For any new point $x_{0}$, the unbiased estimator for the response is \begin{align} y_{0} &= \hat{\theta}_{0} + \hat{\theta}_{1}x_{0}\newline E[y_{0}] &= E[\hat{\theta}_{0}] + E[\hat{\theta}_{1}]E[x_{0}] = \theta_{0} + \theta_{1}x_{0} \end{align} To get the distribution of this mean response, note that \begin{align} Y_{0} &= \hat{\theta}_{0} + \hat{\theta}_{1}x_{0} = \overline{Y} - \hat{\theta}_{1}\bar{x} + \hat{\theta}_{1}x_{0}\newline &= \frac{1}{n}\sum_{i=1}^{n} Y_{i} + (x_{0} - \bar{x})\frac{\sum_{i=1}^{n} (x_{i} - \bar{x})Y_{i}}{\sum_{i=1}^{n} (x-\bar{x})^{2}}\newline &= \sum_{i=1}^{n} \bigg( \frac{1}{n} + \frac{(x_{i} - \bar{x})(x_{0} - \bar{x})}{S_{xx}} \bigg)Y_{i} \end{align} which is a linear combination of independent normally distributed random variables $Y_{i}s$. Thus, the mean response is also a normally distributed random variable and we can get the confidence intervals by considering the mean and variance of this random variable \begin{align} Var(\hat{\theta}_{0} + \hat{\theta}_{1}x_{0}) &= \sum_{i=1}^{n} \bigg( \frac{1}{n} + \frac{(x_{i} - \bar{x})(x_{0} - \bar{x})}{S_{xx}} \bigg)^{2}Var(Y_{i})\newline \hat{\theta}_{0} + \hat{\theta}_{1}x_{0} &\sim \mathcal{N}\bigg(\theta_{0} + \theta_{1}x_{0}, \sigma^{2} \bigg[ \frac{1}{n} + \frac{(x_{0} - \bar{x})^{2}}{S_{xx}} \bigg]\bigg) \end{align} To eliminate $\sigma^{2}$, \begin{align} SS_{R}/\sigma^{2} \sim \chi_{n-2}^{2}\newline \frac{(\hat{\theta}_{0} + \hat{\theta}_{1}x_{0}) - (\theta_{0} + \theta_{1}x_{0})}{\sigma^{2} \bigg[ \frac{1}{n} + \frac{(x_{0} - \bar{x})^{2}}{S_{xx}} \bigg]} \div \sqrt{\frac{SS_{R}}{(n-2)\sigma^{2}}} \sim t_{n-2} \end{align} and the confidence intervals for confidence $1-\alpha$ become \begin{align} (\hat{\theta}_{0} + \hat{\theta}_{1}x_{0}) \pm t_{\alpha/2, n-2} \sqrt{\bigg( \frac{1}{n} + \frac{(x_{0} - \bar{x})^{2}}{S_{xx}} \bigg) \bigg( \frac{SS_{R}}{n-2}\bigg)} \end{align}