Finding The Equation Of A Line Parallel To Another Line
Finding the equation of a line that meets specific criteria is a fundamental concept in algebra and coordinate geometry. This article will delve into the process of determining the equation of a line that passes through a given point and is parallel to another line. We will break down the steps involved, explain the underlying principles, and illustrate the concept with a detailed example.
Understanding Parallel Lines and Slopes
To effectively find the equation of a line parallel to another, it's crucial to grasp the concept of parallel lines and their slopes. Parallel lines are lines in a plane that never intersect, meaning they maintain a constant distance from each other. A fundamental property of parallel lines is that they have the same slope. The slope of a line, often denoted by 'm', represents the steepness and direction of the line. It is defined as the ratio of the change in the vertical coordinate (rise) to the change in the horizontal coordinate (run) between any two points on the line.
The slope is a crucial characteristic of a line, as it dictates its inclination. A positive slope indicates that the line rises as you move from left to right, while a negative slope indicates that the line falls. A slope of zero represents a horizontal line, and an undefined slope represents a vertical line. Understanding the relationship between slopes and parallel lines is the key to solving problems involving parallel lines.
The equation of a line is commonly expressed in two primary forms: slope-intercept form and point-slope form. The slope-intercept form is written as y = mx + b, where 'm' represents the slope and 'b' represents the y-intercept (the point where the line crosses the y-axis). This form is particularly useful when you know the slope and the y-intercept of the line. The point-slope form is written as y - y₁ = m(x - x₁), where 'm' is the slope and (x₁, y₁) is a known point on the line. This form is especially helpful when you know the slope and a point on the line, but not necessarily the y-intercept. Knowing these concepts and forms of linear equations sets the stage for our main task: determining the equation of a line parallel to a given one and passing through a specific point. Mastering these concepts provides a solid foundation for solving a wide range of problems in coordinate geometry and linear algebra.
Determining the Slope of the Given Line
Before we can find the equation of a line parallel to a given line, the first crucial step is to determine the slope of the original line. The slope, denoted as 'm', is the measure of the line's steepness and direction. There are several ways to determine the slope, depending on the information provided about the line. One common method is to use the slope-intercept form of a linear equation, which is y = mx + b, where 'm' represents the slope and 'b' represents the y-intercept. If the equation of the given line is already in slope-intercept form, identifying the slope is straightforward – it is simply the coefficient of the 'x' term. For example, if the equation of the given line is y = 2x + 3, the slope is 2.
However, if the equation of the line is not in slope-intercept form, it will need to be rearranged to isolate 'y' on one side of the equation. This involves algebraic manipulations such as adding or subtracting terms and dividing both sides of the equation by the appropriate coefficient. For instance, if the equation is given in standard form, Ax + By = C, you would subtract Ax from both sides to get By = -Ax + C, and then divide both sides by B to obtain y = (-A/B)x + C/B. In this case, the slope 'm' is -A/B.
Another way to determine the slope is if you are given two points on the line. If the coordinates of these points are (x₁, y₁) and (x₂, y₂), the slope can be calculated using the formula: m = (y₂ - y₁) / (x₂ - x₁). This formula represents the change in the y-coordinates divided by the change in the x-coordinates, which corresponds to the rise over the run. It’s important to ensure that the coordinates are subtracted in the same order in both the numerator and the denominator to obtain the correct slope. For example, if the points are (1, 2) and (3, 6), the slope would be (6 - 2) / (3 - 1) = 4 / 2 = 2.
In the specific question provided, we are told that the line we want to find is parallel to a graphed line. To determine the slope of the graphed line, we would need to identify two distinct points on the line from the graph. Once these points are identified, we can use the slope formula mentioned above to calculate the slope. Accurately determining the slope of the given line is a foundational step in finding the equation of a parallel line, as parallel lines have the same slope. Therefore, any errors in this step will propagate through the rest of the solution. With the slope of the given line in hand, we can proceed to find the equation of the parallel line passing through the specified point.
Using the Point-Slope Form
Once we have determined the slope of the given line, the next step in finding the equation of the parallel line is to utilize the point-slope form of a linear equation. The point-slope form is a powerful tool that allows us to construct the equation of a line when we know its slope and a point it passes through. The point-slope form is expressed as: y - y₁ = m(x - x₁), where 'm' represents the slope of the line, and (x₁, y₁) is the given point through which the line passes. This form is particularly useful in situations like the one we are addressing, where we have the slope (determined from the parallel line) and a specific point.
To apply the point-slope form, we simply substitute the known values into the equation. We replace 'm' with the slope we calculated in the previous step, and we replace x₁ and y₁ with the coordinates of the given point. In the problem presented, the line we seek passes through the point (8, -5). Thus, x₁ = 8 and y₁ = -5. By substituting these values into the point-slope form, we get: y - (-5) = m(x - 8). Simplifying the left side gives us y + 5 = m(x - 8). The next crucial step is to substitute the value of the slope 'm' that we determined earlier from the given line. Since parallel lines have the same slope, the slope 'm' for our desired line is the same as the slope of the line it is parallel to.
Once we substitute the slope into the equation, we will have an equation in the form y + 5 = m(x - 8), where 'm' is a specific numerical value. At this stage, the equation is in point-slope form. However, the goal is often to express the equation in slope-intercept form (y = mx + b) or standard form (Ax + By = C), as these forms are more commonly used and easier to interpret. To convert the equation from point-slope form to slope-intercept form, we need to distribute the slope 'm' across the terms inside the parentheses on the right side of the equation, and then isolate 'y' on the left side. This involves algebraic manipulations such as adding or subtracting terms on both sides of the equation. Converting to slope-intercept form provides the y-intercept directly, which can be useful for graphing or further analysis.
For example, if the slope 'm' is found to be 3/4, then substituting this value into our point-slope equation gives: y + 5 = (3/4)(x - 8). Distributing the 3/4 across the (x - 8) gives y + 5 = (3/4)x - 6. To isolate 'y', we subtract 5 from both sides, resulting in the equation y = (3/4)x - 11. This equation is now in slope-intercept form, where we can easily see that the slope is 3/4 and the y-intercept is -11. The careful and correct application of the point-slope form, combined with accurate algebraic manipulations, is essential for finding the desired equation of the parallel line.
Converting to Slope-Intercept Form
After applying the point-slope form and substituting the given point and the slope, the resulting equation is typically in the form y - y₁ = m(x - x₁). While this form is useful for constructing the equation, it is often more convenient and informative to express the equation in slope-intercept form, which is y = mx + b. The slope-intercept form provides a clear view of the line's slope ('m') and y-intercept ('b'), making it easier to graph the line and understand its properties. Converting the equation from point-slope form to slope-intercept form involves a few straightforward algebraic steps.
The first step in this conversion is to distribute the slope 'm' across the terms inside the parentheses on the right side of the equation. If the equation is in the form y - y₁ = m(x - x₁), distributing the 'm' gives us y - y₁ = mx - mx₁. For example, if our equation is y + 5 = (3/4)(x - 8), distributing the 3/4 across the (x - 8) results in y + 5 = (3/4)x - (3/4)*8, which simplifies to y + 5 = (3/4)x - 6.
Once the distribution is complete, the next step is to isolate 'y' on the left side of the equation. This is achieved by adding or subtracting the constant term on the left side (in this case, y₁) from both sides of the equation. The goal is to have 'y' by itself on one side, with all other terms on the other side. Continuing our example, we have y + 5 = (3/4)x - 6. To isolate 'y', we subtract 5 from both sides of the equation: y + 5 - 5 = (3/4)x - 6 - 5. This simplifies to y = (3/4)x - 11.
Now, the equation is in slope-intercept form, y = mx + b. In this form, it is easy to identify the slope and the y-intercept. The coefficient of 'x' is the slope ('m'), and the constant term is the y-intercept ('b'). In our example, y = (3/4)x - 11, the slope is 3/4, and the y-intercept is -11. This means the line rises 3 units for every 4 units it runs horizontally, and it crosses the y-axis at the point (0, -11). Converting the equation to slope-intercept form not only makes it easier to graph the line but also provides a clear understanding of its characteristics. This form is widely used in various mathematical applications and is essential for further analysis and problem-solving.
Applying the Concepts to the Specific Problem
Now that we have discussed the underlying principles and steps for finding the equation of a line parallel to another and passing through a given point, let's apply these concepts to the specific problem at hand. The problem asks us to find the equation of a line that passes through the point (8, -5) and is parallel to a graphed line. We are also given two options for the solution: A. y = (3/4)x - 11 and B. y = (3/4)x + 1. To solve this problem, we will follow the steps we have outlined: first, determine the slope of the graphed line; second, use the point-slope form to construct the equation of the parallel line; and third, convert the equation to slope-intercept form to match the given options.
Without the actual graph of the line, we will assume that the slope of the graphed line has been determined to be 3/4. This slope is crucial because parallel lines have the same slope. Therefore, the line we are trying to find will also have a slope of 3/4. With the slope determined, we proceed to use the point-slope form of a linear equation, which is y - y₁ = m(x - x₁). We know the line passes through the point (8, -5), so x₁ = 8 and y₁ = -5. We also know the slope, m, is 3/4. Substituting these values into the point-slope form gives us: y - (-5) = (3/4)(x - 8). Simplifying the left side, we get y + 5 = (3/4)(x - 8).
Next, we convert this equation to slope-intercept form (y = mx + b) to compare it with the given options. To do this, we first distribute the 3/4 across the terms inside the parentheses on the right side: y + 5 = (3/4)x - (3/4)*8. Simplifying the multiplication, we get y + 5 = (3/4)x - 6. Now, we isolate 'y' by subtracting 5 from both sides of the equation: y + 5 - 5 = (3/4)x - 6 - 5. This simplifies to y = (3/4)x - 11. This equation is now in slope-intercept form.
Comparing our result, y = (3/4)x - 11, with the given options, we see that it matches option A. Therefore, the equation of the line that passes through the point (8, -5) and is parallel to the graphed line (with an assumed slope of 3/4) is y = (3/4)x - 11. This confirms that option A is the correct answer. Option B, y = (3/4)x + 1, has the same slope but a different y-intercept, so it represents a parallel line but does not pass through the point (8, -5). This step-by-step application of the concepts demonstrates how we can systematically solve such problems by understanding the properties of parallel lines, using the point-slope form, and converting to slope-intercept form.
Conclusion
In summary, finding the equation of a line that is parallel to a given line and passes through a specific point involves a series of well-defined steps. The key concepts to understand are the properties of parallel lines, the slope-intercept form, and the point-slope form of a linear equation. First, it is essential to determine the slope of the given line, as parallel lines share the same slope. This can be done by either reading the slope directly from the slope-intercept form of the equation or by calculating it using two points on the line.
Once the slope is known, the point-slope form, y - y₁ = m(x - x₁), becomes the primary tool for constructing the equation of the parallel line. The given point through which the line must pass is substituted into the equation as (x₁, y₁), and the slope 'm' is the same as that of the given line. This step results in an equation that represents the desired line but is not yet in the most convenient form.
To make the equation more accessible and comparable with standard forms, it is then converted to slope-intercept form, y = mx + b. This conversion involves distributing the slope and isolating 'y' on one side of the equation. The resulting equation provides a clear view of the line's slope and y-intercept, making it easy to graph and analyze.
Applying these steps systematically allows us to solve a variety of problems involving parallel lines. By understanding the relationships between slopes and intercepts, we can efficiently find equations that meet specific criteria. This skill is fundamental in algebra and coordinate geometry, with applications in various fields such as physics, engineering, and computer graphics. The ability to find equations of parallel lines is not only a valuable mathematical skill but also an essential tool for problem-solving in many real-world contexts. Mastering these concepts provides a solid foundation for tackling more complex geometric and algebraic problems.
