Linear Graphs
(Straight Lines)



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Describing Straight Lines:

O x y Vertical Line Horizontal Line Line at Other Angles

Drawing Linear equations

Drawing (xy table)
Drawing using y=m⁢x+c
Practice Questions
Textbook Exercise

Vertical Lines:

0 x y x = 2 x = -3
x=k

where k is some constant, e.g., x=3. This is the equation of a vertical straight line. The more positive k is the more the vertical line is to the right. The more negative, the more the vertical line is to the left.

x=a graphs
Practice Questions
Textbook Exercise

Horizontal Lines:

0 x y y = 4 y = -1
y=k

where k is some constant, e.g., y=-2. This is the equation of a horizontal straight line. The more positive k is the higher the horizontal line is. The more negative, the lower the horizontal line is.

y=a graphs
Practice Questions
Textbook Exercise

Lines at other Angles:

y=m⁢x+c

where m and c are constants, i.e., y=2⁢x+4

m is the gradient of the line.

c is the y-axis intercept [-cm is the x-axis intercept]


Gradient of a Line:

B:(xB, yB) A:(xA, yA) Gradient, m=change in ychange in x =ΔyΔx = yB-yA xB-xA
Gradient of a line
Gradient between points
Practice Questions
Textbook Exercise

The Equation of a Line:

These are some of the ways the equation of a line maybe written:

y=m⁢x+c x-k=0 y-k=0 a⁢x+b⁢y+c=0 a⁢x+b⁢y=c

As seen above, y=m⁢x+c is one of the most useful presentation.

Notice that both x and y have an index one, i.e., there are no other indices in any terms.

y=m⁢x+c
Practice Questions
Textbook Exercise

Length of a line

Graph showing a line segment from point A to point B l2 =(x2-x1)2 +(y2-y1)2
Length of a line
Practice Questions
Textbook Exercise

Mid Point of a Line:

B:(xB, yB) A:(xA, yA) M:(xM, yM) M: ( xM , yM ) = ( xA+xB2 , yA+yB2 )
Midpoint of a line
Practice Questions
Textbook Exercise

Linear Graphs

In real life
Across one method
Practice Questions
Textbook Exercise
Finding the equation of a line
Practice Questions
Textbook Exercise
Equation through 2 points
Practice Questions
Textbook Exercise

Comparing Lines: Parallel and Perpendicular Lines

For two or more lines are three possibilities for any pair. Lines l1 and l2 have equations y=m1⁢x+c1 and y=m2⁢x+c2 can either be parallel to each other (at an angle of 0° to each other), perpendicular to each other (at 90° to each other), or some other angle.

For parallel lines the gradients are equal, i.e., m1=m2.

Parallel lines
Practice Questions
Textbook Exercise

For perpendicular lines the product of the gradients of the lines equal minus one, i.e., m1⁢m2=-1

Perpendicular lines
Practice Questions
Textbook Exercise

Equation of a Perpendicular Bisector

To find the perpendicular bisector of a line segment between two given points:
Find the gradient of the line segment between the two points;
Find the gradient of the perpendicular using m1⁢m2=-1;
Find the mid point of the two given points. The perpendicular bisector goes through this mid point;
To find the y-intercept of the perpendicular bisector substitute the mid point coordinates in the equation for the bisector.

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