Learning cost function for linear regression
There is a worked out example in the Wikipedia page for "simple linear regression"

Just for the sake of it, let me plug in your example into the formulas:

The fitted model should be a straight line with parameters $\alpha$ (value at $x = 0$) and $\beta$ (the slope):

$$f(x) = \alpha + \beta x$$

The values for these parameters that minimize the distance between line and data points are called $\hat{\alpha}$ and $\hat{\beta}$.  They can be computed out of the data point values by using these formulae, derived here:

$$\begin{align}
\hat{\beta} &amp; = \frac{ \overline{xy} - \bar{x}\bar{y} }{ \overline{x^2} - \bar{x}^2 } , 
\\
\\
\hat{\alpha} &amp; = \bar{y} - \hat{\beta}\bar{x}
\end{align}
$$

where an expression with an overline $\overline{xy}$ means the sample average of that expression: $\overline{xy} = \tfrac{1}{n} \sum_{i=1}^n{x_iy_i}$. 

Here are the values I find for the datapoints you have listed in your question:

$$
\begin{align}
\overline{xy} &amp;= \frac{1}{4} \sum{&lt;(1 \times 4000), (2 \times 10000), (3 \times 22000), (4 \times 30000)&gt;} \\
              &amp;= \frac{1}{4}(4000 + 20000 + 66000 + 120000) = 52500, \\
\overline{x} &amp;= \frac{1}{4} \sum{&lt;1, 2, 3, 4&gt;} = 2.5 ,\\
\overline{y} &amp;= \frac{1}{4} \sum{&lt;4000, 10000, 22000, 30000&gt;} = 16500 , \\
\overline{x^2} &amp;= \frac{1}{4} \sum{&lt;1^2, 2^2, 3^2, 4^2&gt;} = 7.5 , \\
\overline{x}^2 &amp;= 2.5^2 = 6.25
\end{align}
$$

and the fitted line should be:

$$
\begin{align}\
\hat{\beta} &amp;= \frac{52500 - 2.5 \times 16500}{7.5 - 6.25} = \frac{41250}{1.25} = 33000 , \\
\hat{\alpha} &amp;= 16500 - 33000 \times 2.5 = - 66000 , \\
\Rightarrow f(x) &amp;= - 66000 + 33000 x 
\end{align}
$$

Therefore, the model would predict, for a house with 10 rooms, a rent of:

$$ f(x) = -66000 + 33000 \times 10 = 264000 $$