1. Presentation on Straight Line Theory
Mr. Lafferty
Higher Still Course
m > 0
m < 0
Presentation on Straight Line Theory
m = 0
y = c
Welcome to this presentation on circles.
It is designed to help students who are at a level
equivalent to the Scottish Higher Still Course.
This presentation constructs the general equation of a circle
by first deriving the special case
“Circle’s with centre the origin”
and then generalizing this result for any circle with centre A,B.
We then rearrange this general equation into a slight different form and then explain the benefits that each form has.
Click button to begin
y = mx+c
x = a
30-Nov-18

2. Essential Information
Mr. Lafferty
Essential Information
y = mx+c
Standard form for a Straight Line
What you need to know to find equation
How to find the gradient (m)
How to find the constant c
Straight Line Theory in a Nutshell
Notice board tips
Self test
m > 0
m < 0
m = 0
y = c
x = a
30-Nov-18

3. Standard form for a Straight Line
Mr. Lafferty
Standard form for a Straight Line
c = y intercept
y-axis
y = mx+c
m = gradient
(0,C)
m = gradient  O
x-axis
We start by find the equation of a circle centre the origin.
First draw set axises x,y and then label the origin O.
Next we plot a point P say, which as coordinates x,y.
Next draw a line from the origin O to the point P and label
length of this line r.
If we now rotate the point P through 360 degrees keep the Origin fixed we trace out a circle with radius r and centre O.
Remembering Pythagoras’s Theorem from Standard grade
a square plus b squared equal c squares we can now write down the equal of any circle with centre the origin.
Note : The angle  made by the line and the positive x-axis is equal to
m = tan()
30-Nov-18

4. Equation of a Straight Line
y = mx+c
To find the equation of a straight line we need to know
Two points that lie on the line
( x1, y1) and ( x2, y2) Or
The gradient and a point on the line
m and (a,b)
30-Nov-18

5. How to find the Gradient m
Mr. Lafferty
How to find the Gradient m
y = mx+c
m > 0
To find the gradient m we need to know 2 point
that lie on the straight line. Then use the rule
m < 0
m = gradient
Gradient can be positive, negative or zero
m = 0
y = c
e.g. What is the gradient of the line passing through
the points (3, -5) and (6,4).
We are now in a position to find the equation of any circle with centre A,B.
All we have to do is repeat the process in shown in slide 2, but this time the centre is chosen to be (a,b).
First plot a point C and label it’s coordinates (a,b), next we plot another point P and label it’s coordinates (x,y).
Next draw a line from C to P and call this length (r).
(r) will be the radius of our circle with centre (a,b).
Again we rotate the point P through 360 degrees keeping the point C fixed.
Using Pythagoras Theorem a squared plus b squared equal c squared we can write down the equation of any circle with centre (a,b) and radius (r).
The equation is (x - a) all squared plus (y-b) all squared equals (r) squared.
Finally to write down the equation of a circle we need to know the co-ordinates of the centre and the length of the radius or co-ordinates of the centre and the co-ordinates of a point on the circumference of the circle.
Solution
x = a
30-Nov-18

6. How to find constant C y = mx+c
To find the constant c, we need to know
the gradient m and a point on the line.
(0,C)
m = gradient
m > 0
y = mx+c
c = y-mx
(a,b) = (x,y)
(0,C) is were the line crosses the y-axis.
c=0
m < 0
(0,C)
e.g. Find the constant c given m = 4 and the point (1,6).
m = 0
Solution
(0,C)
Note : If c = 0 then equation is simply y = mx
30-Nov-18

7. Notice Board Tips y is directly proportional to x y  x y = mx+c or
“If two gradients multiplied
together equal -1 then lines
are perpendicular”
y is directly proportional to x
y  x
y = mx+c
y-b= m(x+a) or
Positive Gradient
m > 0
Negative
Gradient
“Every line cross
the y-axis at (0,C)”
m < 0
“If two equations have
the same gradient then
lines are parallel”
m = 0
y = c
line crosses x-axis at y=0
line crosses y-axis at x = 0 , y = c
30-Nov-18
x = a

8. Online Test y = mx+c Q1. Determine the gradient of the line
Mr. Lafferty
Online Test
y = mx+c
Q Determine the gradient of the line
passing through the points ( 5, 10) and (3, -6)
Click here to enter your answer into spreadsheet
Q Find the value of the constant “c”
for the line in Q1. above.
Determine whether the following equation is the equation of a circle.
Equation
The equation is in the right format to be the equation of a circle
but we have to make sure that radius r > 0 since we cannot have
a circle with radius r = 0 or negative.
We need
working out r gives
hence r=0
since r is not greater than 0
equation is not the equation of a circle.
Use the ALT+TAB keys to navigate to your
Spreadsheet and enter answer
30-Nov-18

9. Online Test
y = mx+c
Q For the values in Q1. and Q2 enter the equation
for the line into your spreadsheet (format y=mx+c)
Use the ALT+TAB keys to navigate to your
Spreadsheet and enter answer
Q Find one mathematical word to describe
each pair of line equations below.
(a) y = 2x+3
y = 2x+30
(b) y = -5x+10
y = 0.2x+7
Use the ALT+TAB keys to navigate to your
Spreadsheet and enter answer
30-Nov-18

10. Online Test y = mx+c Q5. Find the gradient m. Q6. Find the constant c.
y-axis O
x-axis
(2,5)
(4,9)
Q Find the gradient m.
Q Find the constant c.
Q Find the equation. 
Q Find value of angle .
( to 1 decimal place)
Use the ALT+TAB keys to enter answer into spreadsheet
Finally save your spreadsheet, naming it as “name_OTSL”
and to your teacher
e.g. joebloggs_OTSL
30-Nov-18