Thanks for the post, I found it very informative. I stumbled upon your blog when I was looking for alternate ways to a fit a straight line to a set of data points. I currently employ the method of calculating the least squares approximation of the line using linear algebra techniques; I found that my current technique worked for one of my data sets but the methods you posted here did not. I was wondering if you could explain to me why it didn’t work, or if you could please possibly direct me to some other source of information so that I can understand why it doesn’t work. The data set which I mentioned is below:

X_Values = {0.653392, 0.655975, 0.649529, 0.642886, 0.644527, 0.645786, 0.638369, 0.638948, 0.647223, 0.646904, 0.638298, 0.645040, 0.643486, 0.641512, 0.639113, 0.636286, 0.640221, 0.643475, 0.639094, 0.634266, 0.635682, 0.629824, 0.629939, 0.629328, 0.627982, 0.631913, 0.634815, 0.625205, 0.626303, 0.626341, 0.620012, 0.618052, 0.620006, 0.620562, 0.610319, 0.608353, 0.600574};

Y_Values = {0.347414, 0.363613, 0.375006, 0.386285, 0.402745, 0.419378, 0.430585, 0.447396, 0.470235, 0.487477, 0.498693, 0.522343, 0.539949, 0.557658, 0.575460, 0.593347, 0.618254, 0.643475, 0.661801, 0.680168, 0.705996, 0.724530, 0.750733, 0.777155, 0.803780, 0.838577, 0.873748, 0.892885, 0.928532, 0.964481, 0.992226, 1.028611, 1.073882, 1.119524, 1.147842, 1.193959, 1.231358};

The line is almost vertical; and I have a feeling that this is why the method here on your post fails. With my current method I get a gradient of -19.569292 and an intercept of 13.112316; I know that these values are correct for the data set because I plotted the values on Excel and added a trend line, the trend line had the same parameters as the ones I have mentioned above. Thanks in advance for you help.

Kind Regards,
Tony