What does normalizing a straight line mean?

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Good day,
From the first attachment...

Why is "D" divided by "D1"? The book is saying that it is for normalizing "D" by the value of "D1".
It's not quite clear !

Equation: d[D/D1]/dM∞ = (D2 − D1)/D1 ÷ (M2 − M1) = (4,250 − 1,750)/1,750 ÷ (1.0 − 0.9) = 14.3


The whole formula is to find the slope of this graph.

Graph of drag D in lb times 10 to the minus 3 against M infinity; curve dips to a minimum near M 0.5, then rises steeply past M 1, with dashed tangent lines at points 1 and 2.
 
Attachments
  • Equation: d[D/D1]/dM∞ = (D2 − D1)/D1 ÷ (M2 − M1) = (4,250 − 1,750)/1,750 ÷ (1.0 − 0.9) = 14.3
    Screenshot_20180312-185202.jpg
    7.2 KB · Views: 690
  • Graph of drag D in lb times 10 to the minus 3 against M infinity; curve dips to a minimum near M 0.5, then rises steeply past M 1, with dashed tangent lines at points 1 and 2.
    Screenshot_20180312-185151.jpg
    11.5 KB · Views: 783

Discussion

It is simply some scaling applied, e.g. to compare the observed drag (and its change) to some previously considered value D1.
 
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