Finding Equivalent Equations Of A Line Passing Through Two Points

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In mathematics, particularly in coordinate geometry, the equation of a line is a fundamental concept. A line can be represented in several forms, such as slope-intercept form, point-slope form, and standard form. Understanding how to convert between these forms and identify equivalent equations is crucial for solving various problems. This article delves into the process of finding equivalent equations for a line, focusing on the given line passing through the points (−4,10)(-4, 10) and (−1,5)(-1, 5). We will explore how to determine the correct equation of the line and identify alternative equations that represent the same line. This exploration will enhance your understanding of linear equations and their various representations.

Understanding the Problem

At the heart of coordinate geometry lies the line, a fundamental entity defined by its unchanging slope and its constant y-intercept. Understanding the line begins with identifying two points: (−4,10)(-4, 10) and (−1,5)(-1, 5). The task is to find the equation of the line that passes through these points and to determine which of the given options also represent this line. This involves several steps, starting with finding the slope of the line. The slope, often denoted as m, is a measure of the steepness of the line and its direction (positive or negative). It is calculated using the formula: $m = \frac{y_2 - y_1}{x_2 - x_1}$ where (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) are the coordinates of the two given points. Once the slope is determined, we can use the point-slope form of a line, which is given by: $y - y_1 = m(x - x_1)$ This form allows us to create the equation of the line using the slope and one of the points. Subsequently, this equation can be converted into slope-intercept form, which is: $y = mx + b$ where b is the y-intercept. By manipulating the equation, we can identify equivalent forms that represent the same line. This process not only helps in finding the equation of the line but also in understanding the different ways in which a linear equation can be expressed. Therefore, this foundational knowledge is essential for tackling more complex problems in coordinate geometry and linear algebra. In this particular case, the initial step is to calculate the slope using the given points, followed by using the point-slope form to derive an equation. This derived equation can then be manipulated and compared with the given options to identify the equivalent equations. The ability to seamlessly transition between different forms of linear equations is a critical skill in mathematics, enabling a more versatile approach to problem-solving. This exercise will not only reinforce these basic concepts but also sharpen the analytical skills required to identify equivalent mathematical expressions.

Calculating the Slope

To begin, we must calculate the slope (m) of the line passing through the points (−4,10)(-4, 10) and (−1,5)(-1, 5). Calculating the slope is a critical first step because the slope provides essential information about the line's direction and steepness. The formula for the slope, $m = \fracy_2 - y_1}{x_2 - x_1}$, is a cornerstone of coordinate geometry. In this case, we assign (−4,10)(-4, 10) as (x1,y1)(x_1, y_1) and (−1,5)(-1, 5) as (x2,y2)(x_2, y_2). Substituting these values into the slope formula gives us $m = \frac{5 - 10{-1 - (-4)} = \frac{-5}{-1 + 4} = \frac{-5}{3} = -\frac{5}{3}$ Thus, the slope of the line is −53-\frac{5}{3}. This negative slope indicates that the line decreases as we move from left to right on the coordinate plane. This calculation is more than just plugging numbers into a formula; it provides a fundamental understanding of how the line behaves. The slope, −53-\frac{5}{3}, means that for every 3 units we move to the right along the x-axis, the line descends 5 units along the y-axis. This steep negative slope will influence the subsequent steps in determining the equation of the line. With the slope calculated, we can now proceed to use the point-slope form of a line to derive the equation. The accuracy of this slope calculation is paramount, as it serves as the foundation for all further calculations and analyses related to this line. Any error in this initial step would propagate through the rest of the solution, leading to an incorrect equation and potentially wrong answers. Therefore, meticulous attention to detail and a solid understanding of the slope formula are crucial for correctly identifying the line's equation. This precise calculation of the slope is not only a necessary step in this problem but also a fundamental skill in various mathematical and real-world applications involving linear relationships.

Using Point-Slope Form

Having calculated the slope, the next crucial step is to use the point-slope form of a line to derive the equation. Using the point-slope form is an efficient method for constructing the equation of a line when the slope and a point on the line are known. The point-slope form is given by: $y - y_1 = m(x - x_1)$ We already have the slope, m = −53-\frac{5}{3}, and we can use either of the given points, (−4,10)(-4, 10) or (−1,5)(-1, 5), as (x1,y1)(x_1, y_1). Let's use the point (−1,5)(-1, 5) for this calculation. Substituting these values into the point-slope form gives us: $y - 5 = -\frac5}{3}(x - (-1))$ Simplifying this equation, we get $y - 5 = -\frac{5{3}(x + 1)$ This equation is a valid representation of the line, but to compare it with the given options, we need to convert it into a more standard form, such as slope-intercept form (y=mx+by = mx + b). The beauty of the point-slope form is its directness; it allows us to build the equation of the line without needing to explicitly calculate the y-intercept. By using the known slope and a single point, we can capture the line's essential characteristics. This method highlights the fundamental relationship between the slope, a point on the line, and the line's overall equation. The point-slope form is particularly useful in situations where the y-intercept is not immediately obvious or when working with multiple points and slopes. It provides a flexible and reliable way to express a linear relationship. The process of transforming this point-slope equation into other forms will further clarify its nature and allow for easier comparison with other equations. This step is not just about finding an equation; it's about understanding the inherent structure of linear equations and how they can be manipulated to suit different contexts. The correct application of the point-slope form is a key skill in algebra and coordinate geometry, providing a powerful tool for solving a wide range of problems.

Converting to Slope-Intercept Form

After applying the point-slope form, the next step is to convert the equation into slope-intercept form. Converting to slope-intercept form is crucial for making the equation more easily comparable with standard linear equations. The slope-intercept form, $y = mx + b$, explicitly shows the slope (m) and the y-intercept (b), which provides a clear picture of the line's behavior. Starting from the equation derived in the previous step: $y - 5 = -\frac5}{3}(x + 1)$ We need to isolate y on one side of the equation. First, distribute the −53-\frac{5}{3} across the terms inside the parentheses $y - 5 = -\frac{53}x - \frac{5}{3}$ Next, add 5 to both sides of the equation to solve for y $y = -\frac{53}x - \frac{5}{3} + 5$ To combine the constants, we need to express 5 as a fraction with a denominator of 3 $y = -\frac{53}x - \frac{5}{3} + \frac{15}{3}$ Now, combine the fractions $y = -\frac{5{3}x + \frac{10}{3}$ This equation is now in slope-intercept form, where the slope m is −53-\frac{5}{3} and the y-intercept b is 103\frac{10}{3}. This form is particularly useful because it allows for a quick visual interpretation of the line. The coefficient of x gives the slope, and the constant term gives the y-coordinate where the line intersects the y-axis. Converting to slope-intercept form is not just a matter of algebraic manipulation; it's a process that reveals essential characteristics of the line. It allows for a direct comparison with other linear equations and simplifies graphing the line. This form is a standard and widely used representation in mathematics, making it easier to communicate and work with linear relationships. The ability to convert between different forms of linear equations, such as point-slope and slope-intercept, is a fundamental skill in algebra and provides a flexible approach to problem-solving. This conversion highlights the underlying structure of linear equations and their various representations.

Identifying Equivalent Equations

Now that we have the equation in slope-intercept form, we can identify equivalent equations. The equation we derived is: $y = -\frac{5}{3}x + \frac{10}{3}$ This form makes it easy to compare with the given options. Let's analyze each option to determine if it represents the same line.

  • Option 1: y=−53x−2y = -\frac{5}{3}x - 2

    This equation has the same slope, −53-\frac{5}{3}, but a different y-intercept, -2. Therefore, it does not represent the same line.

  • Option 2: y=−53x+103y = -\frac{5}{3}x + \frac{10}{3}

    This equation is identical to the one we derived, so it represents the same line.

  • Option 3: 3y=−5x+103y = -5x + 10

    To check this equation, we can divide both sides by 3 to get it into slope-intercept form: $y = -\frac{5}{3}x + \frac{10}{3}$ This equation is the same as the one we derived, so it represents the same line.

Thus, the equivalent equations are:

  • y=−53x+103y = -\frac{5}{3}x + \frac{10}{3}
  • 3y=−5x+103y = -5x + 10

The process of identifying equivalent equations involves comparing the equations in a common form, such as slope-intercept form. This allows for a direct comparison of the slopes and y-intercepts. Equivalent equations will have the same slope and y-intercept, indicating that they represent the same line on the coordinate plane. This skill is crucial in algebra for simplifying expressions, solving systems of equations, and understanding the relationships between different representations of the same mathematical concept. The ability to recognize and manipulate equations into equivalent forms is a fundamental aspect of mathematical fluency. It demonstrates a deep understanding of the underlying principles and allows for a more flexible and efficient approach to problem-solving. This analysis not only identifies the correct answers but also reinforces the understanding of linear equations and their properties. By comparing different forms and identifying the similarities and differences, we strengthen our grasp of these essential mathematical concepts.

Verifying with Given Points

To ensure our derived equation is correct, we can verify with the given points. This involves substituting the coordinates of the given points, (−4,10)(-4, 10) and (−1,5)(-1, 5), into the equation y=−53x+103y = -\frac{5}{3}x + \frac{10}{3} to see if they satisfy the equation. For the point (−4,10)(-4, 10): $10 = -\frac5}{3}(-4) + \frac{10}{3}$ $10 = \frac{20}{3} + \frac{10}{3}$ $10 = \frac{30}{3}$ $10 = 10$ The equation holds true for the point (−4,10)(-4, 10). For the point (−1,5)(-1, 5) $5 = -\frac{5{3}(-1) + \frac{10}{3}$ $5 = \frac{5}{3} + \frac{10}{3}$ $5 = \frac{15}{3}$ $5 = 5$ The equation also holds true for the point (−1,5)(-1, 5). Since both points satisfy the equation, we can be confident that our equation is correct. Verifying with given points is a crucial step in problem-solving. It serves as a final check to ensure that the derived equation accurately represents the line defined by the given points. This process involves a straightforward substitution and simplification, providing a concrete confirmation of the solution. This verification step is not just a formality; it is an integral part of the mathematical process, reinforcing the connection between the equation and the geometric representation of the line. By plugging in the coordinates and verifying that they satisfy the equation, we gain added assurance in the correctness of our solution. This method also highlights the importance of accuracy in each step of the problem-solving process, from calculating the slope to deriving the equation. If an error were made in any previous step, it would likely be revealed during this verification, underscoring the value of this final check. This rigorous verification process is a hallmark of sound mathematical practice and helps build confidence in the solution.

In this article, we walked through the process of finding equivalent equations for a line passing through two given points. We started by calculating the slope, then used the point-slope form to derive an equation, and finally converted it to slope-intercept form. By comparing the derived equation with the given options, we identified the equivalent equations. Additionally, we verified our solution by substituting the given points into the equation. This comprehensive approach ensures accuracy and a thorough understanding of linear equations. The ability to find and manipulate linear equations is a fundamental skill in mathematics, with applications in various fields, including physics, engineering, and economics. This article has provided a step-by-step guide to solving this type of problem, reinforcing the importance of each step and highlighting the connections between different forms of linear equations. The process of converting between different forms, such as point-slope and slope-intercept, is a powerful tool for problem-solving and provides a deeper understanding of the underlying mathematical concepts. Furthermore, the verification step underscores the importance of accuracy and attention to detail in mathematics. By following these steps, one can confidently solve similar problems and build a strong foundation in linear algebra. This knowledge not only helps in academic pursuits but also equips individuals with the analytical skills necessary for real-world applications. The journey through this problem has reinforced key mathematical principles and provided a practical guide to finding equivalent equations for a line. This understanding will serve as a valuable asset in future mathematical endeavors and beyond.