The first step in determining the regression line is to calculate the mean values of #X# and #Y#.
\[\begin{array}{rcl}
\bar{X}&=&\displaystyle\cfrac{\sum\limits_{i=1}^n{X_i}}{n} = \dfrac{1+2+3+4+5}{5}=\dfrac{15}{5}=3\\\\
\bar{Y}&=&\displaystyle\cfrac{\sum\limits_{i=1}^n{Y_i}}{n} = \dfrac{2+6+9+3+1}{5}=\dfrac{21}{5}=4.2
\end{array}\]
Next, find the values of #(X_i-\bar{X}), (Y_i-\bar{Y}), (X_i-\bar{X})^2# and #(X_i-\bar{X})(Y_i-\bar{Y})# for each pair of data points:

With this information the slope #b_1# and intercept #b_0# can be calculated:
\[\begin{array}{rcl}
b_1 &=& \displaystyle\cfrac{\sum\limits_{i=1}^n{(X_i-\bar{X})(Y_i-\bar{Y})}}{\sum\limits_{i=1}^n{(X_i-\bar{X})^2}}\\\\
&=& \cfrac{4.4-1.8+0-1.2-6.4}{4+1+0+1+4}\\\\
&=& -0.5\\\\
b_0 &=& \bar{Y} - b_1 \cdot \bar{X}\\\\
&=& 4.2 - (-0.5)\cdot3\\\\
&=& 5.7
\end{array}\]
So the regression equation is:
\[\begin{array}{rcl}
\hat{Y} &=& b_0 + b_1X\\\\
&=& 5.7-0.5X
\end{array}\]

Note that the regression line always passes through the mean point #(\bar{X},\bar{Y})#.

In this case, #(\bar{X},\bar{Y})= (3,4.2)# and entering #X=3# into the equation gives:
\[\begin{array}{rcl}
\hat{Y} &=& 5.7 -0.5\cdot 3\\\\
&=& 4.2
\end{array}\]