c# - Plotting a line in a chart given the y intercept and slope -

i've written program calculates line of best fit (intercept/slope) given several input values user. i've plotted each of individual values, unsure of code plot line given slope , y-intercept.

this slope:

double m = ( axy.sum() -             ((levels.sum() * scores.sum()) / 5)) / (newax.sum() - ((powlevels) / 5)); 

the intercept

double b = meany - (m * meanx); 

plotting of points

for (int = 0; < levels.getlength(0); i++) {     chart1.series["series1"].points                             .addxy(levels.getvalue(i), scores.toarray().getvalue(i)); } 

any ideas? no means expert , getting far took fair bit of experimenting..

assuming data plotted scattergraph using charttype.points simplest way add line add 1 extra series charttype.line , set 2 points there.

there are other ways create line on chart, drawing or creating lineannotation, much more complicated!

following this example letter here implementation:

note after creating series line of best fit thing looking last 2 lines..:

private void button1_click(object sender, eventargs e) {     // create 2 series!     chart1.series.clear();     chart1.series.add("data");     chart1.series.add("line of best fit");     chart1.series[0].charttype = seriescharttype.point;     chart1.series[1].charttype = seriescharttype.line;      list<int> levels = new list<int>() { 8, 2, 11, 6, 5, 4, 12, 9, 6, 1};     list<int> scores = new list<int>() { 3, 10, 3, 6, 8, 12, 1, 4, 9, 14};      double minx = levels.tolist().min();     double maxx = levels.tolist().max();     double meanx = 1f * levels.sum() / levels.count;     double meany = 1f * scores.sum() / scores.count;      double st = 0;     double sb = 0;     (int = 0; < levels.count; i++ )     {         st += (levels[i] - meanx) * (scores[i] - meany);         sb += (levels[i] - meanx) * (levels[i] - meanx);     }     double slope = st / sb;     double y0 = meany - slope * meanx;  // y-intercept or y-crossing      (int = 0; < levels.count; i++)     {             chart1.series[0].points.addxy(levels[i], scores[i]);     }     // part creates line of best fit:     chart1.series[1].points.addxy(minx, y0 + minx * slope);     chart1.series[1].points.addxy(maxx, y0 + maxx * slope); } 

if want can add first line point right @ y-axis:

 chart1.series[1].points.addxy(0, y0 ); 

in case may want set minimum x-values shown in chart prevent including -1, maybe this:

 chart1.chartareas[0].axisx.minimum = minx - 1;