I need a slope triangle picking two sets of coordinates that are along the line but not the data and using the formula slope= Rise/Run

Rise is The change in Y ( height)

Run is The Change in X ( Width)

when doing this formula you count the spaces of how high or the height or the rise of the triangle & also the spaces are counted of how wide the triangle is when connecting the points making this invisible triangle to find the slope

I included the graph of the line segment of the data I also am including a picture of choosing two points and directions Of a SLOPE TRIANGLE ALONG THIS LINE BUT NOT THESE COORDINATES THAT ARE PLOTTED WHICH IS THE LINE

PLEASE HELP AND USE CHANGE IN Y THE RISE/CHANGE IN X THE RUN for the solution of the slope

slope=Rise/Run

please label RISE and please label RUN

ALSO PLEASE STATE THE 2 ADDITONALLY COORDINATES PICKED ALONG THE LINE THATS NOT THE DATA

I included a graph of the coordinates also included a picture for an example of the imaginary rise and run slope triangle

since the line is in quadrant one all the coronet pit need to be positive and in quadrant one that are in between the points of the line that was plotted from the DATA Do you not use the data for the slope and slope triangle that's the triangle rise and run will be used to find the slope

please help and please help with step-by-step

\fyaxis 50 ( (12, 50 ) 45 I ( 10 , 4 30 5 ) F ( 10 , 42 ) 35+ H ( 6 , 31 ) 307 E ( 8 , 34 ) 25 D ( 6 , 26 ) 20 . C ( 4, 18 ) 15 A (0, 12 ) 10# . R ( 2 , 10 ) H 15 -12- 10-8- 6 - 4 - 2 2 6 8 10 12 0 - 4 ) - 5 naxis -10 - 20 From the graph the Line - 25 Also passes through - 201 H ( 6 , 31 ) 8 I ( 10 , 9 3 05 ) 35 Let ( 6 , 31 ) = ( 8 1 , y , ) -yo 8 ( 10 , 4 3 . 5 ) = ( 2 2 , 42 ) 50 Slope = 12- 41 = 4305 - 31 2 2- 24 1 10 - 6 = 12.5 = 30125 4