What is the general form of the equation for a line passing through (3, -1) and perpendicular to a given graph?

The general form is y - y₁ = m(x - x₁), where (x₁, y₁) is (3, -1) and m is the negative reciprocal of the slope of the given graph.

How do I find the equation of a line perpendicular to a given curve passing through a specific point?

First, determine the slope of the tangent to the curve at the point, then take its negative reciprocal to find the perpendicular slope, and finally use point-slope form with the point (3, -1).

If the graph's slope at the point (3, -1) is 2, what is the equation of the line passing through that point and perpendicular to the graph?

The perpendicular slope is -1/2. Using point-slope form: y + 1 = -1/2(x - 3), which simplifies to y = -1/2 x + 1/2 - 1 = -1/2 x - 1/2.

Can you provide an example equation of a line passing through (3, -1) and perpendicular to the graph y = 4x + 5?

Yes. The slope of the given line is 4, so the perpendicular slope is -1/4. The line passing through (3, -1) is y + 1 = -1/4(x - 3), which simplifies to y = -1/4 x - 1/4.

What is the step-by-step process to write the equation of a line passing through (3, -1) and perpendicular to any known line?

1. Find the slope of the known line. 2. Calculate the negative reciprocal of that slope for the perpendicular line. 3. Use the point (3, -1) and the perpendicular slope in the point-slope form: y - y₁ = m(x - x₁). 4. Simplify to slope-intercept or standard form as needed.

Which Equation Represents The Line That Passes Through The Point (3,-1) And Is Perpendicular To The Graph

Which Equation Represents The Line That Passes Through The Point (3,-1) And Is Perpendicular To The Graph is a fundamental question in coordinate geometry that involves understanding the relationships between lines, slopes, and points on the Cartesian plane. This topic is essential for students and professionals working in fields such as mathematics, engineering, physics, and computer graphics. In this comprehensive guide, we will explore the concepts necessary to determine the equation of a line passing through a specific point and perpendicular to a given graph, along with practical examples and step-by-step procedures.

Understanding the Basics: Lines, Slopes, and Perpendicularity

What Is a Line Equation?

A line in the coordinate plane can be represented mathematically by an equation. The most common form is the slope-intercept form:

\[ y = mx + b \]

where:

\( m \) is the slope of the line,

\( b \) is the y-intercept (the point where the line crosses the y-axis).

Another common form is the point-slope form:

\[ y - y1 = m(x - x1) \]

which is useful when a point \((x1, y1)\) on the line is known.

Understanding the Slope

The slope \( m \) measures the steepness and direction of the line:

A positive slope means the line rises as it moves from left to right.
A negative slope means it falls.
A zero slope indicates a horizontal line.
An undefined slope (vertical line) occurs when the change in \( x \) is zero.

Perpendicular Lines and Their Slopes

Two lines are perpendicular if they intersect at a 90-degree angle. An important property of perpendicular lines is that their slopes are negative reciprocals of each other:

If one line has a slope \( m \),
The perpendicular line has a slope \( m_{\perp} = -\frac{1}{m} \).

This relationship allows us to find the slope of a perpendicular line once the slope of the original line is known.

Determining the Equation of a Perpendicular Line Through a Specific Point

Step 1: Identify the Slope of the Original Graph

Before finding the perpendicular line, we need to understand the slope of the original graph. The process depends on the type of graph:

If the graph is a line with a known equation: extract the slope directly.
If the graph is a curve or a non-linear function: find the slope at the point of interest using calculus (derivatives).
If only the graph is given visually: estimate the slope by selecting two points close to the point of interest and applying the slope formula.

Example: Suppose the original graph is \( y = 2x + 5 \). The slope is \( m = 2 \).

Step 2: Find the Slope of the Perpendicular Line

Once the original slope \( m \) is known, compute the negative reciprocal:

\[ m_{\perp} = -\frac{1}{m} \]

Continuing the example:
\[ m_{\perp} = -\frac{1}{2} \]

If the original graph is not a straight line but a curve, find the slope of the tangent line at the relevant point using differentiation.

Step 3: Use the Point-Slope Form to Write the Equation

Given the point \((x1, y1) = (3, -1)\) and the slope \( m_{\perp} \):

\[ y - y1 = m{\perp}(x - x_1) \]

Substitute the known values:
\[ y - (-1) = -\frac{1}{2}(x - 3) \]
\[ y + 1 = -\frac{1}{2}(x - 3) \]

This is the equation of the perpendicular line passing through \((3, -1)\).

Step 4: Simplify the Equation (Optional)

Distribute the slope:

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

Subtract 1 from both sides:

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

This is the slope-intercept form of the perpendicular line.

Special Cases and Additional Considerations

Vertical and Horizontal Lines

When the original graph is a vertical or horizontal line, the approach adjusts:

Original line is horizontal (\( y = c \)): the perpendicular line is vertical (\( x = x_0 \)). In this case, the line passing through \((3, -1)\) is:

\[ x = 3 \]

Original line is vertical (\( x = c \)): the perpendicular line is horizontal (\( y = y_0 \)). The line passing through \((3, -1)\) in this case is:

\[ y = -1 \]

Using Derivatives for Curves

If the original graph is a curve, such as \( y = x^3 \), the slope at a point is found via derivatives:

\[ \frac{dy}{dx} = 3x^2 \]

At \( x = 3 \):

\[ m = 3(3)^2 = 27 \]

The perpendicular slope:

\[ m_{\perp} = -\frac{1}{27} \]

Then, the line passing through \((3, -1)\):

\[ y + 1 = -\frac{1}{27}(x - 3) \]

which simplifies accordingly.

Practical Applications of Perpendicular Lines

Understanding and calculating perpendicular lines are crucial in various real-world contexts, including:

Construction and Engineering: Ensuring structures are perpendicular for stability.

Navigation and Mapping: Calculating perpendicular distances and directions.

Computer Graphics: Rendering perpendicular lines to create accurate shapes and models.

Robotics: Programming movement along perpendicular paths for collision avoidance and precision.

Summary of the Process

To find the equation of a line passing through a point and perpendicular to a given graph:

Determine the slope of the original graph at the relevant point (using algebra or calculus).

Calculate the negative reciprocal to find the perpendicular slope.

Use the point-slope form with the point \((3, -1)\) and the perpendicular slope.

Simplify the equation as needed.

Example Walkthrough

Let's walk through a complete example:

Given:

Original graph: \( y = 4x + 1 \)

Point: \( (3, -1) \)

Step 1: Find the slope of the original graph:

\[ m = 4 \]

Step 2: Find the slope of the perpendicular line:

\[ m_{\perp} = -\frac{1}{4} \]

Step 3: Write the equation using point-slope form:

\[ y - (-1) = -\frac{1}{4}(x - 3) \]
\[ y + 1 = -\frac{1}{4}x + \frac{3}{4} \]

Step 4: Simplify to slope-intercept form:

\[ y = -\frac{1}{4}x + \frac{3}{4} - 1 \]
\[ y = -\frac{1}{4}x - \frac{1}{4} \]

Result: The equation of the line passing through \((3, -1)\) and perpendicular to the original graph \( y = 4x + 1 \) is:

\[ y = -\frac{1}{4}x - \frac{1}{4} \]

Conclusion

Knowing how to find the equation of a line that passes through a specific point and is perpendicular to a given graph is a fundamental skill in mathematics. It combines understanding slopes, the properties of perpendicular lines, and the ability to manipulate algebraic equations. Whether working with linear functions or more complex curves, the principles remain consistent. Mastery of this topic enables students and professionals to solve a wide array of geometric and analytical problems with confidence and precision.

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Remember: Always identify the slope of the original graph first, then compute its negative reciprocal for the perpendicular line, and finally use the point-slope form to write the equation passing through the given point. This systematic approach simplifies the process and ensures accuracy in your solutions.