Equation Of Line CD Passing Through Points C(3,-5) And D(6,0) In Standard Form

JU07/28/2025 08, 2025 by THE IDEN 79 views

Finding the Equation of Line CD in Standard Form

In mathematics, determining the equation of a line is a fundamental concept with numerous applications. Today, we will explore how to find the equation of a line in standard form, given two points on the line. Specifically, we will focus on line CD, which passes through points C(3, -5) and D(6, 0). This exercise will not only reinforce our understanding of linear equations but also demonstrate the practical steps involved in converting point-slope form to standard form.

Understanding the Problem

Before diving into the solution, let's break down the problem. We are given two points, C(3, -5) and D(6, 0), and our goal is to find the equation of the line that passes through these points. The equation needs to be in standard form, which is typically represented as Ax + By = C, where A, B, and C are integers, and A is a non-negative integer. This form is particularly useful for quickly identifying intercepts and comparing different linear equations. The process involves several steps, including finding the slope of the line, using the point-slope form, and then converting it to the standard form. Understanding each step is crucial not only for this specific problem but also for tackling various problems related to linear equations. By mastering these concepts, one can efficiently analyze and manipulate linear relationships in various mathematical and real-world scenarios.

Step 1: Calculate the Slope

The first step in finding the equation of line CD is to calculate its slope. The slope, often denoted as m, represents the steepness and direction of the line. It is defined as the change in the y-coordinate divided by the change in the x-coordinate between two points on the line. The formula for the slope m given two points (x₁, y₁) and (x₂, y₂) is:

m = (y₂ - y₁) / (x₂ - x₁)

In our case, we have the points C(3, -5) and D(6, 0). Let's assign these values:

x₁ = 3
y₁ = -5
x₂ = 6
y₂ = 0

Plugging these values into the slope formula, we get:

m = (0 - (-5)) / (6 - 3) = 5 / 3

Thus, the slope of line CD is 5/3. This positive slope indicates that the line rises from left to right. The magnitude of the slope (5/3) tells us that for every 3 units we move to the right along the x-axis, we move 5 units up along the y-axis. Understanding how to calculate the slope is fundamental in linear algebra, as it provides critical information about the line's orientation and rate of change. This calculation forms the basis for all subsequent steps in determining the line's equation.

Step 2: Use the Point-Slope Form

Now that we have calculated the slope m = 5/3, the next step is to use the point-slope form of a linear equation. The point-slope form is a versatile way to represent a line, especially when we know the slope and a point on the line. The general form of the point-slope equation is:

y - y₁ = m( x - x₁)

where (x₁, y₁) is a point on the line and m is the slope. We can use either point C(3, -5) or point D(6, 0) for (x₁, y₁). Let's use point C(3, -5). Substituting the values, we have:

y - (-5) = (5/3)(x - 3)

Simplifying, we get:

y + 5 = (5/3)(x - 3)

This equation is in point-slope form. It accurately describes the line CD and is useful for visualizing the line's behavior. The point-slope form is particularly advantageous because it directly incorporates the slope and a point on the line, making it easy to construct the equation. However, to meet the requirements of the problem, we need to convert this equation into standard form, which is the next crucial step in our solution. This conversion will involve algebraic manipulation to rearrange the equation into the desired format.

Step 3: Convert to Standard Form

The final step is to convert the point-slope equation we obtained in the previous section into standard form. The standard form of a linear equation is Ax + By = C, where A, B, and C are integers, and A is non-negative. Our equation in point-slope form is:

y + 5 = (5/3)(x - 3)

First, we need to eliminate the fraction by multiplying both sides of the equation by 3:

3(y + 5) = 3 * (5/3)(x - 3)

3y + 15 = 5(x - 3)

Next, distribute the 5 on the right side:

3y + 15 = 5x - 15

Now, rearrange the terms to get the standard form. We want the x and y terms on the same side and the constant term on the other side. Subtract 5x from both sides:

-5x + 3y + 15 = -15

Subtract 15 from both sides:

-5x + 3y = -30

To ensure that A is non-negative, multiply the entire equation by -1:

5x - 3y = 30

Thus, the equation of line CD in standard form is 5x - 3y = 30. This standard form is advantageous because it clearly presents the coefficients and constant, making it easier to analyze and compare with other linear equations. Converting to standard form often involves several algebraic steps, but it is a crucial skill for solving linear equations and understanding linear relationships in mathematics.

Verification and Conclusion

To verify our solution, we can substitute the coordinates of points C(3, -5) and D(6, 0) into the equation 5x - 3y = 30 and check if the equation holds true for both points.

For point C(3, -5):

5(3) - 3(-5) = 15 + 15 = 30

For point D(6, 0):

5(6) - 3(0) = 30 - 0 = 30

Since the equation holds true for both points, our solution is correct. We have successfully found the equation of line CD in standard form.

In conclusion, the equation of line CD in standard form is 5x - 3y = 30. This problem demonstrates the essential steps in finding the equation of a line given two points: calculating the slope, using the point-slope form, and converting to standard form. These skills are fundamental in algebra and have broad applications in various fields, including engineering, economics, and computer science. Mastering these concepts allows for a deeper understanding of linear relationships and their mathematical representation. The ability to manipulate and solve linear equations is a critical component of mathematical literacy and problem-solving proficiency.

This process not only provides the answer but also reinforces the understanding of linear equations and their representations. The standard form is particularly useful for quick analysis and comparison of linear equations, making it a valuable skill for any mathematics student or professional. By following these steps, you can confidently tackle similar problems and further enhance your mathematical abilities.