Problem 38 - Algebra A Combined Function

Question

Find an equation of the line passing through each pair of points. Write the equation in the form $A x+B y=C. $$ (0,0) \text { and }\left(-\frac{1}{2}, \frac{1}{3}\right) $$

Step-by-Step Solution

Answer

The equation is \(2x + 3y = 0\).

Step 1: Find the Slope of the Line

To find the slope \( m \) of the line passing through two points \((x_1, y_1)\) and \((x_2, y_2)\), use the formula:\[ m = \frac{y_2 - y_1}{x_2 - x_1} \]For the points \((0,0)\) and \((-\frac{1}{2}, \frac{1}{3})\), we can substitute:\[ m = \frac{\frac{1}{3} - 0}{-\frac{1}{2} - 0} = \frac{\frac{1}{3}}{-\frac{1}{2}} = -\frac{2}{3} \]

Step 2: Use Point-Slope Form to Find Line Equation

The equation of a line in point-slope form is given by:\[ y - y_1 = m(x - x_1) \]Substituting \( m = -\frac{2}{3} \) and the point \((0,0)\), the equation becomes:\[ y - 0 = -\frac{2}{3}(x - 0) \]Thus,\[ y = -\frac{2}{3}x \]

Step 3: Convert to Standard Form \(Ax + By = C\)

To convert \( y = -\frac{2}{3}x \) to the form \(Ax + By = C\), we can multiply through by 3 to eliminate the fraction:\[ 3y = -2x \]Rearrange to obtain the standard form:\[ 2x + 3y = 0 \]

Step 4: Conclusion

The equation of the line passing through the given points \((0,0)\) and \((-\frac{1}{2}, \frac{1}{3})\) in the standard form is:\[ 2x + 3y = 0 \]

Key Concepts

Slope

To determine how steep a line is as it moves across a coordinate plane, we use a concept called "slope." Slope is essentially a measurement of how much a line rises or falls for each unit of movement horizontally. Think of it as the 'tilt' of the line. This can help us understand the direction and the steepness of a line.

The formula to calculate the slope, typically denoted by the letter \( m \), is:

• \( m = \frac{y_2 - y_1}{x_2 - x_1} \)

In this formula, \((x_1, y_1)\) and \((x_2, y_2)\) are two points through which the line passes.

For example, given points \((0,0)\) and \(\left(-\frac{1}{2}, \frac{1}{3}\right)\), using the slope formula gives us \(-\frac{2}{3}\). This negative slope indicates that the line decreases as it moves from left to right.

Point-Slope Form

Once we have the slope, we can use the "point-slope form" to express the equation of a line. This form is very handy when you know the slope of a line and one point through which it passes. The point-slope form is represented by the equation:

• \(y - y_1 = m(x - x_1)\)

In this equation, \(m\) is the slope, and \((x_1, y_1)\) is a known point on the line. Let's use our line's slope, \(-\frac{2}{3}\), and the point \((0,0)\). When we substitute these values, we get:

• \(y - 0 = -\frac{2}{3}(x - 0)\)
• \(\Rightarrow y = -\frac{2}{3}x\)

This form of the equation shows the relationship of \(x\) and \(y\) with respect to the line's slope.

Standard Form

The standard form of a linear equation is useful for a variety of algebraic manipulations. It takes the form \(Ax + By = C\), where \(A\), \(B\), and \(C\) are integers, and \(A\) should be a non-negative integer.

To convert from the point-slope form (or slope-intercept form like \(y = -\frac{2}{3}x\)), we need to rewrite the equation so that there are no fractions and all terms are on one side of the equation. We transform:

• Multiply the entire equation by 3 to clear fractions: \(3y = -2x\)
• Rearrange: \(2x + 3y = 0\)

This is your standard form equation. It's nice and tidy, making it easier to interpret and use in various applications.

Algebra

Algebra is fundamental in mathematics for solving equations and understanding variables and constants. When we find the equation of a line, we're practicing algebra.

Algebra involves operations with symbols (like \(x\) and \(y\)) to find unknowns. Steps like finding the slope, using the point-slope form, and converting to standard form are great examples.

Here’s how algebra simplifies our line equation problem:

• We use substitution to apply specific values for slope and known points.
• We rearrange terms for clarity and simplicity according to the standard form rules.

Understanding these transformations allows us to manipulate equations efficiently, bringing solutions to problems involving lines on graphs much closer.