02.11lineargraphs

# Linear Graphs

For any straight line graph:

The gradient is given by: $m=\dfrac{\textit{rise}}{\textit{run}}=\dfrac{y_2-y_1}{x_2-x_1}$

The gradient can also be found using the angle between the line and the positive direction of the x-axis.

If $\theta$ = the angle with the positive x-axis, then $m = \tan \big(\theta \big)$

The general equation is: $y = mx + c$, where

• m = gradient
• c = y-intercept

The equation of a line can also be found using: … $y - y_1 = m \big( x - x_1 \big)$

## Parallel Lines

Parallel lines have the same gradient … $m_1 = m_2$

## Perpendicular Lines

The product of the gradients of two perpendicular lines is –1 … $m_1 \times m_2 = -1$

## Midpoint of a Line Segment

The midpoint of a line segment can be found using the average of the x values and the average of the y values of the two endpoints.

… … Midpoint between $\big( x_1, \; y_1 \big) \text{ and } \big( x_2 , \; y_2 \big)$

… … is given by … $\Big( \dfrac {x_1 + x_2}{2}, \; \dfrac{y_1 + y_2}{2} \Big)$

## Length of a Line Segment

The length of a line segment can be found by applying Pythagoras to the two endpoints

… … Distance between $\big( x_1, \; y_1 \big) \text{ and } \big( x_2 , \; y_2 \big)$

… …is given by … $D = \sqrt{ \big( x_2 - x_1 \big)^2 + \big( y_2 - y_1 \big)^2 }$

## Domain and Range

• Domain is the set of x values for the function
• Range is the set of y values for the function

For many functions, the domain and range is R .. (the set of all Real numbers)

#### Examples

… … $x \in R$ … indicates the domain for x is all real values

… … $x \in R \backslash \{ 2, 3 \}$ … indicates the domain for x is all real values excluding 2 and 3

… … $y \in [1, 4)$ … indicates the range for y is $1 \leqslant y < 4$

… … see interval notation below

## Interval Notation

… … A square bracket [ ] indicates that the interval includes the endpoints

… … A round bracket ( ) indicates that the interval does not include the enpoints.

… … {a ≤ x ≤ b} is $x \in [a, \; b]$ … {This assumes a is less than b}

… … {a < x < b} is $x \in (a, \; b)$

… … {a ≤ x < b} is $x \in [a, \; b)$

… … {x ≤ a} is $x \in (-\infty, \; a]$

… … {x > b} is $x \in (b, \; \infty)$

Notice that $\infty$ is always ( or ) and never [ or ]

… … R is $(-\infty, \; \infty)$

… … R+ is {x > 0} .. which is the same as .. $x \in (0, \; \infty)$

… … R is {x < 0} .. which is the same as .. $x \in (–\infty, \; 0)$

## Graphing Intervals

< or > is indicated on a graph with an open circle at the end of the line.

< or > is indicated with a closed (filled in) circle.

The circles must be clearly visible and bigger than a single pen dot.

#### Example

… … Sketch .. f: (1, 4] → R, f(x) = 2x – 1

Solution

• Left endpoint is at (1, 1) ..{open circle}
• Right endpoint is at (4, 7) ..{closed circle}

Notice that:

• domain is $x \in (1, 4]$
• co-domain is $y \in R$
• range is $y \in (1, 7]$

## Closest Point to the Origin

For any straight line, the point on the line that is closest to the origin will be the intersection between that line and a perpendicular line that passes through the origin.

To find the coordinates of the point closest to the origin, we need to find the equation of the perpendicular line that goes through the origin.

#### Example

… … Find the coordinates of the point on the line $2x + 4y = 10$ which is closest to the origin.

Solution:

… … Find the gradient of the original line:

… … $2x + 4y = 10$

… … $4y = -2x + 10$

… … $y = \dfrac {-x}{2} + \dfrac {5}{2}$

… … Hence $m_1 = -\dfrac{1}{2}$

… … So $m_2 = 2$$\{ using \; m_1 \times m_2 = -1 \}$

… … Now find the equation of the line through the origin with $m_2 = 2$.

… … $y_2 = 2x$.

… … Now solve the simultaneous equations $2x + 4y = 10 \text{ and } y = 2x$.

… … $2x + 4y = 10$

… … $2x + 4 \big( 2x \big) = 10$

… … $2x + 8x = 10$

… … $10x = 10$

… … $x = 1$

… … $y = 2$

… … Hence coordinates of point closest to origin is (1, 2)

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