Let $W_t$ be a Brownian motion at time $t$ and let $t<t_1<T$. I'm trying to find the variance of $$ k\sigma W_{t_1} + \sigma \left(W_T - W_t\right).$$ I started by letting $$ k\sigma W_{t_1} + \sigma \left(W_T - W_t\right) = k\sigma W_{t_1} + \sigma \left(W_T - W_t\right) -\sigma W_{t_1} + \sigma W_{t_1},$$ which I expressed as $$ k\sigma W_{t_1} + \sigma \left(W_T-W_{t_1}\right) + \sigma \left(W_{t_1}-W_t \right).$$
I know that $W_T-W_{t_1}$ and $W_{t_1}-W_t$ are independent.
Can I use this property to find the variance as $k^2\sigma^2 t_1+ \sigma^2\left(T-t\right)?$