Problem 63 - Precalculus with Limits

Question

In Exercises 63 and 64, find the distance between the parallel lines. \(x + y = 1\) \(x + y = 5\)

Step-by-Step Solution

Answer

The distance between the two parallel lines \(x + y = 1\) and \(x + y = 5\) is approximately 2.83 units.

Step 1: Identify the coefficients and constants

From the two given equations \(x + y = 1\) and \(x + y = 5\), we can identify that A = 1, B = 1 (the coefficients of x and y), C1 = 1 and C2 = 5 (the constants in the equations).

Step 2: Substitute the values into the formula for the distance between two parallel lines

Now let's plug the values into the formula. Distance = \(|C2 - C1| / \sqrt {A^2 + B^2} = |5 - 1|/ \sqrt {1^2 + 1^2} = 4 / \sqrt {1 + 1}\)

Step 3: Calculate the distance

Compute the values under and outside the square root to find the final distance. Distance = 4 / \sqrt {2} = 4/1.41 = 2.83 units.

Key Concepts

Parallel Lines

Parallel lines are lines in a plane that do not intersect or meet, no matter how far they are extended. These lines have the same slope, which means they run alongside each other at an equal distance throughout their stretch. In the context of the given exercise, the lines have been represented by the equations \(x + y = 1\) and \(x + y = 5\). Because both equations can be rearranged to \(y = -x + 1\) and \(y = -x + 5\), they clearly have identical slopes, showing their parallel nature.

• Same slope: This is what makes the lines parallel. For example, in equations like \(y = mx + c\), lines having similar \(m\) (slope) are parallel.
• Never meeting: They can't "dock" or intersect at any given point in a plane, maintaining a consistent space between them.
• Fixed distance apart: This is important when you're asked to calculate the distance between them, which remains constant.

Understanding this idea is crucial because it lays the foundation for many geometric problems requiring the measurement of distances or determining paths in a variety of fields such as physics and engineering.

Distance Formula

The distance formula provides a method to calculate how far apart two parallel lines are from each other. For two parallel lines with equations of the form \(Ax + By = C_1\) and \(Ax + By = C_2\), the distance between these lines is given by \(\text{Distance} = \frac{|C_2 - C_1|}{\sqrt{A^2 + B^2}}\).
This formula arises from coordinate geometry and is specifically useful as it links algebra with geometry, helping to solve spatial problems in the plane. Here's a breakdown of how it works:

• \(C_1\) and \(C_2\) represent the line constants showing how shifted the lines are on their paths.
• \(A\) and \(B\) are coefficients of \(x\) and \(y\), showing slopes in generalized line equations.
• The formula itself is derived from the fact that the perpendicular (and therefore shortest) distance between two lines can be thought of as a "normalized" difference between their intercepts.

This formula simplifies finding distances efficiently, using strictly the algebraic expressions representing the parallel lines.

Coordinate Geometry

Coordinate geometry, or Cartesian geometry, is a branch of geometry where the position of points on the plane is described using an ordered pair of numbers. It's a way of describing geometry using coordinate systems, which can greatly simplify and solve complex problems. In the case of calculating the distance between parallel lines, coordinate geometry helps break down the geometrical concepts into algebraic equations.

• Points are defined by coordinates \((x, y)\) on a two-dimensional plane, providing precision.
• Lines are expressed in the format \(Ax + By = C\), bridging gaps between algebra and geometry.
• Enables visualization of shapes, spatial understanding, and problems involving distance and movement.

By translating geometric problems into algebraic ones, this branch of math helps you understand the relationships between figures and the basic elementary calculus of shapes in planes. It is very useful in fields like physics, engineering, navigation, and even computer graphics.