Problem 18 - College Algebra

Question

Use the given conditions to write an equation for each line in point-slope form and slope-intercept form. Slope \(=-2,\) passing through \((0,-3)\)

Step-by-Step Solution

Answer

The point-slope form of the line is \(y + 3 = -2x\) and the slope-intercept form is \(y = -2x -3\).

Step 1: Formulate the Point-Slope Form

The point-slope form of the line is \(y - y_1 = m(x - x_1)\) where \(m\) is the slope and \((x_1, y_1)\) is the point that the line passes through. Substituting the given values: \(m = -2\) and the point \((0, -3)\) into the equation, we get: \(y - (-3) = -2 * [x-0]\). Which simplifies to \(y + 3 = -2x\).

Step 2: Convert the Point-Slope Form into Slope-Intercept Form

The slope-intercept form of the line is \(y = mx + c\) which, by rearranging the equation from point-slope form \((y + 3 = -2x)\), results in \(y = -2x - 3\).

Key Concepts

Slope-Intercept Form

In mathematics, the slope-intercept form is used to express the equation of a straight line. This form is represented as: \(y = mx + c\), where:

• \(y\) is the dependent variable representing the y-coordinate of a point on the graph.
• \(m\) is the slope of the line, indicating its steepness and direction.
• \(x\) is the independent variable representing the x-coordinate of a point.
• \(c\) is the y-intercept, the point where the line crosses the y-axis.

This form is particularly useful because it readily provides the slope and y-intercept, allowing one to quickly graph the line or understand its behavior. The conversion from point-slope form to slope-intercept form requires simply rearranging the equation, ensuring \(y\) is isolated. For example, from \(y + 3 = -2x\) in the point-slope form, it can be rearranged to \(y = -2x - 3\). Here, the slope \(m\) is \(-2\), and the y-intercept \(c\) is \(-3\).

Linear Equations

Linear equations describe straight lines in algebra. A key feature of these equations is their ability to represent constant relationships between two variables, usually x and y. The general form of a linear equation is \(ax + by = c\), but it can also be presented in slope-intercept form \(y = mx + c\).

• They have a fixed slope (steepness) that never changes along the line.
• Linear equations do not have exponents higher than 1 for the variables.
• The graph of a linear equation is always a straight line.

In practical problems, linear equations help model relationships between quantities, such as speed and time or cost and items bought. Solving a linear equation involves finding the values of variables that make the equation true, often revealing their plotted line's path and intersection with axes.

Coordinates

Coordinates are essential in geometry and algebra as they provide a way to locate points on a graph. Typically expressed in the form \((x, y)\), coordinates contain:

• The x-coordinate, which determines a specific position along the horizontal x-axis.
• The y-coordinate, which determines a position along the vertical y-axis.

This coordinate pair allows for precise positioning of points in a two-dimensional space. For example, the point \((0, -3)\) signifies a location on the y-axis, indicating that this line passes through the y-axis at \(-3\). When working with linear equations like the point-slope form, understanding coordinates aids in plotting the initial point the line passes through, facilitating the graph's creation and analysis.

Algebraic Equations

Algebraic equations represent relationships between variables and constants, often involving addition, subtraction, multiplication, and division. These equations can be simple, like linear equations, or complex with higher-degree polynomials. In the context of this exercise:

• The algebraic equation in point-slope form is \(y - y_1 = m(x - x_1)\).
• Algebraic equations require substitution of known values to find unknowns.
• They are manipulated through operations to solve for variables or simplify expressions.

Algebraic equations are the backbone of problems requiring solutions through manipulation of known and unknown quantities. By substituting the values \(m = -2\), \(x_1 = 0\), and \(y_1 = -3\) into the point-slope equation, we can derive a specific form illustrating the relationship between x and y for that line.