Writing An Equation Of A Perpendicular Line

Let's explore the process of writing the equation of a perpendicular line. This is a fundamental concept in coordinate geometry, essential for understanding the relationships between lines on a plane. We'll break down the steps involved, explain the underlying principles, and provide examples to solidify your understanding.

Understanding Perpendicular Lines

Perpendicular lines are lines that intersect at a right angle (90 degrees). The key characteristic that distinguishes them from other intersecting lines lies in the relationship between their slopes.

• Slope: The slope of a line represents its steepness and direction. It is usually denoted by m. A positive slope indicates an upward trend, while a negative slope indicates a downward trend. A horizontal line has a slope of 0, and a vertical line has an undefined slope.

• Relationship of Slopes: If two lines are perpendicular, the product of their slopes is -1. In other words, if line 1 has a slope of m₁ and line 2 is perpendicular to it with a slope of m₂, then:

  m₁ * m₂ = -1

  This also means that the slope of the perpendicular line (m₂) is the negative reciprocal of the original line's slope (m₁):

  m₂ = -1/m₁

Steps to Write the Equation of a Perpendicular Line

Here’s a step-by-step guide to writing the equation of a line perpendicular to a given line and passing through a specific point:

1. Identify the Slope of the Given Line:

The first step is to determine the slope of the line you are given. The equation of a line can be expressed in several forms, the most common being:

• Slope-Intercept Form: y = mx + b, where m is the slope and b is the y-intercept.
• Standard Form: Ax + By = C, where A, B, and C are constants.
• Point-Slope Form: y - y₁ = m(x - x₁), where m is the slope and (x₁, y₁) is a point on the line.

If the equation is in slope-intercept form, the slope is readily available. If it's in standard form, rearrange it into slope-intercept form to find the slope. If you're given two points on the line, you can calculate the slope using the formula:

m = (y₂ - y₁) / (x₂ - x₁)

Example:

Let's say our given line is y = 3x + 2. Here, the slope m₁ = 3.

2. Calculate the Slope of the Perpendicular Line:

Once you know the slope of the original line (m₁), you can calculate the slope of the perpendicular line (m₂) using the negative reciprocal relationship:

m₂ = -1/m₁

Example (Continuing from Step 1):

Since m₁ = 3, then m₂ = -1/3. This is the slope of our perpendicular line.

3. Use the Point-Slope Form:

You will usually be given a point that the perpendicular line must pass through. Let's call this point (x₁, y₁). The point-slope form of a linear equation is:

y - y₁ = m(x - x₁)

Substitute the slope of the perpendicular line (m₂) and the coordinates of the given point (x₁, y₁) into this equation.

Example (Continuing from Step 2):

Let's say the perpendicular line must pass through the point (2, 1). We know m₂ = -1/3 and (x₁, y₁) = (2, 1). Substituting these values into the point-slope form:

y - 1 = (-1/3)(x - 2)

4. Simplify the Equation (Optional):

The point-slope form is a valid equation for the line, but you might want to simplify it into slope-intercept form (y = mx + b) or standard form (Ax + By = C). To simplify, distribute and rearrange the terms.

Example (Continuing from Step 3):

Let's simplify to slope-intercept form:

y - 1 = (-1/3)x + (2/3) y = (-1/3)x + (2/3) + 1 y = (-1/3)x + (5/3)

This is the equation of the line perpendicular to y = 3x + 2 and passing through the point (2, 1).

5. Express in Standard Form (If Required):

If you need to express the equation in standard form (Ax + By = C), rearrange the equation so that x and y terms are on one side and the constant is on the other. Also, A, B, and C should ideally be integers, with A being positive.

Example (Continuing from Step 4):

To convert y = (-1/3)x + (5/3) to standard form:

Multiply the entire equation by 3 to eliminate fractions: 3y = -x + 5

Add x to both sides: x + 3y = 5

This is the equation in standard form.

Examples with Different Starting Points

Let's work through a few more examples to illustrate the process with different given information.

Example 1: Given a line in Standard Form and a Point

Problem: Find the equation of the line perpendicular to 2x + 5y = 10 and passing through the point (-1, 4).

Solution:

1. Find the slope of the given line:

   Convert the equation to slope-intercept form: 5y = -2x + 10 y = (-2/5)x + 2 So, m₁ = -2/5

2. Find the slope of the perpendicular line:

   m₂ = -1/m₁ = -1/(-2/5) = 5/2

3. Use the point-slope form:

   y - y₁ = m₂(x - x₁) y - 4 = (5/2)(x - (-1)) y - 4 = (5/2)(x + 1)

4. Simplify to slope-intercept form:

   y - 4 = (5/2)x + (5/2) y = (5/2)x + (5/2) + 4 y = (5/2)x + (13/2)

5. Convert to standard form (optional):

   2y = 5x + 13 -5x + 2y = 13 5x - 2y = -13

Therefore, the equation of the perpendicular line is y = (5/2)x + (13/2) (slope-intercept form) or 5x - 2y = -13 (standard form).

Example 2: Given Two Points on the Original Line and a Point

Problem: Find the equation of the line perpendicular to the line passing through points (1, 2) and (4, 8), and passing through the point (3, -1).

Solution:

1. Find the slope of the given line:

   m₁ = (y₂ - y₁) / (x₂ - x₁) = (8 - 2) / (4 - 1) = 6/3 = 2

2. Find the slope of the perpendicular line:

   m₂ = -1/m₁ = -1/2

3. Use the point-slope form:

   y - y₁ = m₂(x - x₁) y - (-1) = (-1/2)(x - 3) y + 1 = (-1/2)(x - 3)

4. Simplify to slope-intercept form:

   y + 1 = (-1/2)x + (3/2) y = (-1/2)x + (3/2) - 1 y = (-1/2)x + (1/2)

5. Convert to standard form (optional):

   2y = -x + 1 x + 2y = 1

Therefore, the equation of the perpendicular line is y = (-1/2)x + (1/2) (slope-intercept form) or x + 2y = 1 (standard form).

Example 3: Dealing with Horizontal and Vertical Lines

This is a special case. Remember that:

• A horizontal line has a slope of 0 and its equation is in the form y = c, where c is a constant.
• A vertical line has an undefined slope, and its equation is in the form x = c, where c is a constant.

Problem: Find the equation of the line perpendicular to y = 3 and passing through the point (5, 2).

Solution:

1. The given line y = 3 is a horizontal line.
2. A line perpendicular to a horizontal line is a vertical line.
3. A vertical line has the equation x = c. Since the perpendicular line passes through (5, 2), the equation is x = 5.

Problem: Find the equation of the line perpendicular to x = -2 and passing through the point (1, -1).

Solution:

1. The given line x = -2 is a vertical line.
2. A line perpendicular to a vertical line is a horizontal line.
3. A horizontal line has the equation y = c. Since the perpendicular line passes through (1, -1), the equation is y = -1.

Common Mistakes to Avoid

• Forgetting the Negative Reciprocal: The most common mistake is simply taking the reciprocal of the slope instead of the negative reciprocal. Remember to change the sign!
• Confusing Point-Slope and Slope-Intercept Forms: Make sure you are using the correct form of the equation and substituting values correctly. Pay attention to which values are x₁, y₁, and m.
• Incorrectly Rearranging Equations: When converting between standard form and slope-intercept form, double-check your algebraic manipulations to avoid errors.
• Not Handling Horizontal and Vertical Lines Correctly: Remember the special cases for horizontal (slope = 0, equation y = c) and vertical (undefined slope, equation x = c) lines.

Why is This Important?

Understanding perpendicular lines is crucial for various applications in mathematics, physics, and engineering:

• Geometry: It's fundamental for understanding shapes, angles, and spatial relationships.
• Calculus: Perpendicular lines are used to find normal lines to curves, which are important in optimization problems.
• Physics: Concepts like normal forces, which act perpendicularly to surfaces, rely on this principle.
• Engineering: Perpendicularity is essential in construction, surveying, and design to ensure stability and accuracy.
• Computer Graphics: Used in rendering 3D objects and creating realistic perspectives.
• Navigation: Calculating shortest distances and optimal routes often involves perpendicular lines and angles.

Advanced Concepts and Extensions

• Distance from a Point to a Line: The shortest distance from a point to a line is along the perpendicular line segment connecting the point to the line. This requires finding the equation of the perpendicular line and then the point of intersection.
• Orthogonal Trajectories: In differential equations, orthogonal trajectories are families of curves that intersect a given family of curves at right angles. Finding these trajectories involves finding the differential equation of the original family, taking the negative reciprocal of the slope, and solving the new differential equation.
• Vectors: The concept of perpendicularity extends to vectors, where two vectors are orthogonal (perpendicular) if their dot product is zero. This is a powerful tool in linear algebra and multivariable calculus.

Conclusion

Understanding how to write the equation of a perpendicular line is a fundamental skill in algebra and geometry. By following these steps and practicing with various examples, you can master this concept. Remember the key relationship between the slopes of perpendicular lines: they are negative reciprocals of each other. This knowledge unlocks a wide range of applications in mathematics, science, and engineering. Consistent practice and careful attention to detail will solidify your understanding and allow you to tackle more complex problems involving perpendicularity with confidence.