How do you evaluate the following expression when x=3 and y=4: for \[{{\left( y-x \right)}^{2}}\]?

Hint: This type of problem is based on the concept of functions with two variables. First, we have to consider the function with the variable x and y and equate it to f(x,y). Use the identity \[{{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}\] to simplify the function f(x,y). Here, we find that a=y and b=x. substitute in the identity and simplify the function. To find the value of f(x,y) at x=3 and y=4, we have to substitute the respective values of x and y in the given function. Do necessary calculations and find the value which is the required answer.

Complete step by step solution:
According to the question, we are asked to find the value of \[{{\left( y-x \right)}^{2}}\] at x=3 and y=4.
We have been given the function \[{{\left( y-x \right)}^{2}}\].
Let us assume the function is \[f\left( x,y \right)={{\left( y-x \right)}^{2}}\]. -----(1)
The given function is a function with two variables x and y.
We know that \[{{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}\]. Using this identity, we can simplify the function (1).
Here, a=y and b=x. substitute the values of a and b.
\[\Rightarrow f\left( x,y \right)={{y}^{2}}-2yx+{{x}^{2}}\] ----------------(2)
Now, we have to find the value of f(x,y) when x=3 and y=4.
When the variables are equated to a constant, we have to substitute the constant in place of the variables in the function to find the required value.
Here, we have to find the value of f(3,4).
To find the value of f(3,4), we have to substitute x equal to 3 and y equal to 4 in the considered function (2).
On substituting x=3 and y=4 in the function (2), we get
\[f\left( 3,4 \right)={{4}^{2}}-2\times 4\times 3+{{3}^{2}}\]
We know that the square of 3 is 9 and the square of 4 is 16.
On substituting the values, we get
\[f\left( 3,4 \right)=16-2\times 4\times 3+9\]
On further simplification, we get
\[f\left( 3,4 \right)=16-8\times 3+9\]
\[\Rightarrow f\left( 3,4 \right)=16-24+9\]
\[\Rightarrow f\left( 3,4 \right)=25-24\]
\[\therefore f\left( 3,4 \right)=1\]
Therefore, the value of the function \[{{\left( y-x \right)}^{2}}\] when x=3 and y=4 is 1.

Note:
We can also solve this problem by another method.
First, we have to substitute x=3 and y=4 in the given function for simplification.
\[\Rightarrow f\left( 3,4 \right)={{\left( 4-3 \right)}^{2}}\]
We know that 4-3=1. We get
\[f\left( 3,4 \right)={{1}^{2}}\]
Therefore, we get f(x,y)=1 when x=3 and y=4.
This method reduces the number of steps of the solution.