersect. No reason to plug it in and get y. However... my solution would be useful if you had two different lines of any given slope.

 

[feaner]

Thank you very much for the help!

 

[javad]

I used another method, copied the x data again then used trend line. The created points can be colored with no fill. You can simply adjust the trend line to fit the plot.