In mathematics, particularly in coordinate geometry, determining the equation of a line is a fundamental concept. Often, we encounter scenarios where we need to find the equation of a line that satisfies specific conditions, such as being parallel to another line and passing through a particular point. This article will delve into the process of finding the equation of a line that is parallel to a given line and passes through a specified point. We will use the example of finding a line parallel to $y = 0.6x + 3$ and passing through the point $(-3, -5)$.
Understanding Parallel Lines
When dealing with parallel lines, the most crucial concept to grasp is that they have the same slope. The slope of a line, often denoted as m, represents the steepness and direction of the line. In the slope-intercept form of a linear equation, $y = mx + b$, m is the coefficient of x and represents the slope, while b represents the y-intercept (the point where the line crosses the y-axis). Therefore, if two lines are parallel, their m values are identical. This understanding is the bedrock of solving problems involving parallel lines.
In our given equation, $y = 0.6x + 3$, we can clearly identify the slope. The coefficient of x is 0.6, which means the slope of this line is 0.6. Any line parallel to this one will also have a slope of 0.6. This immediately narrows down our options when we start constructing the equation of our new line. The slope is a defining characteristic of a line, and knowing it allows us to predict the line's direction and steepness on the coordinate plane.
Furthermore, the y-intercept, represented by b, determines where the line intersects the y-axis. While parallel lines share the same slope, they have different y-intercepts; otherwise, they would be the same line. This distinction is important because it means that while the parallel line we are trying to find will have the same steepness as the given line, it will be positioned differently on the graph. Grasping the relationship between slope and y-intercept is crucial for accurately determining the equation of a line.
In summary, identifying the slope of a given line is the first critical step in finding the equation of a parallel line. Since parallel lines have the same slope, this value becomes a constant in the equation of the new line. The next step involves using the given point to find the unique y-intercept that defines our specific parallel line.
Utilizing the Point-Slope Form
After pinpointing the slope, the next pivotal step involves incorporating the given point through which the line must pass. The point-slope form of a linear equation becomes invaluable here. This form is expressed as $y - y_1 = m(x - x_1)$, where m signifies the slope, and ($x_1$, $y_1$) represents the coordinates of the given point. The point-slope form elegantly allows us to construct the equation of a line using its slope and a single point on the line.
In our scenario, we have established that the slope (m) of the parallel line is 0.6. The point through which the line must pass is given as (-3, -5). Therefore, we can substitute these values into the point-slope form. Substituting m = 0.6, $x_1$ = -3, and $y_1$ = -5, we get: $y - (-5) = 0.6(x - (-3))$. This equation now encapsulates the information about the slope and the specific point, laying the groundwork for finding the equation of the parallel line.
The beauty of the point-slope form lies in its direct applicability. Once the slope and a point are known, the equation can be easily set up, minimizing the chances of errors. The subsequent steps involve simplifying this equation to the slope-intercept form, which is more commonly used and easier to interpret graphically. However, the point-slope form serves as a crucial intermediate step, ensuring the line passes through the intended point.
Moreover, understanding the point-slope form provides a deeper insight into the nature of linear equations. It showcases how the slope and a point collectively define a line's position and orientation on the coordinate plane. This understanding is not only useful for solving specific problems but also for developing a more intuitive grasp of linear relationships in mathematics.
In the next section, we will delve into simplifying the equation we derived from the point-slope form to obtain the slope-intercept form, which will give us the final equation of the line parallel to $y = 0.6x + 3$ and passing through the point (-3, -5).
Converting to Slope-Intercept Form
Having successfully utilized the point-slope form, the next critical step is to transform the equation into the more familiar slope-intercept form, $y = mx + b$. This form provides a clear view of the line's slope (m) and y-intercept (b), making it easier to graph and analyze the line's behavior. The process involves simplifying the equation we obtained from the point-slope form and isolating y on one side.
Our equation from the point-slope form is: $y - (-5) = 0.6(x - (-3))$. The first step in simplification is to address the double negatives, which gives us: $y + 5 = 0.6(x + 3)$. Next, we distribute the 0.6 on the right side of the equation: $y + 5 = 0.6x + 1.8$. This distribution is a crucial algebraic manipulation that allows us to separate the x term and the constant term.
To finally isolate y and achieve the slope-intercept form, we subtract 5 from both sides of the equation: $y = 0.6x + 1.8 - 5$. Performing the subtraction, we get: $y = 0.6x - 3.2$. This is the equation of the line in slope-intercept form, which is parallel to $y = 0.6x + 3$ and passes through the point (-3, -5).
The slope-intercept form provides immediate insights into the line's characteristics. We can see that the slope (m) is 0.6, which confirms that it is parallel to the given line. The y-intercept (b) is -3.2, indicating that the line intersects the y-axis at the point (0, -3.2). This form is particularly useful for graphing the line, as we have two key pieces of information: the steepness of the line and where it crosses the y-axis.
Furthermore, converting to slope-intercept form is a standard practice in linear algebra, making it easier to compare and analyze different lines. It also facilitates solving systems of linear equations, as the slopes and y-intercepts can be readily used in various methods such as substitution and elimination. Understanding how to manipulate equations to achieve this form is therefore a fundamental skill in mathematics.
Verification and Conclusion
After obtaining the equation, it's prudent to verify that it indeed satisfies the given conditions. We found the equation of the line to be $y = 0.6x - 3.2$. To verify, we need to ensure two things: first, the slope is the same as the given line, and second, the line passes through the point (-3, -5).
The slope of our equation is 0.6, which is the same as the slope of the original line, $y = 0.6x + 3$. This confirms that the lines are indeed parallel. Next, we substitute the coordinates of the point (-3, -5) into our equation: $-5 = 0.6(-3) - 3.2$. Calculating the right side, we get: $-5 = -1.8 - 3.2$, which simplifies to $-5 = -5$. This equality confirms that the point (-3, -5) lies on the line we found.
By verifying these two conditions, we can confidently conclude that the equation $y = 0.6x - 3.2$ accurately represents the line that is parallel to $y = 0.6x + 3$ and passes through the point (-3, -5). This process of verification is crucial in mathematics to ensure the correctness of solutions and to catch any potential errors in the calculations.
In conclusion, finding the equation of a line that is parallel to a given line and passes through a specific point involves several key steps. First, identify the slope of the given line, which will be the same for the parallel line. Second, use the point-slope form of a linear equation to incorporate the given point. Third, convert the equation to slope-intercept form to clearly identify the slope and y-intercept. Finally, verify the solution by ensuring the slope is correct and the point lies on the line. This methodical approach not only provides the correct answer but also deepens the understanding of linear equations and their properties. The ability to solve such problems is a fundamental skill in mathematics and has wide-ranging applications in various fields, from physics and engineering to economics and computer science.