Equation Of A Parallel Line Through A Point (4, 7)
Have you ever wondered how to determine the equation of a line that runs parallel to another, especially when you have a specific point it needs to pass through? It's a common question in mathematics, and in this guide, we'll break down the process step-by-step. We'll focus on understanding the core concepts, applying them to a practical example, and ensuring you grasp the method thoroughly. Let's dive in and explore how to find the equation of a line parallel to y = -3x + 5 that includes the point (4, 7).
Understanding Parallel Lines and Their Slopes
To begin, let's clarify what parallel lines actually are. In simple terms, parallel lines are lines that run in the same direction and never intersect. A crucial characteristic of parallel lines is that they have the same slope. The slope of a line, often denoted as 'm', represents its steepness and direction. It tells us how much the line rises or falls for every unit of horizontal change.
The slope-intercept form of a linear equation, which is y = mx + b, is instrumental in identifying the slope. In this form, 'm' represents the slope, and 'b' represents the y-intercept (the point where the line crosses the y-axis). Understanding this form is the cornerstone to solving our problem.
Consider the given line, y = -3x + 5. By comparing it to the slope-intercept form, we can immediately identify the slope. The coefficient of 'x' is -3, so the slope of this line is -3. Therefore, any line parallel to this one will also have a slope of -3. This principle is fundamental to finding the equation of our desired line. Knowing that parallel lines share the same slope allows us to take the first crucial step in determining the new equation.
The slope is the heart of a linear equation, dictating its direction and steepness. When lines are parallel, they move in perfect synchronicity, never meeting, and this shared direction is mathematically represented by their equal slopes. This concept not only simplifies calculations but also offers a clear visual understanding of linear relationships. We will build upon this foundation as we delve deeper into the process of finding the equation of a parallel line, ensuring you have a solid grasp of why and how each step contributes to the final solution.
Utilizing the Point-Slope Form
Now that we know the slope of our parallel line, which is -3, and we have a specific point (4, 7) that the line must pass through, we can employ the point-slope form of a linear equation. This form is particularly useful when you have a point (x₁, y₁) on the line and the slope 'm'. The point-slope form is given by:
**y - y₁ = m(x - x₁) **
This equation directly incorporates the slope and a point on the line, making it a perfect tool for our task. To use it, we simply substitute the known values. In our case, m = -3, x₁ = 4, and y₁ = 7. Plugging these values into the point-slope form, we get:
y - 7 = -3(x - 4)
This equation represents the line we're looking for, but it's not yet in the familiar slope-intercept form (y = mx + b). Our next step is to simplify this equation and convert it to that form.
The point-slope form is an invaluable asset in linear algebra because it bridges the gap between the geometric concept of a line passing through a specific point and the algebraic representation of its equation. It allows us to construct the equation directly from the slope and a single point, bypassing the need for the y-intercept initially. This method highlights the flexibility and power of different forms of linear equations, each serving a unique purpose in problem-solving. By mastering the point-slope form, you gain a versatile tool for tackling a wide range of linear equation problems, further solidifying your understanding of linear relationships.
Converting to Slope-Intercept Form
Our equation is currently in point-slope form: y - 7 = -3(x - 4). To make it more recognizable and easier to interpret, we need to convert it to slope-intercept form (y = mx + b). This involves simplifying the equation and isolating 'y' on one side.
First, distribute the -3 on the right side of the equation:
y - 7 = -3x + 12
Next, to isolate 'y', add 7 to both sides of the equation:
y = -3x + 12 + 7
Finally, combine the constants:
y = -3x + 19
Now our equation is in slope-intercept form. We can clearly see that the slope (m) is -3, which confirms that it is indeed parallel to the original line, and the y-intercept (b) is 19. This means the line crosses the y-axis at the point (0, 19).
Converting to slope-intercept form not only provides a clear view of the line's slope and y-intercept but also makes the equation easier to graph and compare with other linear equations. The process of simplification – distributing, adding, and combining like terms – is a fundamental skill in algebra, and mastering it is crucial for solving more complex equations. By transforming the point-slope form into slope-intercept form, we've not only found the equation of the parallel line but also demonstrated the interconnectedness of different forms of linear equations. This transformation underscores the importance of algebraic manipulation in revealing the underlying properties of linear relationships and solidifies our understanding of how to express the same line in various ways.
Verifying the Solution
To ensure our solution is correct, it's always a good practice to verify it. We have the equation y = -3x + 19, and we know that the line should pass through the point (4, 7). To verify, substitute x = 4 into the equation and see if we get y = 7.
y = -3(4) + 19 y = -12 + 19 y = 7
Since the calculation gives us y = 7, our equation is correct. The line y = -3x + 19 is indeed parallel to y = -3x + 5 and passes through the point (4, 7). This verification step is crucial because it confirms that our algebraic manipulations and substitutions were accurate, providing a high level of confidence in our solution.
Furthermore, we can conceptually verify our solution. The slope of the original line and the new line is the same (-3), confirming they are parallel. The y-intercepts are different (5 and 19), indicating they are distinct lines. The point (4, 7) satisfies the equation we derived, proving that our new line passes through this specified point. This comprehensive verification, combining both numerical substitution and conceptual understanding, solidifies the accuracy of our solution and reinforces the principles of linear equations.
Conclusion
In this guide, we've walked through the process of finding the equation of a line parallel to y = -3x + 5 that includes the point (4, 7). We started by understanding the key property of parallel lines: they have the same slope. We then used the point-slope form to create an initial equation and converted it to the more familiar slope-intercept form. Finally, we verified our solution to ensure accuracy. The final equation of the line is y = -3x + 19.
This problem exemplifies the practical application of linear equations and demonstrates how different forms of equations can be used to solve problems efficiently. By mastering these techniques, you'll be well-equipped to tackle a wide variety of problems involving linear relationships.
To further enhance your understanding of linear equations and related concepts, you can explore resources such as Khan Academy's linear algebra section. This will provide you with additional insights and practice opportunities.
