Determining the Value of k for Parallel Lines

In this article, we will explore the mathematical process required to find the value of k that makes two lines parallel. This involves converting the given linear equations into slope-intercept form, and then equating their slopes. We'll also discuss the significance of the slope in determining whether two lines are parallel.

Introduction to Parallel Lines and Slope

Two lines are considered parallel if they maintain a constant distance apart from each other and never intersect. In the context of linear equations, this means that their slopes must be identical. The slope of a line given in the general form ax by c 0 can be calculated using the formula -a/b. Understanding this concept is the first step towards solving the problem at hand.

Solving with Slope-Intercept Form

First Equation: 3x – 3ky 2

To determine the slope of the first line represented by the equation 3x – 3ky 2, we convert it into the slope-intercept form y mx b.

Starting with the given equation:

3x – 3ky 2

Isolating y:

3ky -3x 2

Dividing by 3k:

y {-1/k}x 2/3k

The slope m1 of the first line is:

m1 -1/k

Second Equation: 2x – 5y 0

Similarly, for the second equation, we rearrange it into the slope-intercept form and solve for the slope.

Starting with the given equation:

2x – 5y 0

Isolating y:

5y -2x

Dividing by 5:

y -2/5x

The slope m2 of the second line is:

m2 -2/5

Equating the Slopes for Parallel Lines

Since the lines are parallel, their slopes must be equal:

m1 m2

Substituting the slopes we found:

-1/k -2/5

Removing the negative signs (optional but simplifies the equation):

1/k 2/5

Cross-multiplying to solve for k:

1 * 5 2 * k

5 2k

k 5/2

Thus, the value of k that makes the lines parallel is:

boxed{5/2}

Alternative Methods to Find the Value of k

While the slope-intercept form method is one effective way to solve for k, there are other algebraic techniques that can be employed:

First Alternative Method: Let's solve for y directly:

3x – 3ky 2 → 3ky -3x 2 → y -x/3k 2/3k

Slope of the first line: m1 -1/3k

2x – 5y 0 → 5y -2x → y -2/5x

Slope of the second line: m2 -2/5

Since m1 m2, we have: -1/3k -2/5

Solving for k: k 5/6

Second Alternative Method: Multiply the first equation by 2 and the second by 3:

3x – 3ky 2 → 6x – 6ky 4

2x – 5y 0 → 6x – 15y 0

Comparing the coefficients of y: -6k -15

Solving for k: k 15/6 5/2

Third Alternative Method: Using the slope directly:

3x – 2ky 2 → y -3/2kx 3/2k

2x – 5y 0 → y -2/5x - 1/5

Equating slopes: -2/5 -3/2k

Solving for k: 4k 15 → k 15/4 3.75

Conclusion

The value of k that makes the lines parallel can be found through various mathematical methods, including the slope-intercept form, direct algebraic manipulation, and coefficient comparison. The key takeaway is that the slopes of the lines must be identical for them to be parallel. Understanding and applying these methods will help in solving similar problems in the future.