... "negative slope" or "negative intercept"

technically, as soon as the slope is too much positive then the intercept may turn out to be negative

Steftcho P. Dokov I'm extremely sorry for the mistake. Its negative intercept. I have some sets of data obtained from UV titraion in which absorbance gradually increases upon addition of probe. But only one data set gives both intercept and slope positive. Others are giving negative intercept though having comparable values. Both x and y are positive.. please explain why intercept become negative and how to solve the issue. Thanks.

Both, the slope and the intercept are coming/ derived (calculated) from the data points in hand (from the given data points). So, if data points exhibit/ show too positive trend (slope) then the intercept might appear negative (can be negative). Just check the data points for validity, i.e., see if the data points are all valid without any typos / errors etc.

Steftcho P. Dokov Thanks for the explanation. All data are valid and spectroscopically obtained. In 2 sets of data, X/Y1 gives a realistic slope and intercept, while X/Y2 results negative intercept. it may be due to being too positive. data is highly linear by which nonlinear fit is not possible. How to address this? some are fixing the intercept to positive but I think fixing is not appropriate. Manual calculation gives a positive value but linear fit in Excel, and origin produces a negative intercept. Please let me know the possible solution. Thanks.

3.60606 data point is repeated / seen twice in the last column ... is it really the case with calculations ... given that "absorbance gradually increases"

Steftcho P. Dokov ya got it, this is the raw data with X: 0.301023 0.150989 0.101 0.075999 0.061002 0.051 0.043858 0.0385 0.034334 0.031

Y : 25.34854 12.26512 7.774418 5.692006 4.412576 3.695806 3.133343 2.394682 2.151134 1.534453. give y=87.19x-0.9084

It appears that the rate at which absorption gradually increases is much less for y1 than for Y. Somehow the smallest value of Y is smaller than its counterpart in y1, yet the largest value of Y is larger than the largest y1. That is, should it be the case that the rate of change of Y is greater than that of y1?

Both are different probes. Thus absorbance rate of increasing differ. Thus slope and intercept will give information for binding constant which is different for both. Thing is that how to overcome negative intercept value? Software may calculate below x or y which give negative. Is there any method other than fixing intercept?

Well, constrained optimization can do the job. That is, the usual linear regression which minimizes the sum of squares (sum of squared errors as an objective function) can be extended with one (or with many constraints) on the variables, i.e., on the slope and intercept. See, e.g.

Article Bounded-Variable Least-Squares: an Algorithm and Applications