How do I find the equation of a line in standard form with a slope of -1/6 passing through a specific point?

Use point-slope form: y - y₁ = m(x - x₁), then rearrange to standard form Ax + By = C.

What is the process to convert a line's equation from slope-intercept form to standard form?

Rearrange the slope-intercept form y = mx + b into Ax + By = C by moving all terms to one side and clearing fractions if necessary.

If a line has a slope of -1/6 and passes through (x₁, y₁), how do I write its equation in standard form?

Start with point-slope form: y - y₁ = (-1/6)(x - x₁), then multiply through by 6 to clear fractions and rearrange to Ax + By = C.

Can you give an example of finding the standard form of a line with slope -1/6 passing through (2, 3)?

Yes. Using point-slope form: y - 3 = (-1/6)(x - 2). Multiply both sides by 6: 6(y - 3) = -1(x - 2). Simplify: 6y - 18 = -x + 2. Rearranged: x + 6y = 20.

What are common mistakes to avoid when deriving the standard form from a slope and a point?

Avoid forgetting to clear fractions by multiplying through by the denominator, and ensure all terms are moved to one side with proper signs.

How does knowing the slope (-1/6) help in constructing the standard form of the line?

The slope is used in the point-slope form to create the initial equation, which can then be rearranged into standard form with proper algebraic manipulation.

Is the standard form of a line unique for a given slope and point, and how do I ensure the coefficients are integers?

Yes, the standard form is unique up to multiplication by a non-zero constant. To have integer coefficients, multiply the entire equation by the least common denominator if necessary.

What is the general formula for the standard form of a line with a given slope passing through a specific point?

Start with y - y₁ = m(x - x₁), then multiply both sides to clear fractions, and rearrange to Ax + By = C, ensuring A, B, C are integers.

How do I verify that my standard form equation correctly represents the line with the given slope and point?

Substitute the given point into your equation to check if it satisfies the equation, and verify that the slope derived from your equation matches the given slope.

Find The Equation Of The Line In Standard Form Ax+By=C That Has A Slope Of (-1)/(6) And Passes Through

Find The Equation Of The Line In Standard Form Ax+By=C That Has A Slope Of (-1)/(6) And Passes Through

Understanding how to determine the equation of a straight line given a specific slope and a point it passes through is a fundamental concept in coordinate geometry. The standard form of a line’s equation, Ax + By = C, provides a concise way to represent lines and analyze their properties. In this article, we will explore how to derive the standard form of a line with a given slope and passing through a particular point, focusing on the slope of -1/6. We will break down the process step-by-step, illustrate with examples, and discuss various methods to arrive at the desired equation.

Understanding the Components of a Line Equation

What Is the Standard Form of a Line?

The standard form of a linear equation is written as:

Ax + By = C

where:

A, B, and C are real numbers,

A and B are not both zero,

The coefficients are usually chosen to be integers with no common factors (for simplicity).

This form is particularly useful for quickly identifying intercepts, and for analyzing the relationship between x and y in a linear relationship.

What Is the Slope of a Line?

The slope (m) of a line indicates its steepness and direction. It is calculated as:

m = (change in y) / (change in x)

Given the slope m = -1/6, it means:

For every 6 units moved horizontally to the right, the line drops 1 unit vertically.

The line is decreasing from left to right.

Given Data and Objective

Suppose the line passes through a specific point, say, \( (x1, y1) \). Our goal is to find the equation of the line in standard form, Ax + By = C, that:

Has a slope \( m = -\frac{1}{6} \),

Passes through \( (x1, y1) \).

The process involves:

Using the point-slope form of the line equation.

Rearranging into standard form.

Method 1: Using the Point-Slope Form

Step 1: Write the Point-Slope Equation

The point-slope form of a line is:

y - y₁ = m(x - x₁)

Given:

\( m = -\frac{1}{6} \),

a point \( (x1, y1) \).

The equation becomes:

\[ y - y1 = -\frac{1}{6}(x - x1) \]

Step 2: Clear the Fraction

Multiply both sides by 6 to eliminate the denominator:

\[ 6(y - y1) = - (x - x1) \]

which simplifies to:

\[ 6y - 6y1 = -x + x1 \]

Step 3: Rearrange into Standard Form

Bring all terms to one side:

\[ x + 6y = x1 + 6 y1 \]

This is the line in standard form:

\[ Ax + By = C \]

where:

\( A = 1 \),

\( B = 6 \),

\( C = x1 + 6 y1 \).

Note: If specific point coordinates are provided, substitute them into the above to find the exact C.

Method 2: General Approach with a Specific Point

Suppose the line passes through the point \( (x1, y1) = (2, 3) \). Let's find the equation step-by-step.

Step 1: Write the point-slope form

\[ y - 3 = -\frac{1}{6}(x - 2) \]

Step 2: Multiply through by 6 to clear fractions

\[ 6(y - 3) = -(x - 2) \]

\[ 6y - 18 = -x + 2 \]

Step 3: Rearrange to standard form

\[ x + 6y = 20 \]

This is the standard form of the line passing through (2, 3) with slope -1/6.

General Process for Any Point

If the point is \( (x1, y1) \), the general formula becomes:

\[ x + 6 y = x1 + 6 y1 \]

which directly provides the standard form once the specific point is substituted.

Additional Considerations

Ensuring the Standard Form is in the Correct Format

Usually, A should be positive. If A is negative, multiply the entire equation by -1 to make A positive.
Coefficients A, B, and C should be integers with no common factors for simplicity, if possible.

Example: Line Passing Through Multiple Points

Suppose you are asked to find the line passing through two points \( (x1, y1) \) and \( (x2, y2) \) with the given slope:

Find the slope between the two points.
Confirm that it matches the given slope (-1/6).
Use the point-slope form with either point.
Convert to standard form.

Practice Problem and Solution

Problem: Find the equation of a line in standard form that has a slope of -1/6 and passes through the point \( (4, -2) \).

Solution:

Write point-slope form:

\[ y - (-2) = -\frac{1}{6}(x - 4) \]

\[ y + 2 = -\frac{1}{6}x + \frac{4}{6} \]

\[ y + 2 = -\frac{1}{6}x + \frac{2}{3} \]

Multiply through by 6:

\[ 6 y + 12 = -x + 4 \]

Rearrange to standard form:

\[ x + 6 y = -12 + 4 \]

\[ x + 6 y = -8 \]

This is the standard form of the line.

Summary and Key Takeaways

To find the equation of a line in standard form given a slope and a point, start with the point-slope form.
Clear fractions by multiplying through by the denominator.
Rearrange the resulting equation to the form Ax + By = C.
Ensure the coefficients are simplified and A is positive.

Conclusion

Deriving the equation of a line in standard form from a given slope and point involves a systematic process that starts with the point-slope form, simplifies to eliminate fractions, and rearranges into the desired standard form. Mastery of this method enables students and mathematicians to quickly analyze line equations, find intersections, and solve geometric problems efficiently. Whether working with theoretical problems or practical applications, understanding these steps is essential for a solid foundation in analytic geometry.