# Examples of the Two Points Form of the Equation of a Line

__Example 1__**:**

A milkman can sell 650 liters of milk at $3.15 per liter and 800 liters of milk at $3.00 per liter. Assuming the graph of the sale price and the milk sold to be a straight line, find the number of liters of milk that the milkman can sell at $2.50 per liter.

__Solution__**:**

Let $$x$$ be the number of liters of milk sold and $$y$$ be the price per liter. The given information can be written in the form of points $$\left( {650,3.15} \right)$$ and $$\left( {800,3.00} \right)$$.

Since the graph of the sale price and the milk sold is a straight line, we find the equation of a straight line through the points $$\left( {650,3.15} \right) = \left( {{x_1},{y_1}} \right)$$ and $$\left( {800,3.00} \right) = \left( {{x_2},{y_2}} \right)$$ as follows:

Using the two points form of the equation of a straight line

\[\begin{gathered} \frac{{y – {y_1}}}{{{y_2} – {y_1}}} = \frac{{x – {x_1}}}{{{x_2} – {x_1}}} \\ \Rightarrow \frac{{y – 3.15}}{{3.00 – 3.15}} = \frac{{x – 650}}{{800 – 650}} \\ \Rightarrow \frac{{y – 3.15}}{{ – 0.15}} = \frac{{x – 650}}{{150}} \\ \end{gathered} \]

In order to find the number of liters of milk that the milkman can sell at $$2.50$$, we put $$y = 2.50$$ in the above equation, and we get

\[\begin{gathered} \Rightarrow \frac{{2.50 – 3.15}}{{ – 0.15}} = \frac{{x – 650}}{{150}} \\ \Rightarrow x – 650 = 4.34 \times 150 \\ \Rightarrow x = 1301 \\ \end{gathered} \]

This result shows that the milkman can sell 1301 liters of milk at $2.50 per liter.

__Example 2__**:**

Find the equation of straight line passing through the points $$A\left( {0,8} \right)$$ and $$B\left( {2,3} \right)$$.

Consider the points $$A\left( {0,8} \right) = \left( {{x_1},{y_1}} \right)$$ and $$B\left( {2,3} \right) = \left( {{x_2},{y_2}} \right)$$. Now using these points in the two point form of the equation of straight line, we get

\[\begin{gathered} \frac{{y – {y_1}}}{{{y_2} – {y_1}}} = \frac{{x – {x_1}}}{{{x_2} – {x_1}}} \\ \Rightarrow \frac{{y – 8}}{{3 – 8}} = \frac{{x – 0}}{{2 – 0}} \\ \Rightarrow \frac{{y – 8}}{{ – 5}} = \frac{x}{2} \\ \Rightarrow 2\left( {y – 8} \right) = – 5x \\ \Rightarrow 5x + 2y – 16 = 0 \\ \end{gathered} \]

This is the required equation of a straight line.