Given An Equation Of A Line Find Equations For Parallel Line

Given An Equation Of A Line Find Equations For Lines Parallel Or Perp

Given an equation of a line, find equations for lines parallel or perpendicular to it passing through specified points. Simplify your equations into slope-intercept form.

Write the equation of a line parallel to y = -½ x + 2 passing through (6, 3).

Write the equation of a line perpendicular to y = -½ x + 2 passing through (6, 3).

Paper For Above instruction

Introduction

In coordinate geometry, understanding the relationships between lines—particularly their slopes—is essential for deriving equations of lines that are either parallel or perpendicular. When given a specific line and a point, the task involves calculating the equations of new lines that maintain the specified geometric relationship with the original line and pass through the given point. This paper elucidates the process of transforming these geometric relationships into explicit algebraic equations in slope-intercept form.

Analyzing the Given Line

The original line provided is y = -½ x + 2. The slope (m) of this line is -½. This information is crucial because the slope determines the nature of the lines that are either parallel or perpendicular to it. Any line parallel to this one must have the same slope, whereas any line perpendicular will have a slope that is the negative reciprocal of -½, which is 2.

Equation of a Parallel Line through a Point

To find the equation of a line parallel to the original line passing through point (6, 3), we keep the slope the same: m = -½.

Using the point-slope form of a line equation: y - y₁ = m(x - x₁).

Substituting the point (6, 3) and the slope:

y - 3 = -½(x - 6).

Expanding the right side:

y - 3 = -½ x + 3.

Adding 3 to both sides:

y = -½ x + 6.

This is the equation of the line parallel to the original, passing through (6, 3), expressed in slope-intercept form.

Equation of a Perpendicular Line through a Point

The slope of the perpendicular line is the negative reciprocal of -½, which is 2.

Using the point-slope form with point (6, 3):

y - 3 = 2(x - 6).

Expanding:

y - 3 = 2x - 12.

Adding 3 to both sides:

y = 2x - 9.

This equation represents the line perpendicular to the original and passing through the point (6, 3).

Implications and Applications

Understanding how to quickly find equations of parallel and perpendicular lines is crucial in various fields such as architecture, engineering, and computer graphics. These calculations support designing structures, plotting routes, and modeling real-world phenomena where geometric relationships are pivotal.

Conclusion

By analyzing the slope of the original line and applying the principles of point-slope form, we can efficiently derive equations of lines parallel and perpendicular to any given line passing through specified points. Mastery of these techniques enhances spatial reasoning and mathematical problem-solving competencies fundamental to advanced mathematics and practical applications.

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