Problem 67 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 67 and 68, find the (perpendicular) distance between the given parallel lines. Hint: First find a point on one of the lines. \(2 x+4 y=7,2 x+4 y=5\)

Step-by-Step Solution

Answer

The perpendicular distance between the lines is approximately 0.447 units.

Step 1: Identify the General Form of the Lines

Both given lines are in the form \(2x + 4y = C\). The first line is \(2x + 4y = 7\) and the second is \(2x + 4y = 5\). These are parallel because they have the same coefficients for \(x\) and \(y\).

Step 2: Simplify the Line Equation

Divide each equation by 2 to simplify them: \(x + 2y = 3.5\) and \(x + 2y = 2.5\). Both lines are parallel.

Step 3: Find a Point on One Line

Choose an easy value for \(x\). Let \(x = 0\) for the first line, then solve for \(y\): \(2(0) + 4y = 7\), which gives \(y = 1.75\). Thus, the point \((0, 1.75)\) is on the first line.

Step 4: Use the Distance Formula Between a Point and a Line

The formula for the perpendicular distance from point \((x_0, y_0)\) to the line \(Ax + By + C = 0\) is \(\frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}\). Convert the second line, \(x + 2y = 2.5\), to the standard form \(x + 2y - 2.5 = 0\).

Step 5: Calculate the Distance

Using the point \((0, 1.75)\) and the line \(x + 2y - 2.5 = 0\), substitute into the formula: \(\frac{|0 + 2(1.75) - 2.5|}{\sqrt{1^2 + 2^2}} = \frac{|3.5 - 2.5|}{\sqrt{5}} = \frac{1}{\sqrt{5}}\approx 0.447\).

Step 6: Conclusion: Interpret the Result

The perpendicular distance between the two parallel lines is approximately 0.447 units.

Key Concepts

Equations of Parallel Lines

Parallel lines have the same slope, meaning that they never intersect each other. The equations of parallel lines will look similar as they will have identical coefficients for both the variables. Consider the lines given by equations:

• The first line: \(2x + 4y = 7\)
• The second line: \(2x + 4y = 5\)

These lines are clearly parallel, as they share the same coefficients of 2 for \(x\) and 4 for \(y\). This results in both their slopes being equal. To simplify, dividing both equations by 2 gives the same form \(x + 2y = C\), maintaining their parallel nature.
While simplifying, watch for constants on different sides of the equation because they affect the intercepts but not the slope.

Distance Formula for a Point to a Line

Calculating the perpendicular distance from a point to a line is a useful skill. The formula used for this purpose is derived from geometry and calculus. Given a point \((x_0, y_0)\) and a line in the form \(Ax + By + C = 0\), the distance is calculated as:

• \(\frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}\)

This formula allows us to find the shortest distance, which is perpendicular, from the point to the line. Continuing with our problem, we substitute the point \((0, 1.75)\) and the line equation \(x + 2y - 2.5 = 0\) into the formula. After computing, the perpendicular distance turns out to be approximately 0.447 units. This method is reliable for finding distances efficiently and is a vital tool in analytical geometry.

Solving Linear Equations

Solving linear equations often involves finding the value of variables that satisfy the equation conditions. In our exercise, we took the equations for the lines and made them simpler. For example, considering \(2x + 4y = 7\), dividing through by 2 lets us solve for \(y\), easily finding values like when \(x = 0\) resulting in \(y = 1.75\).
The concept here underscores how finding a point \((0, 1.75)\) on the line can be simple once the equation is in reduced form. This process is part of a larger skill set in algebra: restructuring equations to not only find specific values but also to provide insights into relationships of the variables, such as the slope or intercepts. Recognizing these relationships can help in graphing equations, determining parallels, and assessing the critical points in geometry problems.