Calculating the Distance Between Parallel Straight Lines

In mathematics, determining the distance between parallel lines is an important concept in geometry and calculus. Parallel lines maintain a constant distance apart, and this distance can be calculated using specific formulas and steps. Here, we will show how to find the distance between the lines given by the equations 3x 4y - 3 0 and 6x 8y - 1 0. We will also discuss how to use the standard form and slope-intercept form to solve similar problems.

Standard Form and Coefficients

To find the distance between two parallel lines, it's helpful to express their equations in the standard form:

Ax By C 0

Step 1: Express Equations in Standard Form

For the first line: 3x 4y - 3 0, the coefficients are: A1 3 B1 4 C1 -3 For the second line: 6x 8y - 1 0, the coefficients are: A2 6 B2 8 C2 -1

Notice that since these lines are parallel, the ratio (frac{A_1}{A_2} frac{B_1}{B_2} frac{1}{2}). This confirms that the two lines are indeed parallel.

Distance Formula for Parallel Lines

The distance d between two parallel lines Ax By C1 0 and Ax By C2 0 is given by:

d (frac{|C_2 - C_1|}{sqrt{A^2 B^2}})

Step 2: Simplify the Second Line Equation

First, we simplify the second equation to match the standard form with the same coefficients for A and B. By dividing the entire equation by 2:

6x 8y - 1 0 becomes 3x 4y - (frac{1}{2}) 0

Now, we identify:

C2 -(frac{1}{2}) C1 -3

Step 3: Calculate the Distance

Using the distance formula:

d (frac{-(frac{1}{2}) - (-3)}{sqrt{3^2 4^2}})

d (frac{5/2}{5})

d (frac{1}{2})

Thus, the distance between the two parallel lines is (frac{1}{2}) units.

Slopes and Distance Calculation

Alternatively, we can also use the slope-intercept form to find the distance between parallel lines. Let's rewrite the equations in this form:

3x 4y - 3 0 can be written as y -(frac{3}{4})x (frac{3}{4})

6x 8y - 1 0 can be written as y -(frac{3}{4})x (frac{1}{8})

Using the formula for the distance between parallel lines in slope-intercept form:

d (frac{|C_2 - C_1|}{sqrt{1 (text{slope})^2}})

Here, the slope is (-(frac{3}{4}) and C2 (frac{1}{2}), C1 3:

d (frac{|(frac{1}{2}) - 3|}{sqrt{1 (-(frac{3}{4}))^2}})

d (frac{|(frac{1}{2}) - (frac{6}{2})|}{sqrt{1 (frac{9}{16})}})

d (frac{|-(frac{5}{2})||}{sqrt{(frac{25}{16})}})

d (frac{(frac{5}{2})}{frac{5}{4})

d (frac{1}{2})

Using Trigonometry for Verification

We can also use trigonometry to verify the distance. First, we find the slope of the line 3x 4y - 3 0, which is -(frac{3}{4}). The length of one segment of the perpendicular distance can be calculated using the tangent and sine of the angle:

(tan(text{angle ABC}) frac{4}{3})

(sin(text{angle ABC}) frac{4}{5})

The length of the segment AC (the perpendicular distance) is:

AC (frac{1}{2}) (units)

This value matches our previous calculation, thus confirming the distance is (frac{1}{2}) units.

The distance between parallel lines is a fundamental concept in geometry, and understanding how to calculate it using different methods (standard form, slope-intercept form, and trigonometry) is essential for solving related problems.