Problem 23 - College Algebra and Calculus: An Applied Approach

Question

Use the point on the line and the slope of the line to find three additional points through which the line passes. \(\begin{array}{ll}\text { Point } & \text { Slope }\end{array}\) \((-6,-1)\) \(m=\frac{1}{2}\)

Step-by-Step Solution

Answer

The three additional points the line passes through are (-4, 0), (-2, 1) and (0, 2).

Step 1: Find the First Point

Starting from the point \(-6, -1\), use the slope to find the next point. Since the slope is \(m=\frac{1}{2}\), it means for every 2 units you go to the right (increasing x by 2), you go up by 1 unit (increase y by 1). So if we add 2 to the x-coordinate -6, we get -4. And if we add 1 to the y-coordinate -1, we get 0. Therefore, the next point is \(-4, 0\).

Step 2: Find the Second Point

Apply the same principle starting from point \(-4, 0\). Adding 2 to -4 gives -2. Adding 1 to 0 gives 1. So, the second additional point the line passes through is \(-2, 1\).

Step 3: Find the Third Point

Continuing the method, starting from the point \(-2, 1\), by adding the slope \(m=\frac{1}{2}\) gives us another new point. Adding 2 to -2 will give 0, and adding 1 to 1 will give 2. The third additional point is \(0, 2\).

Key Concepts

Slope-Intercept Form

The slope-intercept form is fundamental in understanding linear equations and graphing lines. It has the form
\[ y = mx + b \]
where \( m \) is the slope of the line, and \( b \) is the y-intercept; that is, the point where the line crosses the y-axis. The slope represents how steep the line is, showing the rise over run. For a slope of \( \frac{1}{2} \), like in the original exercise, the line rises 1 unit for every 2 units it runs horizontally. To identify additional points on the line using this form, you start with one known point, such as \( (-6, -1) \), and apply the slope to find the next points. The y-intercept in this form also quickly provides us with a specific point, where the line crosses the y-axis, which is at \( (0, b) \). This format allows for easy graphing and analysis of linear relationships.

Linear Equations

Linear equations form the basis of a wide range of mathematical and real-world applications. These are algebraic equations where each term is either a constant or the product of a constant and a single variable. Linear equations describe straight lines when graphed on a coordinate plane. The general form of a linear equation in two variables is \[ ax + by = c \], which can be rearranged into the slope-intercept form we previously discussed. To solve linear equations for different values of \( x \), you simply plug in those values into the equation and solve for \( y \), giving you the coordinates of various points that lie on the line described by the equation. The points found in the original exercise, \( (-4, 0), (-2, 1), \) and \( (0, 2) \), are all solutions to the linear equation derived from the given point and slope.

Coordinate Geometry

Coordinate geometry, also known as analytic geometry, seamlessly combines algebra and geometry. It involves the use of a coordinate system to geometrically represent algebraic equations. In this system, two lines (usually perpendicular) intersect at a point called the origin, defining the x-axis (horizontal) and y-axis (vertical). Each point in the plane is determined by an ordered pair of numbers, the coordinates: \( (x, y) \). In the context of the original exercise, \( (-6, -1) \) is one such point. Using coordinate geometry, we define the position of lines, shapes, and curves. When we talk about the slope of a line in this context, we're discussing how the line moves through this coordinate space. The method used in the exercise, moving from one point to another by following the slope, reflects the practical application of coordinate geometry in finding the points on a linear path. This approach is crucial in various fields, such as physics, engineering, and computer graphics.