Higher Mathematics Unit 1.1
The Straight Line
Contents
1.
1.
2.
3.
4.
5.
6.
7.
8.
9.
Definition of Gradient.
Finding gradients and intercepts.
y  b  m( x  a ) .
Parallel and perpendicular lines.
Mid-point of a line segment/perpendicular bisector.
Distance between two points.
Medians, Altitudes and points of intersection.
Some general points.
References/resources.
Definition of Gradient
It is best to define gradient by m  tan  , where  is the angle that the line makes with the
positive direction of the x – axis.
y  y1
The definition m  2
follows immediately from this.
x2  x1
y
Q( x2 , y2 )
P( x1 , y1 ) 

O
x
y 2  y1
.
x2  x1
It is important that pupils are proficient with both definitions of gradient.
From the diagram it can be easily seen that tan   m and tan  
Examples should be done where m  0
m undefined
(horizontal line)
(vertical line)
Take care when angle is obtuse, e.g.
y

O

x
y  2 x  5
Here tan  2 and  is obtuse. Calculator gives  63.4 , i.e. the angle  in the diagram.
So   (180  63.4)  116.6 , to one decimal place.
2.
Finding Gradient and Intercepts, given the equation of the line
Some pupils will be confident at this from Credit Level, but it should be gone over thoroughly,
because weaker pupils will need much revision.
Pupils should know that the equation y  mx  c represents a straight line, gradient m, with y
intercept (0, c).
Do examples such as 4 x  5 y  10 , where the equation has to be rearranged to find the value of m
and c.
Do examples such as “ 3 x  5 y  15 . Find the coordinates of the points where this straight line
crosses the coordinate axes.”
3.
Finding the Equation of a Straight Line
y  b  m( x  a) . This is the most common technique in “H” Mathematics: Pupils will get
practice at this in several different Outcomes.
It is worthwhile to do a “proof” as follows:
y
Q(x,y)
P(a,b)
O
x
The straight line passes through the point P(a, b) and has gradient m.
Let Q ( x, y ) be another point on the line .
y b
 m , i.e. y  b  m( x  a) .
Then, by definition of gradient,
xa
The above proof is not a full one:
Strictly speaking, we should show that every point on the line satisfies the equation.
But at this level the above argument is enough.
Question
a)
Find the equation of the straight line through the points A(-1,5) and B(3,1).
b)
Find the size of the angle which AB makes with the positive direction of the
x-axis.
1.
2.
3.
4.
m  1
y  5  1( x  1) or y  1  1( x  3)
Tan(angle) = -1
angle = 135o
Some Notes
(a)
Weak candidates will be weak at clearing fractions, so give them lots of practice.
Working may be set out as follows:
(2)
(2)
y 3
  23 ( x  5)
3( y  3)  3( 23 )( x  5)
3 y  9  2( x  5)
3 y  9  2 x  10
2 x  3 y  19
More able candidates will not need to show this amount of working.
4.
(b)
Encourage pupils to rearrange the equation to the form ax  by  c (or ax  by  c  0 )
since it is often required to solve simultaneous linear equations in a later part of the
question.
(c)
Do examples where two points are given, but no gradient, e.g.
“Find the equation of the straight line passing through the points (-2,3) and (4,7).”
(d)
Give the pupils practice at finding the equations of horizontal lines (zero gradient) and
vertical lines (gradient undefined).
Parallel and Perpendicular Lines
(a)
Parallel lines have the same gradient. Draw sketches to illustrate this .
Do examples such as: “Find the equation of the straight line passing through the point (2,5)
which is parallel to the line with equation 3x  4 y  12 .”
Find the equation of the straight line which is parallel to the line with equation
2 x  3 y  5 and which passes through the point (2,-1).
1.
2.
3.
(b)
(3)
2
5
x 
3
3
2
mline  
3
2
y  (1) 
( x  2)
3
y
Perpendicular Lines
If  1 has gradient m1 and  2 has gradient m2
Then  1   2  m1m2  1
This is a key result, which can be got in either (or both) of the following ways:
(i)
1
y x
2
3
2
y   x , y  x , etc.
2
3
Consideration of gradients of each pair of lines should lead to the conjecture.
Drawing graphs of pairs of lines such as
y  2x ,
Do examples such as “Find the equation of the straight line through the point (5,2) which
is perpendicular to the line with equation 3x  4 y  12 .”
Notes
(i)
Pupils must give reasons for perpendicular/parallel lines; e.g. in the above example:
3
y   x3
(after rearrangement)
4
3
4
m1  
 m2 
( m1m2  1 for  lines)
4
3
etc,etc.
5.
(ii)
Do examples involving horizontal/vertical lines.
(iii)
Working should be clearly shown, e.g. if pupil is calculating gradient of AB then mAB
should be used , not just m.
Mid point of a Line Segment/Perpendicular Bisector
(a)
Mid point
Do numerical examples first, and make the pupils draw sketches.
For example; “M is the mid point of AB, where A is the point (1,1) and B is (7,13)”
The coordinates of M may be got as follows: “Think of the x direction first: From A to B
we go 6 units to the right, so at M we will be 3 units to the right.”
(And a similar argument for y.)
Therefore M is the point (4,7).
This colloquial, “common sense”, approach should be mastered before a more rigorous
approach is considered.
 x  x 2 y1  y 2 
Some pupils will see the general result M  1
,
 and a simple approach could
2 
 2
be:
B( x2 , y2 )
M
y2  y1
A( x1 , y1 )
x2  x1
xM  x1  12 ( x2  x1 ) 
x1  x2
2
yM  y1  12 ( y2  y1 ) 
y1  y2
2
(b)
Perpendicular Bisector of a line segment
Defined as a line which cuts another line in half and is perpendicular to that line.
This will be picked up quite easily.
Pupils should draw a sketch, if one is not given.
Working may be set out as follows:
B(-7,4)
y
M
A(1,2)
O
x
42
1

 7 1
4
  7 1 4  2 
M
,
 , i.e. M (3,3)
2 
 2
m AB 
 m  4 (m1m2  1 for  lines) .
[And then do the usual y  b  m( x  a) stuff.]
6.
Distance Between two Points
Best to do numerical examples first. For example “Find the distance between the points
(1,1) and (5,4).”
y
(5,4)
d
3
(1,1)
O
4
x
By P.T. , d 2  4 2  32 , etc, etc .
Pupils will rapidly become adept at this. Once mastery has been achieved the distance
formula can be derived if it is considered appropriate.
The distance formula can be got by considering the following diagram:
(Some weaker pupils may struggle with this).
( x2 , y 2 )
d
y2  y1
( x1 , y1 )
x2  x1
By P.T., d 2  ( x2  x1 ) 2  ( y 2  y1 ) 2 , etc , etc .
7.
Medians, Altitudes and Intersections
(a)
Medians
Draw sketches to illustrate these.
A
P
N
C
M
B
The Medians of a triangle are concurrent; they meet at the centroid.
The centroid divides each median in the ratio 2:1.
Proofs of these facts are not required.
But it is important that the meaning of the word “concurrent” is known.
(b)
Altitudes
Draw sketches to illustrate these.
It is best to sketch an acute angled triangle, for the altitudes intersect inside the triangle.
The altitudes are concurrent; they meet at the Orthocentre . (No need to prove this.)
(c)
The perpendicular bisectors of a triangle meet at the circumcentre (no need to prove this)
(d)
Intersection Points
Do examples such as “Triangle ABC has A(2,9), B(-11,-1) and C(9,-5).
(i)
(ii)
(iii)
Find the equation of the median through A.
Find the equation of the altitude through B.
Find the coordinates of the point of intersection
of the two lines.”
NB Coordinates to be given!
Always do a reasonably accurate sketch, if one is not already provided.
(3)
(3)
(3)
Answers
(e)
1.
MID BC (-1,-3)
2.
MA MID =
3.
Equation of MEDIAN y  9  4( x  2)
4.
MAC =
5.
MPERP =
6.
Equation of Altitude y  1 
1
( x  11)
2
7.
y  9  4( x  2) and y  1 
1
( x  11) or equivalent.
2
8.
For x  1
9.
For y  5
12
4
3
14
 2
7
1
2
(m1m2 = -1)
Collinear
To prove points A, B and C are collinear, pupils must show e.g. MAB is a multiple of MAC
explain lines are parallel then state this and mention a common point – thus collinear. (can also
use vectors from unit 3).
8.
Some General Points
(i) Some candidates, even able ones, will make careless mistakes when obtaining the
equations of straight lines. Give them a great deal of encouragement not to accept bizarre
results, but to work through the question again, locating and correcting the mistake. (The
use of negative signs and fractions especially worth checking).
(ii) When all the topics of the outcome have been taught, give the pupils practice at “mixed
examples”.
(iii) S.Q.A. past papers contain many excellent examples.
(iv) Weaker candidates are likely to need practice at solving simple linear equations.
(v) It is often useful to demonstrate more than one way of doing a question, for example
proving that a triangle is right angled by either
Converse of P.T.
or by
Using gradients.
(vi) Pupils must know the meaning of the terms “collinear” and “concurrent”.
Do examples showing three lines to be concurrent by finding the coordinates of the
intersection point of two of the lines and then determining whether the intersection point
lies on the third line.
9.
References/Resources
1.
2.
“H.H.M.” pp 1-21.
MIA pp 2 - 18