Equation of a Line: Slope-Intercept Form

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✖️ 1. Slope-intercept form vs. Point-slope form

📐 Slope-intercept form vs. Point-slope form

Slope-intercept: $y = mx + b$ where $m$ is slope and $b$ is y-intercept.
Use slope-intercept when you know the slope and where the line crosses the y-axis.
Point-slope: $y - y_1 = m(x - x_1)$ where $(x_1, y_1)$ is a known point.
Use point-slope when you know the slope and any point on the line.
Both forms describe the same line, just different starting information.

Example: Line with slope 3 through point (2, 5) is $y - 5 = 3(x - 2)$ in point-slope form, or $y = 3x - 1$ in slope-intercept form.

💡 Slope-intercept shows the y-intercept directly; point-slope plugs in any point you know.

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1. Slope-intercept form vs. Point-slope form

Slope-intercept form vs. Point-slope form

Slope-intercept form $y = mx + b$ expresses a line using its slope $m$ and $y$ -intercept $b$ . Point-slope form $y - y_1 = m(x - x_1)$ expresses a line using its slope $m$ and a known point $(x_1, y_1)$ on the line.

Intuition: Slope-intercept form is ideal when the $y$ -intercept is known or needed directly; point-slope form is preferred when working with a specific point and slope, especially when the $y$ -intercept is not immediately available.

Core Rules:

In $y = mx + b$ , $m$ is the slope and $b$ is the $y$ -coordinate where the line crosses the $y$ -axis.
In $y - y_1 = m(x - x_1)$ , $(x_1, y_1)$ is any point on the line and $m$ is the slope.
Both forms describe the same line; conversion between them is algebraic manipulation.
Point-slope form is undefined for vertical lines (slope undefined), while slope-intercept form cannot represent vertical lines.

Consequence: Choosing the appropriate form simplifies calculations and aligns with given information.

Example: Given slope $m = 2$ and point $(3, 5)$ : point-slope gives $y - 5 = 2(x - 3)$ , which simplifies to $y = 2x - 1$ in slope-intercept form.

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✖️ 2. Converting equations into standard form

🔄 Converting equations into standard form

Standard form: $Ax + By = C$ where $A$ , $B$ , $C$ are integers and $A$ is non-negative.
Move all $x$ and $y$ terms to the left side, constant to the right.
Clear fractions by multiplying every term by the least common denominator.
Make $A$ positive by multiplying the entire equation by $-1$ if needed.
Convention: $A$ , $B$ , $C$ should have no common factors (reduce if possible).

Example: Convert $y = \frac{2}{3}x + 4$ to standard form. Multiply by 3: $3y = 2x + 12$ . Rearrange: $-2x + 3y = 12$ . Make $A$ positive: $2x - 3y = -12$ .

💡 Standard form hides slope but shows both intercepts cleanly.

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2. Converting equations into standard form

Converting equations into standard form

Standard form of a linear equation is $Ax + By = C$ , where $A$ , $B$ , and $C$ are integers, and by convention $A \geq 0$ . If $A = 0$ , then $B > 0$ .

Intuition: Standard form emphasizes integer coefficients and symmetry between $x$ and $y$ , making it useful for certain algebraic techniques and for representing vertical lines (where $B = 0$ ).

Core Rules:

Eliminate fractions by multiplying through by the least common denominator.
Move all variable terms to one side and the constant to the other.
Ensure $A$ is non-negative; if $A < 0$ , multiply the entire equation by $-1$ .
Coefficients $A$ , $B$ , $C$ should be integers with no common factor greater than 1.

Consequence: Standard form provides a canonical representation suitable for systems of equations and intercept calculations.

Example: Convert $y = \frac{2}{3}x + 4$ to standard form. Multiply by 3: $3y = 2x + 12$ . Rearrange: $-2x + 3y = 12$ . Multiply by $-1$ : $2x - 3y = -12$ .

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✖️ 3. Finding the equation given two points or parallel/perpendicular conditions

🎯 Finding the equation given conditions

Two points: Find slope $m = \frac{y_2 - y_1}{x_2 - x_1}$ , then use point-slope with either point.
Parallel lines: Use the same slope as the given line, then plug in your new point.
Perpendicular lines: Use the negative reciprocal of the given slope, then plug in your new point.
If given slope is $m$ , perpendicular slope is $-\frac{1}{m}$ .

Example: Line through (1, 2) and (3, 8) has slope $m = \frac{8-2}{3-1} = 3$ . Equation: $y - 2 = 3(x - 1)$ or $y = 3x - 1$ .

💡 Parallel means copy the slope; perpendicular means flip and negate.

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3. Finding the equation given two points or parallel/perpendicular conditions

Finding the equation given two points or parallel/perpendicular conditions

Given two points $(x_1, y_1)$ and $(x_2, y_2)$ , the slope is $m = \frac{y_2 - y_1}{x_2 - x_1}$ (provided $x_2 \neq x_1$ ); then use point-slope form with either point. Parallel lines share the same slope; perpendicular lines have slopes that are negative reciprocals ( $m_1 \cdot m_2 = -1$ , provided neither is vertical).

Intuition: Two points uniquely determine a line; parallel/perpendicular conditions constrain the slope, requiring only one additional point to fix the line.

Core Rules:

Two points: Compute slope, then apply point-slope form.
Parallel condition: Use the same slope as the given line.
Perpendicular condition: Use the negative reciprocal of the given slope.
Vertical/horizontal cases: Vertical lines have undefined slope; horizontal lines have slope 0.

Consequence: These methods systematically construct equations from geometric constraints.

Example: Find the line through $(1, 2)$ perpendicular to $y = 3x + 1$ . Perpendicular slope: $m = -\frac{1}{3}$ . Point-slope: $y - 2 = -\frac{1}{3}(x - 1)$ , simplifying to $y = -\frac{1}{3}x + \frac{7}{3}$ .

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✖️ 4. Applications: Modeling constant-rate depletion and linear depreciation

💰 Applications: Modeling depletion and depreciation

Linear depreciation: Asset loses fixed value each time period (slope is negative).
Fuel consumption: Tank starts full, decreases at constant rate per hour or mile.
Slope represents the rate of change (negative for depletion).
Y-intercept is the starting amount (initial fuel or asset value).
X-axis is time or distance; y-axis is remaining quantity or value.

Example: Car worth 20000 dollars depreciates 2000 dollars per year. Equation: $V = -2000t + 20000$ where $t$ is years. After 5 years: $V = -2000(5) + 20000 = 10000$ dollars.

💡 Negative slope means something is running out or losing value over time.

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4. Applications: Modeling constant-rate depletion and linear depreciation

Applications: Modeling constant-rate depletion and linear depreciation

Constant-rate depletion (e.g., fuel consumption) and linear depreciation (e.g., asset value over time) are modeled by linear equations where the dependent variable decreases at a fixed rate. The slope $m$ represents the rate of change (negative for depletion/depreciation), and the $y$ -intercept $b$ represents the initial quantity or value.

Intuition: Linear models capture scenarios where a quantity changes uniformly over time, simplifying prediction and analysis.

Core Rules:

Identify the rate: Slope $m$ equals the change per unit time (negative for depletion).
Identify the initial value: $y$ -intercept $b$ is the starting amount at time $t = 0$ .
Formulate: Use $y = mt + b$ where $t$ is time and $y$ is the remaining quantity or value.
Interpret domain: Ensure $y \geq 0$ for physical quantities; solve $mt + b = 0$ to find depletion time.

Consequence: Linear models enable straightforward forecasting and decision-making in finance and resource management.

Example: A machine worth 10000 dollars depreciates linearly by 1200 dollars per year. Equation: $V = -1200t + 10000$ . After 5 years: $V = -1200(5) + 10000 = 4000$ dollars.

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Equation of a line

✖️ 1. Slope-intercept form vs. Point-slope form

📐 Slope-intercept form vs. Point-slope form

1. Slope-intercept form vs. Point-slope form

Slope-intercept form vs. Point-slope form

WARN: PRACTICE_BLOCK_EMPTY

✖️ 2. Converting equations into standard form

🔄 Converting equations into standard form

2. Converting equations into standard form

Converting equations into standard form

WARN: PRACTICE_BLOCK_EMPTY

✖️ 3. Finding the equation given two points or parallel/perpendicular conditions

🎯 Finding the equation given conditions

3. Finding the equation given two points or parallel/perpendicular conditions

Finding the equation given two points or parallel/perpendicular conditions

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✖️ 4. Applications: Modeling constant-rate depletion and linear depreciation

💰 Applications: Modeling depletion and depreciation

4. Applications: Modeling constant-rate depletion and linear depreciation

Applications: Modeling constant-rate depletion and linear depreciation

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