This is the classic total least square problem. Sometimes it is called the errors in variables problem, sometimes known as orthogonal regression.
Lacking your data, I can't show how to do it using your data. But it is not difficult to do. Assume you have data in the form of vectors x and y. I'll make some up, here:
t = rand(100,1);
x = t/10 + randn(size(t))/10;
y = 5*t + randn(size(t))/10;
plot(x,y,'o')
axis equal
The actual slope of the curve should have been 50. What happens when we use polyfit?
P1 = polyfit(x,y,1)
The problem is, polyfit always produces biased estimates in these problems, where the slope will be biased towards zero. Do you see the polyfit extimated slope is very small? Instead of 50, we got a number around 2.9081 from polyfit.
The trick is to use total least squares. We can get the coefficients of the line from:
mux = mean(x);
muy = mean(y);
[~,~,V] = svd([x(:)-mux,y(:)-muy],0);
coef = V(:,end)
AEst = -coef(1)/coef(2)
So despite the rather high presence of noise on our data, we get a slope estimate 56.7846, instead of the 2.9 value that polyfit predicted. That seems pretty reasonable.
The constant term in a regression model will be:
BEst = muy - AEst*mux
The line of best orthogonal fit is:
y = 0.0527 + 56.7846*x
