Transcript The Laws Of Surds.

NEW Higher
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APP 1.1
Unit 1 Applications 1.1
Prior Knowledge
Distance Formula
The Midpoint Formula
m = tan θ
Gradients of Perpendicular Lines
Median, Altitude & Perpendicular Bisector
Collinearity
Exam Type Questions
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Mindmap
Circle
Recurrence
Relations
Vector
Where is
Straight Line Theory
Used in Higher
Logs &
Exponentials
Differentiation
Distance between
2 points
m<0
m = undefined
x  x y  y 
Mid   1 2 , 1 2 
2 
 2
Terminology
D  (x2  x1 )2  (y2  y1 )2
Median – midpoint
Bisector – midpoint
Perpendicular – Right Angled
Altitude – right angled
m=0
m>0
m1.m2 = -1
Possible values
for gradient
Straight Line
y = mx + c
Form for finding
line equation
y – b = m(x - a)
(a,b) = point on line
Parallel lines
have same
gradient
For Perpendicular lines
the following is true.
m1.m2 = -1
m = gradient
m
y2  y1
x2  x1
c = y intercept (0,c)
m = tan θ
θ
Straight Line Facts
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APP 1.1
y = mx + c
y2 - y1
Gradient =
x2 - x1
Y – axis
Intercept
Another version of the straight line formula is
ax + by + c = 0
Questions
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APP 1.1
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
The Equation of the Straight Line
y – b = m (x - a)
APP 1.1
Demo
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The equation of any line can be found if we know
the gradient and one point on the line.
y
y
P (x, y)
m
b
O
m=
y -- bb
y-b
(x – a)
A (a, b)
xx –- aa
a
x
x
Gradient, m Point (a,
y–b=m(x–a)
b)
Point on the line ( a, b )
Outcome 1
Higher
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Sloping left to right up has +ve gradient
m>0
Sloping left to right down has -ve gradient
m<0
Horizontal line has zero gradient.
m=0
y=c
Vertical line has undefined gradient.
x=a
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7
Outcome 1
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Higher
Lines with the same gradient
m>0
means lines are Parallel
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8
Straight Line Theory
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APP 1.1
Straight Line Theory
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APP 1.1
Straight Line Theory
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APP 1.1
Straight Line Theory
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APP 1.1
Typical Exam Questions
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APP 1.1
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) .
Find gradient of given line:
2
3
3 y  2 x  5  y   x  5  m  
2
3
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)
3 y  2x 1  0
m
2
3
Distance Formula
Length of a straight line
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APP 1.1
(AB)2 =(AC)2 +(BC)2
y
B(x2,y2)
This is just
y2 – y1
A(x1,y1)
Pythagoras’ Theorem
x2 – x1
O
C
x
Distance Formula
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APP 1.1
The length (distance ) of ANY line
can be given by the formula :
DAB  (x2  x1 )2  (y2  y1 )2
Demo
Just
Pythagoras
Theorem in
disguise
DAB  (x2  x1 )  (y2  y1 )
2
2
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APP 1.1
2√26
Isosceles Triangle !
2√26
8√2
Demo
Mid-Point of a line
APP 1.1
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The mid-point (Median)
y
between 2 points is given by
y2
Simply add both x
coordinates together
and divide by 2.
Then do the same with
the y coordinates.
MAB
M
A(x1,y1)
y1
O
 x1  x 2 y1  y 2 

,

2 
 2
B(x2,y2)
x1
x2
Demo
x
 x1  x2 y1  y2 
M 
,
,
2
 2

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APP 1.1
Outcome 1
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Higher
The gradient of a line is ALWAYS
equal to the tangent of the angle
θ
made with the line and the positive x-axis
m = tan θ
H
θ
A
O
m=
V
H
=
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O
A
17-Jul-15
0o ≤ θ < 180o
tan θ =
O
A
Demo
19
m = tan θ
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APP 1.1
60o
60o
m = tan 60o = √3
m = tan θ
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APP 1.1
y = -2x
m = tan θ
θ = tan-1 (-2)
θ = 180 – 63.4
θ = 116.6o
Exam Type Questions
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APP 1.1
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.
2  (1)
3


Find gradient of the line: m 
3 3 0
3 3
Use m  tan 
tan  
Use table of exact values
1
3
  tan
1
1
  30
3
1
3
Typical Exam Questions
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APP 1.1
The line AB makes an angle of 60o with
60o
the y-axis, as shown in the diagram.
Find the exact value of the gradient of AB.
Find angle between AB and x-axis:90o
Use m  tan 
m  tan 30o
Use table of exact values
1
m
3
 60o  300 (x and y axes are perpendicular.)
APP 1.1
Typical Exam
Questions
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The lines y  2 x  4
72o
63o
and x  y  13
45o
135o
make angles of a and b with the positive direction of the xaxis, 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 gradient of
y  2x  4
m2
Find a°
tan a  2  a  63
Find gradient of
x  y  13
m  1
Find b°
tan b  1  b  135
Find supplement of b
 180 135  45
Use angle sum triangle = 180° angle between two lines
72°
Gradient of perpendicular lines
APP 1.1
Investigation
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If 2 lines with gradients m1 and m2 are perpendicular then m1 × m2 = -1
When rotated through 90º about the origin A (a, b)
y
B(-b,a)
mOB
a-0
a

-b - 0
b
-a
-b
-b
mOA  mOB
→ B (-b, a)
A(a,b)
O
mOA
a
b-0 b


a-0 a
x
b a
ab
 
 -1
a -b
ab
Demo
Conversely:
If m1 × m2 = -1 then the two lines with gradients m1 and m2 are perpendicular.
Outcome 1
Higher
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Median means a line
from a vertex to
the midpoint of the base.
A
A
B
D
D
Demo
C
Altitude means a perpendicular line
B
C
Demo
from a vertex to the base.
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Outcome 1
Higher
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Perpendicular bisector - a line that cuts another line
into two equal parts at an angle of 90o
A
B
D
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C
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Demo
27
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
Collinearity
APP 1.1
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Points are said to be collinear
if they lie on the same straight.
y
C
The coordinates A,B C are
collinear since they lie on
the same straight line.
B
D,E,F are not collinear they
do not lie on the same
straight line.
A
E
F
D
O
x1
x2
x
DPQ=√2
DPQ=2√2
Straight Line TheoryDPQ:DPQ
APP 1.1
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Ratio
1:2
Since mPQ = mQR and
y  y1
m 2
x2  x1
mPQ 
2 1
1
54
they have a point in common Q
mQR 
4 2
1
7 5
PQR are collinear.
Common Straight Strategies for Exam Questions
Finding the Equation of an Altitude
A
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:
• Substitute the gradient
and the point C into
y – b = m( x – a)
maltitude
=
Important
Write final equation in the form
Ax + By + C = 0
with A x positive.
–1
mAB
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.
M =
(
x2 + x1
2
,
y2 + y1
2
)
O
• Calculate the gradient of the median OM.
• Substitute the gradient and either
point on the line (O or M) into
y – b = m( x – a)
Important
Write answer in the form
Ax + By + C = 0
with A x positive.
Typical Exam Questions
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APP 1.1
Find the equation of the line which passes through the point (-1, 3)
and is perpendicular to the line with equation 4 x  y  1  0
Find gradient of given line:
4 x  y  1  0  y  4 x  1  m  4
Find gradient of perpendicular:
Find equation:
m
1
(using formula m  m  1)
1
2
4
1
y 3

 x  1  4( y  3)  x  1  4 y  12
4 x( 1)
4 y  x  13  0
Exam Type Questions
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APP 1.1
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 gradient of the AB: m 
Find mid-point of AB
1, 2 
3
4
Find equation of AB 4 y  3x  5
4
Gradient of AB (perp): m  
3
Use y – b = m(x – a) and point ( 1, 2) to obtain line of perpendicular
bisector of AB we get
3 y  4 x  10  0
Typical Exam Questions
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APP 1.1
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
 x1  x2 y1  y2 
,
Find mid-point of BC: (2, 1) Using M 

2
2


Find gradient of median AM m  2 Using m 
y2 - y1
x2 - x1
Find equation of median AM y  2 x  5 Using y - b  m( x - a)
Typical Exam Questions
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APP 1.1
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: m 
y -y
1
Using m  2 1
2
x2 - x1
Find gradient of PS (perpendicular to QR)
m   2 (m1  m2  1)
Find equation of altitude PS
y  2x  3  0
Using y  b  m( x  a)
Exam Type
Questions
APP 1.1
p
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Triangle ABC has vertices
q
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.
Find mid-point of AB (-2, 2)
Find equation of p
y2
Find mid-point of BC
(1, 0)
Find gradient of q
m  2
Find gradient of p
Find gradient of BC
Find equation of q
Solve p and q simultaneously for intersection (0, 2)
m0
m
1
2
y  2 x  2
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APP 1.1
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
 7, 2
Find mid-point AC
Gradient AC perp.
Point of intersection
Perpendicular bisector AB
x7
2
(5, 4) Find gradient of AC m 
3
3
m
Equation of perp. bisector AC 2 y  3x  23
2
(7, 1)
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APP 1.1
Exam Type
Questions
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
 4, 2
Gradient CM (median)
Equation of median CM using y – b = m(x – a)
Gradient BC
m2
m  3
y  3x  14  0
Gradient of perpendicular AD
Equation of AD using y – b = m(x – a)
2y  x  2  0
Solve simultaneously for point of intersection (6,
-4)
m
1
2
Exam Type
Questions
M
APP 1.1
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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
m2
Product of gradients
Gradient BC
2  
1
 1
2
1
m
2
Hence AB is perpendicular to BC, so B = 90°
1
Equation AD 3 y  x  6  0
3
4
2,

3
m


Equation AD 3 y  4 x  1  0
Mid-point AC 
 Gradient of median BE
3
5

Solve simultaneously for M, point of intersection  1,  
3

Mid-point BC  3, 1 Gradient of median AD m 
Distance between
2 points
m<0
m = undefined
x  x y  y 
Mid   1 2 , 1 2 
2 
 2
Terminology
D  (x2  x1 )2  (y2  y1 )2
Median – midpoint
Bisector – midpoint
Perpendicular – Right Angled
Altitude – right angled
m=0
m>0
m1.m2 = -1
Possible values
for gradient
Straight Line
y = mx + c
Form for finding
line equation
y – b = m(x - a)
(a,b) = point on line
Parallel lines
have same
gradient
For Perpendicular lines
the following is true.
m1.m2 = -1
m = gradient
m
y2  y1
x2  x1
c = y intercept (0,c)
m = tan θ
θ
Are you on Target !
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APP 1.1
•
Update you log book
•
Make sure you complete and correct
ALL of the Straight Line questions in
the past paper booklet.