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 distance between the parallel lines is \(\frac{\sqrt{5}}{5}\).

Step 1: Understand the Problem

Two lines are given by the equations \(2x + 4y = 7\) and \(2x + 4y = 5\). These lines are parallel because they have the same slope.

Step 2: Express Lines in Slope-Intercept Form

First, we convert both equations to the slope-intercept form (\(y = mx + c\)) to confirm their slopes:For the equation \(2x + 4y = 7\):\[ 4y = -2x + 7 \ y = -\frac{1}{2}x + \frac{7}{4} \]For the equation \(2x + 4y = 5\):\[ 4y = -2x + 5 \ y = -\frac{1}{2}x + \frac{5}{4} \] Both lines have a slope of \(-\frac{1}{2}\), reaffirming they are parallel.

Step 3: Choose a Point on One of the Lines

Let's find a point on the line \(2x + 4y = 7\). We set \(x = 0\) to simplify our calculation:\[ 2(0) + 4y = 7 \ y = \frac{7}{4} \ \text{Point on the line: } (0, \frac{7}{4}) \]

Step 4: Find Distance Formula Between a Point and a Line

The distance \(d\) from a point \((x_1, y_1)\) to the line \(Ax + By + C = 0\) is given by:\[ d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \]

Step 5: Set Up the Distance Formula

The second line can be written as:\[ 2x + 4y - 5 = 0 \]Substitute \(A = 2, B = 4, C = -5\) and the point \((0, \frac{7}{4})\):\[ d = \frac{|2(0) + 4(\frac{7}{4}) - 5|}{\sqrt{2^2 + 4^2}} \]

Step 6: Calculate the Distance

First, calculate the numerator:\[ 4 \times \frac{7}{4} = 7 \|7 - 5| = 2 \]Now calculate the denominator:\[ \sqrt{2^2 + 4^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5} \]Plug these into the distance formula:\[ d = \frac{2}{2\sqrt{5}} = \frac{1}{\sqrt{5}} \]Therefore, the distance is \(\frac{\sqrt{5}}{5}\) after rationalizing the denominator.

Key Concepts

Slope-Intercept Form

The slope-intercept form of a linear equation is a popular way to express straight lines. It is written as \(y = mx + c\), where \(m\) represents the slope and \(c\) is the y-intercept. This form is especially useful when you want to quickly identify the slope of a line. To convert an equation, like \(2x + 4y = 7\), into this form, you must solve for \(y\). Start by isolating \(y\) on one side of the equation, which gives:

• \(4y = -2x + 7\)
• \(y = -\frac{1}{2}x + \frac{7}{4}\)

This formula shows that the slope is \(-\frac{1}{2}\) and the y-intercept is \(\frac{7}{4}\). By converting both lines to this form, we can confirm their parallel nature by checking if they have identical slopes.

Distance Formula

To find the distance between a point and a line, or between two parallel lines, the distance formula is a handy tool. For a point \((x_1, y_1)\) and a line given by \(Ax + By + C = 0\), the distance \(d\) is calculated by:

• \(d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}\)

This formula measures the perpendicular distance, which is often needed in mathematics for exact measurements. By plugging the values into the formula appropriately, like using the point \((0, \frac{7}{4})\) and the line \(2x + 4y - 5 = 0\), you can determine the precise distance from the point to the line. This results in the calculation \(d = \frac{1}{\sqrt{5}}\). It's a versatile tool for solving geometric problems.

Parallel Lines

Parallel lines are lines in a plane that never intersect, no matter how far they are extended. A key property of parallel lines is that they have the same slope when expressed in slope-intercept form \(y = mx + c\). This means they rise and fall at the same rate. In our problem, both lines \(2x + 4y = 7\) and \(2x + 4y = 5\) have a slope of \(-\frac{1}{2}\). By confirming the slopes are equal, we verify the lines are parallel. Understanding this helps to recognize that any distance calculated between these lines is constant, strengthening our geometric intuition.

Rationalizing the Denominator

Rationalizing the denominator is a process used to eliminate irrational numbers from the denominator of a fraction. It involves multiplying both the numerator and the denominator by a suitable form to make the denominator a rational number (like a whole number or a fraction without a square root). For example, if you have \(\frac{1}{\sqrt{5}}\), you multiply both the numerator and denominator by \(\sqrt{5}\) to get:

• \(\frac{1}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = \frac{\sqrt{5}}{5}\)

This technique simplifies calculations and makes expressions easier to comprehend. It is a crucial algebraic step that ensures mathematical expressions are presented in a standard form.