# Linear Interpolation Method Using C++ with Output

To interpolate value of dependent variable y at some point of independent variable x using Linear Interpolation, we take two points i.e. if we need to interpolate y corresponding to x which lies between x0 and x1 then we take two points [x0, y0] and [x1, y1] and constructs Linear Interpolants which is the straight line between these points i.e.

```y - y0 = ((y1 - y0)/(x1 - x0)) * (x - x0)
```

For more detail algorithm of this method, we encourage you to read article Linear Interpolation Method Algorithm. In this article we are going to implement Linear Interpolation using C++ and output is also provided.

``````
#include<iostream>
#include<conio.h>

using namespace std;

int main()
{
float x0,y0,x1,y1,xp,yp;

/* Inputs */
cout<<"Enter first point (x0,y0):"<< endl;
cin>>x0>>y0;
cout<<"Enter second point (x1,y1):"<< endl;
cin>>x1>>y1;
cout<<"Enter interpolation point: ";
cin>>xp;

/* Linear Interpolation */
yp = y0 + ((y1-y0)/(x1-x0)) * (xp - x0);

/* Displaying Output */
cout<<"Interpolated value at "<< xp<<" is "<< yp;

return 0;
}
``````

## Output

Consider we have interpolation problem stated as: "From some observation it is found that pressure recorded at temperature 35°C is 5.6KPa and at 40°C is 7.4 KPa. Later it is required to use pressure at 37°C which is not in observation table. So pressure value at 37°C need to be Interpolated and this can be calculated using above program as:"

```Enter first point (x0,y0):
35
5.6
Enter second point (x1,y1):
40
7.4
Enter interpolation point: 38
Interpolated value at 38 is 6.68
```