1
$\begingroup$

Just want to ask if this scenario is still possible in linear interpolation. If not, what interpolation would I need to get the missing value:

X1 = 50 Y1 = 35 Z1 = 60

X2 = 70 Y2 = 40 Z3 = 70

X3 = 55 Y3 = 30 Z3 = ?

Thanks

1 Answers 1

1

Use can use linear interpolation, as follows: Using the ansatz \[ z = ax + by \] We get the linear system \begin{align*} 50a + 35b &= 60\\ 70a + 40b &= 70 \end{align*} which has the solution $a = \frac 19$, $b = \frac{14}9$. So \[ Z_3 \approx\frac 19 \cdot \left( 1 \cdot 55 + 14 \cdot 30\right) = 52\frac 79. \]