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.