Proof $C$= {$(x, y) \in \mathbb{R}^2 | x+y=b $} is convex

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In an optimization problem, I have constraints of the form $x+y=b$. In order to prove that the solution is unique, I proved that the criterion is strictly convex and now I need to show that the set of constraints :

$C$= {$(x, y) \in \mathbb{R}^2 | x+y=b $} is convex.

I wanted to go by the definition and prove that $z=tx+(1-t)y$ $\in C$ for $ t ∈ [0,1]$ but I don't see how I can get to the result ?

Thank you

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This is what you have to show:

Suppose $(x_1,y_1)\in C$, that is$$x_1 + y_1 = b$$

and $(x_2,y_2)\in C$, $$x_2+y_2=b$$

Show that $(\lambda x_1 +(1-\lambda )x_2,\lambda y_1 +(1-\lambda) y_2) \in C$, that is

$$(\lambda x_1+(1-\lambda)x_2)+(\lambda y_1+(1-\lambda)y_2)=b$$

where $\lambda \in [0,1]$.