@dilip's answer is sufficient, but I just thought I'd add some details on how you get to the result. We can use the method of characteristic functions. For any d-dimensional multivariate normal distribution X∼Nd(μ,Σ) where μ=(μ1,…,μd)T and Σjk=cov(Xj,Xk)j,k=1,…,d, the characteristic function is given by:
φX(t)=E[exp(itTX)]=exp(itTμ−12tTΣt)
=exp(i∑j=1dtjμj−12∑j=1d∑k=1dtjtkΣjk)
For a one-dimensional normal variable Y∼N1(μY,σ2Y) we get:
φY(t)=exp(itμY−12t2σ2Y)
Now, suppose we define a new random variable Z=aTX=∑dj=1ajXj. For your case, we have d=2 and a1=a2=1. The characteristic function for Z is the basically the same as that for X.
φZ(t)=E[exp(itZ)]=E[exp(itaTX)]=φX(ta)
=exp(it∑j=1dajμj−12t2∑j=1d∑k=1dajakΣjk)
If we compare this characteristic function with the characteristic function φY(t) we see that they are the same, but with μY being replaced by μZ=∑dj=1ajμj and with σ2Y being replaced by σ2Z=∑dj=1∑dk=1ajakΣjk. Hence because the characteristic function of Z is equivalent to the characteristic function of Y, the distributions must also be equal. Hence Z is normally distributed. We can simplify the expression for the variance by noting that Σjk=Σkj and we get:
σ2Z=∑j=1da2jΣjj+2∑j=2d∑k=1j−1ajakΣjk
This is also the general formula for the variance of a linear combination of any set of random variables, independent or not, normal or not, where Σjj=var(Xj) and Σjk=cov(Xj,Xk). Now if we specialise to d=2 and a1=a2=1, the above formula becomes:
σ2Z=∑j=12(1)2Σjj+2∑j=22∑k=1j−1(1)(1)Σjk=Σ11+Σ22+2Σ21