01Dlineargraphs

toc = Linear Graphs =

For any straight line graph:
The gradient is given by: math . \qquad m=\dfrac{\textit{rise}}{\textit{run}}=\dfrac{y_2-y_1}{x_2-x_1} math

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(theta)

The **__general equation__** is: //y = mx + c//, where
 * m = gradient
 * c = y-intercept

The __**equation of a line**__ can be found using: ... ... // y – y // 1 // = m(x – x // 1 // ) //

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 × 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.

math . \qquad \qquad \text{Midpoint between } \big( x_1, \; y_1 \big) \text{ and } \big( x_2, \; y_2 \big) \text { is given by } \qquad. \\ . \\ . \qquad \qquad \Big( \dfrac {x_1 + x_2}{2}, \; \dfrac{y_1 + y_2}{2} \Big) \qquad. math


 * Length of a Line Segment **


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

math . \qquad \qquad \text{Distance between } \big( x_1, \; y_1 \big) \text{ and } \big( x_2, \; y_2 \big) \text { is given by } \qquad. \\ . \\ . \qquad \qquad \sqrt{ \big( x_2 - x_1 \big)^2 + \big( y_2 - y_1 \big)^2 } \qquad. math

**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**
math \\ . \qquad x \in R \qquad \textit{ indicates the domain for x is all real values} \\. \\ . \qquad x \in R \backslash \{ 2, 3 \} \quad \textit{ indicates the domain for x is all real values excluding 2 and 3} \\. \\ . \qquad y \in [1, 4) \qquad \textit{ indicates the range for y is } 1 \leqslant y < 4 \; \{ \textit{see interval notation below} \} math

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 Î [a, b] ... {This assumes a is less than b}

... ... {a < x < b} is x Î (a, b)

... ... {a ≤ x < b} is x Î [a, b)

... ... {x ≤ a} is x Î (– ¥, a]

... ... {x > b} is x Î (b, ¥ )


 * Notice that ¥ is always ( or ) and never [ or ]

... ... R is (– ¥, ¥ )

... ... R + is {x > 0} .. which is the same as .. x Î (0, ¥ )

... ... R – is {x < 0} .. which is the same as .. x Î (– ¥, 0)

Graphing Intervals[[image:01Dline3.gif width="308" height="319" align="right"]]

 * 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


 * Left endpoint is at (1, 1) .. {open circle}


 * Right endpoint is at (4, 7) .. {closed circle}

Notice that:
 * domain is x Î (1, 4]
 * co-domain is y Î ** R **
 * range is y Î (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.

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

... ... ... Find the gradient of the original line: math . \qquad \qquad 2x + 4y = 10 \qquad. \\ . \\ . \qquad \qquad \quad 4y = -2x + 10 \qquad. \\ . \\ . \qquad \qquad \quad y = \dfrac {-x}{2} + \dfrac {5}{2} math
 * Solution:**

... ... ... Hence m 1 = -1/2 ... ... ... So m 2 = 2 .. ... ... .... ... {using m 1 × 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 and y = 2x.

math . \qquad \qquad 2x + 4y = 10 \qquad. \\ . \\ . \qquad \qquad 2x + 4(2x) = 10 \qquad. \\ . \\ . \qquad \qquad 2x + 8x = 10 \qquad. \\ . \\ . \qquad \qquad 10x = 10 \qquad. \\ . \\ . \qquad \qquad x = 1 \\. \\ . \\ . \qquad \qquad y = 2 math

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

Go to top of page flat

.