# How to get only those output values which lie under a given curve!

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am on 24 Dec 2014
Answered: Shoaibur Rahman on 24 Dec 2014 Hi everyone, I have a graph with tabulated values (x1-xn, y1-yn and z1-zn). I used interpolation to get required values of 'Z' w.r.t x and y. However i also need to use only those values of Z which lie under a given line ( limiting curve ) having four coordinates ( as mentioned in fig). I don't have any equation for this line then how can i make sure my output value always remain under this given curve or satisfy this condition!
TiA for all the good work you do! regards

Shoaibur Rahman on 24 Dec 2014
Perhaps, you unconsciously mistyped the x coordinate of third point. I use 3800 instead of 2100, but use what the actual value is. Also, I use random Z data here for test purposes.
x = 1000:4500;
z = 11+randn(1,length(x)); % use the data that you got
plot(x,z,'om'), hold on
x1 = 1000:2100;
y1 = 8+(8-13.2)/(1000-2100)*(x1-1000); % first boundary, straight line equation
z1 = z(1:length(y1)); % z values in the given range x1
z1(z1 >= y1) = NaN; % discard z values that are above the boundary
plot(x1,y1,'r'), plot(x1,z1,'o','MarkerSize',12)
x2 = 2100:3800;
y2 = 13.2+(13.2-14.6)/(2100-3800)*(x2-2100); % second boundary
z2 = z(length(y1)+1:length(y1)+length(y2)); % z values in the given range x2
z2(z2 >= y2) = NaN; % discard z values that are above the boundary
plot(x2,y2,'r'), plot(x2,z2,'o','MarkerSize',12)
x3 = 3800:4500;
y3 = 14.6+(14.6-12)/(3800-4500)*(x3-3800); % third boundary
z3 = z(length(y1)+length(y2)-1:end); % remaining z values
z3(z3 >= y3) = NaN; % discard z values that are above the boundary
plot(x3,y3,'r'), plot(x3,z3,'o','MarkerSize',12)