Equation Of A Perpendicular Line: Step-by-Step Guide
Introduction
Hey guys! In this article, we're going to dive into a super common problem in algebra: finding the equation of a line that's perpendicular to a given line and passes through a specific point. This might sound a bit intimidating at first, but don't worry, we'll break it down step by step so it's crystal clear. We'll tackle the problem: Write the equation of a line that is perpendicular to $y=\frac{7}{5} x+6$ and that passes through the point (2,-6).
Understanding perpendicular lines is crucial in various fields, from architecture and engineering to computer graphics and even everyday tasks like aligning picture frames. So, let's get started and unravel this mathematical puzzle together!
Understanding Slope and Perpendicular Lines
Okay, first things first, let's talk about slope. Slope is basically a measure of how steep a line is. It tells us how much the line rises (or falls) for every unit it runs horizontally. Mathematically, we represent slope as "m," and it's calculated as the change in y divided by the change in x (rise over run). The equation $y=\frac{7}{5} x+6$ is in slope-intercept form, which is written as $y = mx + b$, where m is the slope and b is the y-intercept (the point where the line crosses the y-axis). In our given equation, the slope (m) is $\frac{7}{5}$. This means for every 5 units we move to the right along the x-axis, the line goes up 7 units along the y-axis. Understanding this concept is fundamental because the slope of a line dictates its direction and steepness, which are key elements when dealing with perpendicularity.
Now, what about perpendicular lines? Perpendicular lines are lines that intersect at a right angle (90 degrees). The relationship between their slopes is super important: the slopes of perpendicular lines are negative reciprocals of each other. This means if one line has a slope of m, a line perpendicular to it will have a slope of $- \frac{1}{m}$. To get the negative reciprocal, you flip the fraction and change the sign. For example, if a line has a slope of 2 (which can be written as $\frac{2}{1}$), a line perpendicular to it will have a slope of $- \frac{1}{2}$. If the original slope is a negative fraction, the perpendicular slope will be positive. This inverse relationship ensures the lines meet at a perfect right angle, a cornerstone principle in geometry and design. The concept of negative reciprocals might seem abstract, but it’s the magic behind creating perpendicular lines. Recognizing and applying this relationship correctly is crucial for solving problems like the one we're tackling today.
So, in our case, the given line has a slope of $\frac{7}{5}$. To find the slope of a line perpendicular to it, we need to find the negative reciprocal of $\frac{7}{5}$. Flip the fraction and change the sign, and we get $- \frac{5}{7}$. This is the slope of our new line! Remember, this negative reciprocal relationship is the key to ensuring the two lines intersect at a 90-degree angle, which is the very definition of perpendicularity. Understanding this principle makes it possible to visualize and calculate the direction of lines that meet at right angles, a crucial skill in various mathematical and real-world applications.
Point-Slope Form
Alright, now that we know the slope of our perpendicular line, we need to find its equation. To do this, we'll use something called the point-slope form. The point-slope form is a super handy way to write the equation of a line when you know a point on the line and its slope. The formula looks like this: $y - y_1 = m(x - x_1)$, where m is the slope, and $(x_1, y_1)$ is the given point. This form is derived directly from the definition of slope and provides a straightforward method for constructing the equation of a line based on specific known conditions.
The point-slope form is particularly useful because it bypasses the need to first determine the y-intercept, which can sometimes be a more roundabout process. Instead, it directly incorporates the slope and a point through which the line passes, allowing for a quick and efficient equation formulation. The formula reflects the consistent rate of change (slope) across the line, ensuring that any point (x, y) satisfying the equation will align with the specified slope and the given point. Mastering the point-slope form is essential for anyone studying linear equations, as it provides a flexible and powerful tool for solving a wide range of problems involving lines. It's not just a formula to memorize; it's a way of thinking about how lines are defined by their slope and the points they pass through.
We're given the point (2, -6), which means $x_1 = 2$ and $y_1 = -6$. We also found that the slope of the perpendicular line is $m = - \frac5}{7}$. Now we just plug these values into the point-slope form{7}(x - 2)$. See how we're just substituting the numbers we know into the formula? This is the beauty of the point-slope form – it makes the process super systematic and clear. By inserting the given coordinates and the calculated slope, we've set the stage to reveal the equation of our perpendicular line. It's like filling in the blanks in a template, which demystifies the process and makes it accessible even if you're new to linear equations.
Converting to Slope-Intercept Form
Okay, we've got our equation in point-slope form: $y + 6 = - \frac{5}{7}(x - 2)$. But sometimes, it's helpful to have the equation in slope-intercept form ($y = mx + b$) because it makes it easy to see the slope and y-intercept at a glance. To convert from point-slope form to slope-intercept form, we just need to do a little bit of algebra. We'll start by distributing the $- \frac{5}{7}$ on the right side of the equation.
This distribution step is crucial because it removes the parentheses, allowing us to isolate y and rearrange the equation into the familiar slope-intercept format. When we distribute, we're essentially applying the slope to every point along the line, ensuring that the equation accurately reflects the line's behavior. This process might seem like a simple algebraic manipulation, but it's a fundamental technique for transforming equations into a more usable and interpretable form. Mastering this step is key to understanding and working with linear equations effectively, as it allows us to easily identify the line's slope and y-intercept, two critical parameters that define its position and direction on a graph. The ability to move between different forms of an equation is a powerful tool in algebra, providing flexibility in problem-solving and a deeper understanding of the underlying concepts.
So, let's do it: $y + 6 = - \frac{5}{7}x + \frac{10}{7}$. Notice how we multiplied $- \frac{5}{7}$ by both x and -2. This is super important to get right! Distributing correctly ensures that we maintain the equality of the equation and accurately represent the line's properties. It's a bit like making sure every ingredient is added in the right proportion when baking a cake – if you miss a step or miscalculate, the final result won't be what you expect. In this case, accurate distribution leads us closer to the final, simplified form of the equation, where the slope and y-intercept are clearly visible.
Next, we need to isolate y on the left side. To do this, we'll subtract 6 from both sides of the equation. Remember, whatever you do to one side of the equation, you have to do to the other to keep it balanced. This principle of equality is a cornerstone of algebra and ensures that we're not changing the fundamental meaning of the equation as we manipulate it. Subtracting 6 from both sides is a strategic move that brings us closer to the desired $y = mx + b$ format, where y is expressed in terms of x and a constant. This step simplifies the equation and makes it easier to interpret, revealing the relationship between x and y that defines our line. By maintaining balance and performing the same operation on both sides, we're carefully transforming the equation while preserving its integrity.
So, we get: $y = - \frac5}{7}x + \frac{10}{7} - 6$. Now, we need to combine the constant terms. To do this, we need a common denominator. We can rewrite 6 as $\frac{42}{7}$, so the equation becomes{7}x + \frac{10}{7} - \frac{42}{7}$. Finding a common denominator is a fundamental skill when adding or subtracting fractions, and it's essential for simplifying expressions like this one. By expressing both constants with the same denominator, we create a level playing field for the subtraction, allowing us to combine them accurately. This step is a bit like converting measurements to the same unit before adding them – it ensures that we're comparing apples to apples. In the context of our equation, combining the constants helps us to finalize the y-intercept, which is a crucial piece of information for understanding the line's position on the graph.
Finally, we combine the fractions: $y = - \frac{5}{7}x - \frac{32}{7}$. Ta-da! We've got our equation in slope-intercept form. This final step is where all our previous work comes together, resulting in a clear and concise representation of the line's equation. The slope-intercept form is like the final, polished version of a document, where all the information is organized and easy to read. In this case, we can immediately see that the slope is $- \frac{5}{7}$ (which we already knew) and the y-intercept is $- \frac{32}{7}$. This means the line crosses the y-axis at the point (0, $- \frac{32}{7}$). This form is super useful for graphing the line or comparing it to other lines. The journey from the initial problem to this final equation highlights the power of algebraic manipulation and the systematic approach to solving mathematical problems.
Conclusion
So, there you have it! We've successfully found the equation of a line that is perpendicular to $y=\frac{7}{5} x+6$ and passes through the point (2, -6). The equation is $y = - \frac{5}{7}x - \frac{32}{7}$. We did it by understanding the relationship between the slopes of perpendicular lines, using the point-slope form, and converting to slope-intercept form. You guys rock!
Remember, the key to mastering these kinds of problems is practice. The more you work with linear equations, the more comfortable you'll become with the concepts and the steps involved. So keep practicing, and you'll be a pro in no time! Happy solving!
