Finding The Equation Of A Line With Slope And A Point

Have you ever wondered how to pinpoint the exact equation of a line when you know its slope and a point it passes through? It's a fundamental concept in mathematics, and mastering it opens doors to understanding more complex topics in algebra and calculus. In this guide, we'll walk you through the process step by step, using a friendly and accessible approach.

Understanding Slope-Intercept Form

Before we dive into solving the problem, let's quickly recap the slope-intercept form of a linear equation. This form is expressed as:

y = mx + b

Where:

• y represents the dependent variable (usually plotted on the vertical axis)
• x represents the independent variable (usually plotted on the horizontal axis)
• m represents the slope of the line, indicating its steepness and direction
• b represents the y-intercept, the point where the line crosses the y-axis

The beauty of the slope-intercept form is its simplicity. It allows us to easily visualize and understand the characteristics of a line. The slope (m) tells us how much the y-value changes for every unit change in the x-value. A positive slope indicates an upward trend, while a negative slope indicates a downward trend. The y-intercept (b) gives us a fixed point on the line, which serves as a reference for plotting the entire line.

Grasping the Significance of Slope

Let's delve deeper into the concept of slope. Imagine you're hiking up a mountain. The slope represents the steepness of the trail. A steep slope means you'll be climbing quickly, while a gentle slope means you'll be ascending gradually. Similarly, in the context of a line, the slope tells us how rapidly the y-value changes as the x-value changes.

A line with a slope of 2, for instance, rises 2 units for every 1 unit increase in the x-value. Conversely, a line with a slope of -1 descends 1 unit for every 1 unit increase in the x-value. A horizontal line has a slope of 0, indicating no change in the y-value, while a vertical line has an undefined slope, as the change in the x-value is zero.

Unveiling the Y-Intercept

The y-intercept, denoted by b, is the point where the line intersects the y-axis. It's the value of y when x is equal to 0. Think of it as the starting point of the line on the y-axis. It provides a crucial anchor point for drawing the line and understanding its position on the coordinate plane. For example, if the y-intercept is 3, the line passes through the point (0, 3).

Problem Setup: Slope and a Point

Now that we have a solid understanding of slope-intercept form, let's tackle the problem at hand. We're given the slope of a line, which is $\frac{1}{2}$, and a point it passes through, which is (6, -3). Our mission is to find the equation of this line in slope-intercept form.

In essence, we know the value of m (the slope) and a pair of x and y coordinates that satisfy the equation. What we need to find is the value of b (the y-intercept). Once we have both m and b, we can plug them into the slope-intercept form to get the equation of the line.

The point-slope form is a powerful tool for finding the equation of a line when you know a point on the line and the slope. Guys, this form is expressed as:

y - y₁ = m(x - x₁)

Where:

• (x₁, y₁) is a known point on the line
• m is the slope of the line

This form directly incorporates the slope and a point on the line, making it a convenient starting point for our problem. Let's see how we can use it to find the equation of the line.

Applying the Point-Slope Form

To apply the point-slope form, we'll substitute the given values into the equation. We know that the slope m is $\frac{1}{2}$ and the point (6, -3) lies on the line. So, we can plug in x₁ = 6, y₁ = -3, and m = \frac{1}{2} into the point-slope form:

y - (-3) = \frac{1}{2}(x - 6)

Simplifying the equation, we get:

y + 3 = \frac{1}{2}(x - 6)

This is the equation of the line in point-slope form. However, our goal is to express the equation in slope-intercept form, which is y = mx + b. So, we need to do some algebraic manipulation to transform the equation.

Transforming to Slope-Intercept Form

To transform the equation to slope-intercept form, we need to isolate y on one side of the equation. Let's start by distributing the $\frac{1}{2}$ on the right side:

y + 3 = \frac{1}{2}x - 3

Next, we subtract 3 from both sides to isolate y:

y = \frac{1}{2}x - 3 - 3

Combining the constants, we get:

y = \frac{1}{2}x - 6

Eureka! We've successfully transformed the equation into slope-intercept form. Now we can clearly see the slope and the y-intercept of the line.

The Solution: Slope-Intercept Form

The equation of the line with a slope of $\frac{1}{2}$ that passes through the point (6, -3) in slope-intercept form is:

y = \frac{1}{2}x - 6

This equation tells us that the line has a slope of $\frac{1}{2}$ and a y-intercept of -6. We can now easily graph this line by plotting the y-intercept (0, -6) and using the slope to find another point on the line. For example, since the slope is $\frac{1}{2}$, we can move 2 units to the right and 1 unit up from the y-intercept to find another point (2, -5) on the line. Connecting these two points will give us the graph of the line.

Verifying the Solution

To ensure our solution is correct, we can plug the given point (6, -3) into the equation and see if it satisfies the equation:

-3 = \frac{1}{2}(6) - 6

-3 = 3 - 6

-3 = -3

The equation holds true, which confirms that the point (6, -3) lies on the line represented by the equation y = \frac{1}{2}x - 6. This verification step is a crucial habit to develop in mathematics, as it helps prevent errors and builds confidence in your solutions.

Alternative Approach: Direct Substitution

While the point-slope form provides a structured approach, there's another way to solve this problem using direct substitution. We know the slope-intercept form is y = mx + b, and we're given the slope m = \frac{1}{2}. So, we can write:

y = \frac{1}{2}x + b

Now, we know that the point (6, -3) lies on the line, which means its coordinates must satisfy the equation. So, we can substitute x = 6 and y = -3 into the equation:

-3 = \frac{1}{2}(6) + b

Simplifying the equation:

-3 = 3 + b

Subtracting 3 from both sides, we get:

b = -6

Now that we have the value of b, we can plug it back into the equation y = \frac{1}{2}x + b:

y = \frac{1}{2}x - 6

Voila! We arrived at the same solution using a different method. This demonstrates that there can be multiple paths to the correct answer in mathematics, and it's beneficial to explore different approaches to deepen your understanding.

Key Takeaways

Let's recap the key concepts and steps involved in finding the equation of a line when given its slope and a point:

1. Understand the slope-intercept form (y = mx + b) and its components (slope m and y-intercept b).
2. Utilize the point-slope form (y - y₁ = m(x - x₁)) as a starting point when you know a point and the slope.
3. Transform the point-slope form into slope-intercept form by isolating y.
4. Alternatively, use direct substitution by plugging the given slope and point into the slope-intercept form and solving for b.
5. Verify your solution by plugging the given point into the equation and ensuring it holds true.

By mastering these techniques, you'll be well-equipped to tackle various problems involving linear equations. Remember, practice makes perfect, so don't hesitate to work through numerous examples to solidify your understanding. Guys, keep exploring the fascinating world of mathematics!

Further Exploration

If you're eager to delve deeper into linear equations, here are some avenues for further exploration:

• Explore different forms of linear equations, such as standard form and intercept form.
• Investigate parallel and perpendicular lines and their relationships in terms of slopes.
• Apply linear equations to real-world problems, such as modeling linear relationships in physics, economics, and other fields.
• Delve into systems of linear equations and methods for solving them.

Mathematics is a vast and interconnected field, and the more you explore, the more you'll appreciate its beauty and power. So, keep learning, keep questioning, and keep pushing the boundaries of your mathematical knowledge!