NEW Higher

1
NEW Higher
Unit 1 Applications 1.1
Prior Knowledge
Distance Formula
The Midpoint Formula
m tan ?
Gradients of Perpendicular Lines
Median, Altitude Perpendicular Bisector
Collinearity
Mindmap
Exam Type Questions
2
Circle
Recurrence Relations
Vector
Where is Straight Line Theory Used in Higher
Logs Exponentials
Differentiation
3
Possible values for gradient
Straight Line y mx c
Parallel lines have same gradient
For Perpendicular lines the following is
true. m1.m2 -1
m tan ?
4
Straight Line Facts
y mx c
Y axis Intercept
Another version of the straight line formula is
ax by c 0
5
Questions
Find the gradient and y intercept for equations.
(a) 4x 2y 10 0
m -2 c -5
(b) 3x - 5y 1 0
m 3/5 c 1/5
(c) 4 - x 3y 0
m - 1/3 c 4/3
Demo
6
The Equation of the Straight Liney b m (x -
a)
Demo
The equation of any line can be found if we know
the gradient and one point on the line.
y - b
y

m
(x a)
y - b
b
x a
Gradient, m
a
x
Point (a, b)
y b m ( x a )
Point on the line ( a, b )
7
Gradient Facts
Sloping left to right up has ve gradient
m gt 0
Sloping left to right down has -ve gradient
m lt 0
Horizontal line has zero gradient.
m 0
y c
Vertical line has undefined gradient.
x a
8
Gradient Facts
Lines with the same gradient means lines are
Parallel
m gt 0
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Straight Line Theory
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Straight Line Theory
11
Straight Line Theory
12
Straight Line Theory
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Typical Exam Questions
Find the equation of the straight line which is
parallel to the line with equation
and which passes through the point (2,
1) .
Find gradient of given line
Knowledge Gradient of parallel lines are the
same. Therefore for line we want to find has
gradient
Find equation Using y b m(x - a)
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Distance FormulaLength of a straight line
B(x2,y2)
This is just Pythagoras Theorem
y2 y1
A(x1,y1)
C
x2 x1
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Distance Formula
The length (distance ) of ANY line can be given
by the formula
Just Pythagoras Theorem in disguise
Demo
16
2v26
2v26
Isosceles Triangle !
8v2
Demo
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Mid-Point of a line
The mid-point (Median) between 2 points is given
by
B(x2,y2)
y2
Simply add both x coordinates together and divide
by 2. Then do the same with the y coordinates.
M
A(x1,y1)
y1
x1
x2
Demo
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Straight Line Facts
The gradient of a line is ALWAYS equal to the
tangent of the angle made with the line and the
positive x-axis
?
m tan ?
0o ? lt 180o
tan ?
Demo
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m tan ?
m tan 60o v3
21
m tan ?
y -2x
m tan ?
? tan-1 (-2)
? 180 63.4
? 116.6o
22
Exam Type Questions
Find the size of the angle a that the
line joining the points A(0, -1) and B(3?3, 2)
makes with the positive direction of the x-axis.
Find gradient of the line
Use table of exact values
23
Typical Exam Questions
The line AB makes an angle of 60o with the
y-axis, as shown in the diagram. Find the exact
value of the gradient of AB.
60o
Find angle between AB and x-axis
(x and y axes are perpendicular.)
Use table of exact values
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Typical Exam Questions
72o
135o
63o
45o
The lines and
make angles of a? and b? with the
positive direction of the x-axis, as shown in the
diagram. a) Find the values of a and
b b) Hence find the acute angle between the two
given lines.
Find a
Find b
Find supplement of b
72
angle between two lines
Use angle sum triangle 180
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Gradient of perpendicular lines
Investigation
If 2 lines with gradients m1 and m2 are
perpendicular then m1 m2 -1
When rotated through 90º about the origin A (a,
b) ? B (-b, a)
y
B(-b,a)
-a
A(a,b)
-b
a
O
x
-b
Demo
Conversely If m1 m2 -1 then the two lines
with gradients m1 and m2 are perpendicular.
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Terminology
A
Median means a line from a vertex to the
midpoint of the base.
B
C
D
Demo
Altitude means a perpendicular line from a vertex
to the base.
Demo
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Terminology
Perpendicular bisector - a line that cuts another
line into two equal parts at an angle of 90o
A
C
B
D
Demo
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Any number of lines are said to be concurrent if
there is a point through which they all pass.
For three lines to be concurrent, they must all
pass through a single point.
Demo
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Collinearity
Points are said to be collinear if they lie on
the same straight.
The coordinates A,B C are collinear since they
lie on the same straight line. D,E,F are not
collinear they do not lie on the same straight
line.
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Straight Line Theory
Ratio DPQDPQ 12
DPQ2v2
DPQv2
Since mPQ mQR and they have a point in common
Q PQR are collinear.
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Common Straight Strategies for Exam Questions
A
Finding the Equation of an Altitude
B
To find the equation of an altitude
Find the gradient of the side it
is perpendicular to ( ).
mAB
C
To find the gradient of the altitude, flip the
gradient of AB and change from positive to
negative
maltitude
1
mAB
Substitute the gradient and the point C into
Important
Write final equation in the form
y b m ( x a )
A x B y C 0
with A x positive.
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Common Straight Strategies for Exam Questions
Q
Finding the Equation of a Median

M
To find the equation of a median

P
Find the midpoint of the side it bisects, i.e.
y2 y1
x2 x1
O


( )
,
M
2
2
Calculate the gradient of the median OM.
Important
Substitute the gradient and either point on
the line (O or M) into
Write answer in the form
A x B y C 0
y b m ( x a )
with A x positive.
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Typical Exam Questions
Find the equation of the line which passes
through the point (-1, 3) and is perpendicular to
the line with equation
Find gradient of given line
Find gradient of perpendicular
Find equation
34
Exam Type Questions
A and B are the points (3, 1) and (5, 5). Find
the equation of a) the line AB. b) the
perpendicular bisector of AB
Find equation of AB
Find gradient of the AB
Gradient of AB (perp)
Find mid-point of AB
Use y b m(x a) and point ( 1, 2) to obtain
line of perpendicular bisector of AB we get
35
Typical Exam Questions
A triangle ABC has vertices A(4, 3), B(6, 1) and
C(2, 3) as shown in the diagram. Find the
equation of AM, the median from B to C
Find mid-point of BC
Find gradient of median AM
Find equation of median AM
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Typical Exam Questions
P(4, 5), Q(2, 2) and R(4, 1) are the
vertices of triangle PQR as shown in the diagram.
Find the equation of PS, the altitude from P.
Find gradient of QR
Find gradient of PS (perpendicular to QR)
Find equation of altitude PS
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Exam Type Questions
p
Triangle ABC has vertices A(1, 6), B(3, 2)
and C(5, 2) Find a) the equation of the line
p, the median from C of triangle ABC. b) the
equation of the line q, the perpendicular
bisector of BC. c) the co-ordinates of the
point of intersection of the lines p and q.
q
Find mid-point of AB
Find gradient of p
(-2, 2)
Find equation of p
(1, 0)
Find gradient of BC
Find mid-point of BC
Find gradient of q
Find equation of q
(0, 2)
Solve p and q simultaneously for intersection
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Exam Type Questions
l2
l1
Triangle ABC has vertices A(2, 2), B(12, 2) and
C(8, 6). a) Write down the equation of l1, the
perpendicular bisector of AB b) Find the
equation of l2, the perpendicular bisector of AC.
c) Find the point of intersection of lines l1
and l2.
Mid-point AB
Perpendicular bisector AB
(5, 4)
Find mid-point AC
Find gradient of AC
Equation of perp. bisector AC
Gradient AC perp.
(7, 1)
Point of intersection
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Exam TypeQuestions
A triangle ABC has vertices A(4, 1), B(12,3) and
C(7, 7). a) Find the equation of the median
CM. b) Find the equation of the altitude
AD. c) Find the co-ordinates of the point
of intersection of CM and AD
Mid-point AB
Gradient CM (median)
Equation of median CM using y b m(x a)
Gradient of perpendicular AD
Gradient BC
Equation of AD using y b m(x a)
(6, -4)
Solve simultaneously for point of intersection
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Exam Type Questions
M
A triangle ABC has vertices A(3, 3), B(1, 1)
and C(7,3). a) Show that the triangle ABC
is right angled at B. b) The medians AD and
BE intersect at M. i) Find the equations of
AD and BE. ii) Find find the co-ordinates of M.
Gradient AB
Gradient BC
Product of gradients
Hence AB is perpendicular to BC, so B 90
Gradient of median AD
Mid-point BC
Equation AD
Gradient of median BE
Mid-point AC
Equation AD
Solve simultaneously for M, point of intersection
41
Possible values for gradient
Straight Line y mx c
Parallel lines have same gradient
For Perpendicular lines the following is
true. m1.m2 -1
m tan ?
42
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