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Anybody who can help me to make a linear fit if I have a specific number of intersection point (x0,y0), so instead of using this y=mx I have to use this y-y0=m(x-x0),x0 = 0.3459; y0 = 0.4478 so I need to get a straight line from fit function that can cross the point(x0,y0). Please look at my code below and let me know how I can modify fit function to work with the above equation .

     range = {{0.3389`, 0.44079999999999997`}, {0.3389`, 
     0.4415`}, {0.3389`, 0.4422`}, {0.3389`, 
     0.44289999999999996`}, {0.3396`, 0.4436`}, {0.34099999999999997`, 
     0.4443`}, {0.3417`, 0.44499999999999995`}, {0.34309999999999996`, 
    0.4457`}, {0.3438`, 0.44639999999999996`}, {0.3452`, 
    0.4471`}, {0.3459`, 0.4478`}, {0.34659999999999996`, 
    0.44849999999999995`}, {0.348`, 0.4492`}, {0.3487`, 
    0.44989999999999997`}, {0.35009999999999997`, 0.4506`}, {0.3508`, 
    0.4513`}, {0.35219999999999996`, 0.45199999999999996`}, {0.3529`, 
    0.4527`}, {0.3529`, 0.45339999999999997`}, {0.3529`, 
    0.45409999999999995`}, {0.3529`, 0.4548`}};
  p = ListLinePlot[range, BaseStyle -> Red, 
  PlotRange -> {{0.342, 0.349}, {0.435, 0.455}}]
   x = range[[All, 1]]
  y = range[[All, 2]];
  drop = Drop[Drop[range, 4], -4]
  ListPlot[drop]
  x0 = 0.3459;
 y0 = 0.4478;
 yy = Fit[drop, {0, xx}, xx]
  pnew = Plot[yy, {xx, .335, .35}, PlotStyle -> {Black}]

Thanks in advance!

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    $\begingroup$ Please take another look at your code and your question. Should drop = Drop[Drop[data, 4], -4] be drop = Drop[Drop[range, 4], -4] ? I see no zeros but I see "duplicate" x-values at the beginning and the end of the list of data. $\endgroup$
    – JimB
    Mar 12 '18 at 1:29
  • $\begingroup$ Here is short cut for your function {x, y} = Transpose@range and drop = Take[range, {5, -5}] $\endgroup$
    – OkkesDulgerci
    Mar 12 '18 at 13:06
  • $\begingroup$ Thanks for your response. I corrected my question, so from my code I got a line that is not a straight line, but it is wavy and it doesn't cross the point (x0,y0). How I can modify the fit function to get a straight line that crosses the point (xo,y0) $\endgroup$
    – Ghady
    Mar 12 '18 at 13:12
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$\begingroup$ 
 ClearAll["Global`*"]

 range = {{0.3389`, 0.44079999999999997`}, {0.3389`, 
    0.4415`}, {0.3389`, 0.4422`}, {0.3389`, 
    0.44289999999999996`}, {0.3396`, 0.4436`}, {0.34099999999999997`, 
    0.4443`}, {0.3417`, 0.44499999999999995`}, {0.34309999999999996`, 
    0.4457`}, {0.3438`, 0.44639999999999996`}, {0.3452`, 
    0.4471`}, {0.3459`, 0.4478`}, {0.34659999999999996`, 
    0.44849999999999995`}, {0.348`, 0.4492`}, {0.3487`, 
    0.44989999999999997`}, {0.35009999999999997`, 0.4506`}, {0.3508`, 
    0.4513`}, {0.35219999999999996`, 0.45199999999999996`}, {0.3529`, 
    0.4527`}, {0.3529`, 0.45339999999999997`}, {0.3529`, 
    0.45409999999999995`}, {0.3529`, 0.4548`}};
{x, y} = Transpose@range;
drop = Take[range, {5, -5}];
x0 = 0.3459;
y0 = 0.4478;

 model[t_] := m (t - x0) + y0

 fit = FindFit[drop, model[t], m, t]

{m -> 0.6875}

pnew = 
 Plot[model[t] /. fit, {t, .335, .35}, PlotStyle -> {Black}]
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  • $\begingroup$ Thank you so much!! Do you have an idea how to use that for quadratic fit? $\endgroup$
    – Ghady
    Mar 12 '18 at 13:57
  • $\begingroup$ Fit[drop, {1, t,t^2}, t] I assume all coefficient are unknown. $\endgroup$
    – OkkesDulgerci
    Mar 12 '18 at 16:50

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