2.4-2.5 Linear Equations in Two Variables and Linear Functions

Linear Equations in Two Variables and Linear Functions

Slope Definition

• Let $P = (x<em>1, y</em>1)$ and $Q = (x<em>2, y</em>2)$ be two distinct points.

• If $x<em>1 \neq x</em>2$, the slope $m$ of the nonvertical line $L$ containing $P$ and $Q$ is defined by the formula:
  $m = \frac{y<em>2 - y</em>1}{x<em>2 - x</em>1} = \frac{\Delta y}{\Delta x}$

• The symbol $\Delta$ (Greek uppercase letter delta) means "change in".

• The expression $\frac{\Delta y}{\Delta x}$ is also called the average rate of change of y with respect to x.

• If $x<em>1 = x</em>2$, then $L$ is a vertical line, and the slope of $L$ is undefined.

Example 2: Computing Slopes

1. $L<em>1$: Points $(2, 3), (-1, -2)$
   $m</em>1 = \frac{-2 - 3}{-1 - 2} = \frac{-5}{-3} = \frac{5}{3}$

2. $L<em>2$: Points $(2, 3), (3, -1)$
   $m</em>2 = \frac{-1 - 3}{3 - 2} = \frac{-4}{1} = -4$

3. $L<em>3$: Points $(2, 3), (5, 3)$ $m</em>3 = \frac{3 - 3}{5 - 2} = \frac{0}{3} = 0$

   • This indicates a horizontal line.

4. $L<em>4$: Points $(2, 3), (2, 5)$ $m</em>4 = \frac{5 - 3}{2 - 2} = \frac{2}{0}$ (undefined)

   • This indicates a vertical line.

Example 3: Drawing a Graph from a Point and Slope

To draw a graph of the line containing the point $(3, 2)$ and having a slope of $-\frac{4}{5}$.

1. Plot the given point $(3, 2)$.

2. From $(3, 2)$, use the slope $m = -\frac{4}{5} = \frac{\text{change in y}}{\text{change in x}}$:

   • A change in y of $-4$ (move down 4 units).

   • A change in x of $5$ (move right 5 units).

3. This leads to a new point $(3+5, 2-4) = (8, -2)$.

4. Draw a line through $(3, 2)$ and $(8, -2)$.

Example 4: Graphing a Vertical Line

To graph the equation $x = 3$:

• This is a vertical line where the x-coordinate of every point on the line is $3$.

• It passes through points like $(3, 0)$, $(3, 1)$, $(3, -2)$, etc.

Forms of Linear Equations

Point-Slope Form of an Equation of a Line

• An equation of a nonvertical line with slope $m$ that contains the point $(x<em>1, y</em>1)$ is given by:
  $y - y<em>1 = m(x - x</em>1)$

Example 5: Finding the Equation of a Horizontal Line

To find an equation of the horizontal line containing the point $(3, 2)$.

• A horizontal line has a slope of $m = 0$.

• Using the point-slope form: $y - 2 = 0(x - 3)$

• Simplifying: $y - 2 = 0 \implies y = 2$

Slope-Intercept Form of an Equation of a Line

• An equation of a line with slope $m$ and y-intercept $b$ (i.e., it passes through the point $(0, b)$) is given by:
  $y = mx + b$

Example 6: Finding Slope and Y-intercept, then Graphing

To find the slope $m$ and y-intercept $b$ of the equation $2x + 4y = 8$ and graph it.

1. Rewrite in slope-intercept form ($y = mx + b$):
   $4y = -2x + 8$
   $y = \frac{-2x}{4} + \frac{8}{4}$
   $y = -\frac{1}{2}x + 2$

2. From this form, the slope is $m = -\frac{1}{2}$ and the y-intercept is $b = 2$. This means the line crosses the y-axis at $(0, 2)$.

3. To graph:

   • Plot the y-intercept $(0, 2)$.

   • Use the slope $m = -\frac{1}{2}$ (down 1 unit, right 2 units) to find another point: $(0+2, 2-1) = (2, 1)$.

   • Draw the line through $(0, 2)$ and $(2, 1)$.

Example 7: Finding the Equation from Two Points

To find an equation of the line containing the points $(2, 3)$ and $(-4, 5)$ and graph the line.

1. Calculate the slope $m$:
   $m = \frac{5 - 3}{-4 - 2} = \frac{2}{-6} = -\frac{1}{3}$

2. Use the point-slope form with one of the points (e.g., $(2, 3)$):
   $y - 3 = -\frac{1}{3}(x - 2)$

3. Convert to slope-intercept form (optional, but good for graphing if asked):
   $y - 3 = -\frac{1}{3}x + \frac{2}{3}$
   $y = -\frac{1}{3}x + \frac{2}{3} + 3$
   $y = -\frac{1}{3}x + \frac{2}{3} + \frac{9}{3}$
   $y = -\frac{1}{3}x + \frac{11}{3}$

4. To graph: Plot the two given points $(2, 3)$ and $(-4, 5)$ and draw a line through them.

General Form (or Standard Form) of a Line

• Definition 8: The equation of a line is in general form when it is written as:
  $Ax + By = C$
  where $A$, $B$, and $C$ are real numbers, and $A$ and $B$ are not both $0$.

Example 9: Graphing by Finding Intercepts

To graph the equation $2x + 4y = 8$ by finding its intercepts.

1. Find the x-intercept: Set $y = 0$ and solve for $x$.
   $2x + 4(0) = 8$
   $2x = 8 \implies x = 4$
   The x-intercept is $(4, 0)$.

2. Find the y-intercept: Set $x = 0$ and solve for $y$.
   $2(0) + 4y = 8$
   $4y = 8 \implies y = 2$
   The y-intercept is $(0, 2)$.

3. To graph: Plot the x-intercept $(4, 0)$ and the y-intercept $(0, 2)$, then draw a line through them.

Parallel Lines

• Definition: Two lines (in the plane) are parallel if they do not intersect (they have no points in common).

• Theorem 10: Two nonvertical lines are parallel if and only if their slopes are equal ($m<em>1 = m</em>2$) and they have different y-intercepts.

  • If their slopes are equal and their y-intercepts are also equal, the lines are identical, not just parallel.

Example 11: Finding an Equation of a Parallel Line

To find an equation for the line that contains the point $(2, -3)$ and is parallel to the line $2x + y = 6$.

1. Find the slope of the given line: Convert $2x + y = 6$ to slope-intercept form:
   $y = -2x + 6$
   The slope of the given line is $m_{given} = -2$.

2. Determine the slope of the parallel line: Since the new line is parallel, its slope $m<em>{new}$ must be equal to $m</em>{given}$.
   $m_{new} = -2$

3. Use the point-slope form with the new slope and the given point $(2, -3)$:
   $y - (-3) = -2(x - 2)$
   $y + 3 = -2x + 4$
   $y = -2x + 1$

Perpendicular Lines

• Definition: Two lines are perpendicular if they intersect at a right angle ($90^\circ$).

• Theorem 12: Two nonvertical lines are perpendicular if and only if the product of their slopes is $-1$ ($m<em>1 m</em>2 = -1$).

  • This also means that their slopes are negative reciprocals of each other ($m<em>2 = -\frac{1}{m</em>1}$).

  • Horizontal lines ($m=0$) are perpendicular to vertical lines (undefined slope).

Example 13: Finding an Equation of a Perpendicular Line

To find an equation of the line that contains the point $(1, -2)$ and is perpendicular to the line $x + 3y = 6$, and graph both lines.

1. Find the slope of the given line: Convert $x + 3y = 6$ to slope-intercept form:
   $3y = -x + 6$
   $y = -\frac{1}{3}x + 2$
   The slope of the given line is $m_{given} = -\frac{1}{3}$.

2. Determine the slope of the perpendicular line: The slope $m<em>{new}$ must be the negative reciprocal of $m</em>{given}$.
   $m_{new} = -\frac{1}{(-1/3)} = 3$

3. Use the point-slope form with the new slope and the given point $(1, -2)$:
   $y - (-2) = 3(x - 1)$
   $y + 2 = 3x - 3$
   $y = 3x - 5$

4. To graph:

   • Graph the first line $y = -\frac{1}{3}x + 2$ (y-intercept $(0, 2)$, slope down 1, right 3).

   • Graph the second line $y = 3x - 5$ (y-intercept $(0, -5)$, slope up 3, right 1). Confirm they intersect at a right angle.

Linear Functions

• Definition 14: A linear function is a function of the form:
  $f(x) = mx + b$

• The graph of a linear function is a line with slope $m$ and y-intercept $b$.

• Its domain is the set of all real numbers ($(-\infty, \infty)$).

Example 15: Graphing a Linear Function

To graph the linear function $f(x) = -3x + 7$. What are the domain and the range of $f$?

1. Identify slope and y-intercept: From $f(x) = -3x + 7$, the slope is $m = -3$ and the y-intercept is $b = 7$.

2. Graph:

   • Plot the y-intercept $(0, 7)$.

   • Use the slope $m = -3 = -\frac{3}{1}$ (down 3 units, right 1 unit) to find another point: $(0+1, 7-3) = (1, 4)$.

   • Draw the line through $(0, 7)$ and $(1, 4)$.

3. Domain: For any linear function, the domain is all real numbers, $(-\infty, \infty)$.

4. Range: For any non-constant linear function ($m \neq 0$), the range is also all real numbers, $(-\infty, \infty)$.

Average Rate of Change of a Linear Function

• Theorem 16: Linear functions have a constant average rate of change. That is, the average rate of change of a linear function $f(x) = mx + b$ is:
  $\frac{\Delta y}{\Delta x} = m$