1. Sketch the graph.
a.) y = log2x (1 mark)
b.) y = log2(2)x (2 marks)
2. f(x) = 3 - x and g(x) = 3/x , x ≠ 0
a.) Find p(x) where p(x) = f(g(x)) (2 marks)
b.) If q(x) = 3 / (3-x) , x ≠ 0, find p(q(x)) in its simplest form. (3 marks)
3. Using the graph y = log2x from question 1.
a.) If the coordinates were (a,0) and (8,b), state the values of a and b. (1 mark)
b.) Sketch the graph of y = log2(x +1) - 3 (3 marks)
4. Using the graph y = sinx, sketch the graphs of:
a.) y = sin(x+1) (1 mark)
b.) y = -2sin(x) (1 mark)
5. f(x) = 2x - 1, g(x) = 3 - 2x and h(x) = 1/4 (5 - x)
a.) Find a formula for k(x) where k(x) = f(g(x)) (2 marks)
b.) Find a formula for h(k(x)) (2 marks)
c.) What is the connection between the functions h and k? (1 mark)
Monday, 7 November 2011
Unit 1, Outcome 2 - Functions and Graphs
1. Sets
1. List all the numbers in the set P = { x ∈ ℕ : 1 < x < 5 }.
P = { 2, 34 }
2. Functions
1. A function g is defined by g (x) = x - ( 6 / (x+4) ). Define a suitable domain for g.
{ x ∈ R : x ≠ 4 }.
2. A function f is defined by f (x) = sinx° for x ∈ R. Identify its range.
y = sinx°
range = -1 < x < 1
3. Composite Functions
1. Functions f and g are defined by f(x) = 2x and g(x) = x - 3. Find:
a.) f(2)
f(2) = 2x
= 2 x 2
= 4
b.) f ( g(x) )
f ( g(x) ) = f ( x - 3)
= 2(x - 3)
= 2x - 6
c.) g ( f(x) )
g ( f(x) ) = g ( 2x )
= 2x - 3
2. Functions f and g are defined on suitable domains by f(x) = x³ + 1 and g(x) = 1/x. Find formulae for h(x) = f ( g(x) ) and k(x) = g ( f(x) ).
h(x) = f ( 1/x )
= (1/x)³ + 1
k(x) = g ( x³ + 1)
= 1 / ( x³ + 1 )
4. Inverse Functions
1. f(x) = 5x so g(x) = (1/5)x
2. h(x) = 1/2 (1 - 5x) so k(x) = (1- 2x) / 5
5. Exponential Functions
6. Introduction to Logarithms
1. Sketch the curve with equation y = log6x
x = 1, y = log6(1) = 0 (1,0)
x = 6, y = log6(6) = 1 (6,1)
7. Radians
1. Convert from degrees to radians.
a.) 50° = 50 x (π/180) = (5/18)π radians
b.) 60° = 60 x (π/180) = π/3 radians
2. Convert from radians to degrees.
a.) π/4 = 180/4 = 45°
b.) π/6 = 180/6 = 30°
8. Exact Values
1. Find sinx + cosx, when x = π/6 radians.
sinπ/6 + cosπ/6 = (1/2) + (√3/2)
= (1 + √3)/2
10. Graph Transformations
1. Sketch the graph of y = -f(x) - 2
Firstly reflect in the x-axis then move down by 2.
2. Sketch the graph of y = 5cos(2x°) when 0 ≤ x ≤ 360.
NOTE:
TRANSLATION
y = f(x) + a THEN if it's positive move up, negative move down.
y = f(x + a) THEN if it's positive move left, negative move right.
REFLECTION
y = - f(x) THEN flip in the x-axis
y = f(-x) THEN flip in the y-axis.
SCALING
y = kf(x) THEN stretches/compresses vertically.
If k > 1 THEN stretch.
If 0 < k < 1 THEN compress
y = f(kx) THEN stretches/compresses horizontally.
If k > 1 THEN stretch.
If 0 < k < 1 THEN compress
Always reflect before you translate, but scale before you reflect.
1. List all the numbers in the set P = { x ∈ ℕ : 1 < x < 5 }.
P = { 2, 34 }
2. Functions
1. A function g is defined by g (x) = x - ( 6 / (x+4) ). Define a suitable domain for g.
{ x ∈ R : x ≠ 4 }.
2. A function f is defined by f (x) = sinx° for x ∈ R. Identify its range.
y = sinx°
range = -1 < x < 1
3. Composite Functions
1. Functions f and g are defined by f(x) = 2x and g(x) = x - 3. Find:
a.) f(2)
f(2) = 2x
= 2 x 2
= 4
b.) f ( g(x) )
f ( g(x) ) = f ( x - 3)
= 2(x - 3)
= 2x - 6
c.) g ( f(x) )
g ( f(x) ) = g ( 2x )
= 2x - 3
2. Functions f and g are defined on suitable domains by f(x) = x³ + 1 and g(x) = 1/x. Find formulae for h(x) = f ( g(x) ) and k(x) = g ( f(x) ).
h(x) = f ( 1/x )
= (1/x)³ + 1
k(x) = g ( x³ + 1)
= 1 / ( x³ + 1 )
4. Inverse Functions
1. f(x) = 5x so g(x) = (1/5)x
2. h(x) = 1/2 (1 - 5x) so k(x) = (1- 2x) / 5
5. Exponential Functions
1. Sketch the curve with equation y = log6x
x = 1, y = log6(1) = 0 (1,0)
x = 6, y = log6(6) = 1 (6,1)
1. Convert from degrees to radians.
a.) 50° = 50 x (π/180) = (5/18)π radians
b.) 60° = 60 x (π/180) = π/3 radians
2. Convert from radians to degrees.
a.) π/4 = 180/4 = 45°
b.) π/6 = 180/6 = 30°
8. Exact Values
1. Find sinx + cosx, when x = π/6 radians.
sinπ/6 + cosπ/6 = (1/2) + (√3/2)
= (1 + √3)/2
10. Graph Transformations
1. Sketch the graph of y = -f(x) - 2
Firstly reflect in the x-axis then move down by 2.
2. Sketch the graph of y = 5cos(2x°) when 0 ≤ x ≤ 360.
NOTE:
TRANSLATION
y = f(x) + a THEN if it's positive move up, negative move down.
y = f(x + a) THEN if it's positive move left, negative move right.
REFLECTION
y = - f(x) THEN flip in the x-axis
y = f(-x) THEN flip in the y-axis.
SCALING
y = kf(x) THEN stretches/compresses vertically.
If k > 1 THEN stretch.
If 0 < k < 1 THEN compress
y = f(kx) THEN stretches/compresses horizontally.
If k > 1 THEN stretch.
If 0 < k < 1 THEN compress
Always reflect before you translate, but scale before you reflect.
Straight Line Revision Questions
1. A is the point ( 7, 0 ), B is ( -3, -2 ) and C ( -1, 8 ). The Median CE and the altitude BD intersect at J.
a.) Find the equations of CE and BD. (6 marks)
b.) Find the co-ordinates of J. (2 marks)
2. A triangle ABC has vertices A ( 4, 8 ) , B ( 1, 2 ) and C ( 7, 2 ).
a.) Show that the triangle is isosceles. (2 marks)
b.) (i) The altitudes AD and BE intersect at H, where D and E lie on BC and CA reprectively. Find the coordinates of H. (7 marks)
(ii) Hence show that H lies one quarter of the way up DA. (1 mark)
3. Find the equation of the perpendicular bisector of the line joining A ( 2, -1 ) and B ( 8, 3 ). (4 marks)
4. P ( -4, 5 ) , Q ( -2, -2 ) and R ( 4, -1 ) are the vertices of triangle PQR. Find the equation of PS, the altitude from P. (3 marks)
5. A circles passes through A ( -2, 3 ) and B ( 4, -1 ). Find the equation of the perpendicular to the chord AB. ( 4 marks)
6. ABCD is a square. A is the point with coordinates ( 3, 4 ) and ODC has equation y = (1/2)x.
a.) Find the equation of the line AD. (3 marks)
b.) Find the coordinates of D. (3 marks)
c.) Find the area of the square ABCD (2 marks)
a.) Find the equations of CE and BD. (6 marks)
b.) Find the co-ordinates of J. (2 marks)
2. A triangle ABC has vertices A ( 4, 8 ) , B ( 1, 2 ) and C ( 7, 2 ).
a.) Show that the triangle is isosceles. (2 marks)
b.) (i) The altitudes AD and BE intersect at H, where D and E lie on BC and CA reprectively. Find the coordinates of H. (7 marks)
(ii) Hence show that H lies one quarter of the way up DA. (1 mark)
3. Find the equation of the perpendicular bisector of the line joining A ( 2, -1 ) and B ( 8, 3 ). (4 marks)
4. P ( -4, 5 ) , Q ( -2, -2 ) and R ( 4, -1 ) are the vertices of triangle PQR. Find the equation of PS, the altitude from P. (3 marks)
5. A circles passes through A ( -2, 3 ) and B ( 4, -1 ). Find the equation of the perpendicular to the chord AB. ( 4 marks)
6. ABCD is a square. A is the point with coordinates ( 3, 4 ) and ODC has equation y = (1/2)x.
a.) Find the equation of the line AD. (3 marks)
b.) Find the coordinates of D. (3 marks)
c.) Find the area of the square ABCD (2 marks)
Unit 1, Outcome 1 - Straight Lines
1. The Distance Between Points
1. Calculate the distance between the points ( -7, -3 ) and ( 16, -3 )
( -7, -3 ) and ( 16, -3 )
x1 = -7 x2 = 16 y1 = -3 y2 = -3
Distance (d) = x2 - x1
= 16 - ( -7 )
= 16 + 7
= 23
2. A is the point ( -2, 4 ) and B ( 3, 1). Calculate the length of the line AB.
( -2, 4 ) and ( 3, 1 )
x1 = -2 x2 = 3 y1 = 4 y2 = 1
Distance (d) = √ ( x2 - x1 )² + ( y2 - y1)²
= √ ( 1 - 4 )² + ( 3 - (-2) )²
= √ ( 9 + 25)
= √ 34
= 5.83
3. Calculate the distance between the points ( 1/2, - 15/4 ) and ( -1, -1 )
Distance (d) = √ ( x2 - x1 )² + ( y2 - y1)²
= √ ( (-1) - 1/2 )² + ( (-1) - (-15/4) )²
= √ (-3/2)² + (11/4)²
= √ 9/4 + 121/16
= √ 36/16 + 121/16
= √ 157/16
= (√157) / 4
2. The Midpoint Formula
1. Calculate the midpoint of the points ( 1, -4 ) and ( 7, 8 )
x1 = 1 x2 = 7 y1 = -4 y2 = 8
Midpoint (m) = ( (x1 + x2)/2 , (y1 + y2)/2 )
= ( ( 1 + 7 )/2 , ( (-4) + 8 )/2 )
= ( 4, 2 )
2. In the diagram below, A ( 9, -2 ) lies on the circumference of the circle with centre C ( 17, 12 ), and the line AB is a diameter of the circle. Find the coordinates of B.
Let ( x, y ) be B.
Midpoint C, ( 17, 12 ) = ( ( x + 9 )/2 , ( y + (-2) )/2 )
( x + 9 ) / 2 = 17 ( y - 2 ) / 2 = 12
x + 9 = 17 x 2 y - 2 = 12 x 2
x + 9 = 34 y - 2 = 24
x = 34 - 9 y = 24 + 2
x = 25 y = 26
Therefore the Midpoint C is ( 25, 36 )
3. Gradients
1. Calculate the gradient of the straight line shown in the diagram below.
m = tan32° = 0.62
2. Find the angle that the line joining P ( -2, -2 ) and Q ( 1, 7 ) makes with the positive direction of the x-axis.
Gradient (Mpq) = ( 7 - (-2) ) / ( 1 - (-2) )
= 9 / 3
= 3
tanX = 3 X = tan -¹ (3) = 71.6
The angle X = 71.6°
3. Find the size of the angle X shown in the diagram below.
Tan x° = gradient(m) = 5
x° = tan-¹ (5)
x° = 78.7
X = 90° - 78.7
X = 11.3°
4. Collinearity
1. Show that the points P ( -6, -1) , Q ( 0, 2 ) and R ( 8, 6 ) are collinear.
Gradient (Mpq) = ( 2 - (-1) ) / ( 0 - (-6) )
= 3 / 6
= 1 / 2
Gradient (Mqr) = ( 6 - 2 ) / ( 8 - 0 )
= 4 / 8
= 1 / 2
Mpq = Mqr so P, Q and R are Collinear.
2. The points A ( 1, -1 ) , B ( -1, k ) and C ( 5, 7 ) are collinear. Find the value of k.
Mab = Mbc
( k - (-1) ) / ( (-1) - 1 ) = ( 7 - k ) / ( 5 - (-1) )
( k + 1 ) / (-2) = ( 7 - k ) / 6
-2 ( 7 - k ) = 6 ( k + 1 )
-14 + 2k = 6k + 6
-20 = 4k
k = -5
5. Gradients of Perpendicular Lines
1. Given that T is the point ( 1, -2 ) and S i s ( -4, 5), find the gradient of a line perpendicular to ST.
Gradient (ST) = ( 5 - (-2) ) / ( (-4) - 1 )
= - ( 7 / 5 )
Perpendicular Gradient (ST) = 5 / 7
2. Triangle MOP has vertices M ( -3, 9 ) , O ( 0, 0 ) and P ( 12, 4 ). Show that the triangle is right-angled.
Gradient (MO) = ( 9 - 0 ) / ( (-3) - 0 )
= -3
Gradient (OP) = ( 4 - 0 ) / ( 12 - 0 )
= 1 / 3
Since the Gradient (MO) x Gradient (OP) = -1 then Gradient (MO) is perpendicular to Gradient (OP) so the triangle MOP is right-angled.
6. The Equation of a Straight Line
1. Find the equation of the line with gradient 1/3 passing through the point ( 3, -4 ).
m = 1/3 a = 3 b = -4
y - b = m ( x - a )
y - (-4) = 1/3 ( x - 3 )
y + 4 = 1/3 ( x - 3 )
3y + 12 = 1 ( x - 3 )
3y = x - 3 - 12
3y = x - 15
x - 3y - 15 = 0
2. Find the equation of the line passing through A ( 3, 2 ) and B ( -2, 1 )
x1 = 3 x2 = - 2 y1 = 2 y2 = 1
Gradient (AB) = ( y2 - y1 ) / ( x2 - x1 )
= ( 1 - 2 ) / ( (-2) - 3 )
= -1 / -5
= 1 / 5
m = 1/5 a = 3 b = 2
y - b = m ( x - a )
y - 2 = 1/5 ( x - 3 )
5y - 10 = 1 ( x - 3 )
5y = x - 3 + 10
x - 5y + 7 = 0
3. Find the equation of the line passing through ( -3/5, 4 ) and ( -3/5, 5 ).
x1 = -3/5 x2 = -3/5 y1 = 4 y2 =5
Gradient (M) = ( y2 - y1 ) / ( x2 - x1 )
= ( 5 - 4 ) / ( (-3/5) - (-3/5) )
= 1 / 0
= undefined
so equation is x = -3/5
4. Find the gradient of the line with equation 3x + 2y + 4 = 0
3x + 2y + 4 = 0
For y = mx + c
y = (-3/2x) - 4/2
m = -3/2
5. The line through points A ( 3, -3 ) and B has equation 5x - y - 18 = 0. Find the equation of the line through A which is perpendicular to AB.
5x - y - 18 = 0
y = 5x - 18
m = 5
perpendicular m = -1/5
A ( 3, -3 ) a = 3 b = -3 m = -1/5
y - b = m ( x - a )
y - (-3) = -1/5 ( x - 3 )
5y + 15 = -1 ( x - 3 )
5y + 15 = -x + 3
x + 5y + 12 = 0
7. Medians
1. Triangle ABC has vertices A ( 4, -9 ) , B ( 10, 2 ) and C ( 4, -4 ). Find the equation of the median from A.
Midpoint BC = ( ( x1 + x2 ) / 2 ) + ( ( y1 + y2 ) / 2 )
= ( ( 10 + 4 ) / 2 ) + ( ( 2 + (-4) ) / 2 )
= ( 14 / 2 ) + ( (-2) / 2 )
= ( 7, -1 )
Gradient (AM) = ( (-1) - (-9) ) / ( 7 - 4 )
= 8 / 3
m = 8/3 a = 4 b = -9
y - b = m ( x - a )
y - (-9) = 8/3 ( x - 4 )
3y + 27 = 8x - 32
8x - 3y - 59 = 0
8. Altitudes
1. Triangle ABC has vertices A ( 3, -5 ) , B ( 4, 3 ) and C ( -7, 2 ). Find the equation of the altitude from A.
B ( 4, 3 ) and C ( -7, 2 )
x1 = 4 x2 = -7 y1 = 3 y2 = 2
Gradient (BC) = ( y2 - y1 ) / ( x2 - x1 )
= ( 2 - 3 ) / ( (-7) - 4 )
= (-1) / (-11)
= 1/11
Gradient perpendicular = - 11
m = -11 a = 3 b = -5
y - b = m ( x - a )
y - (-5) = -11 ( x - 3 )
y + 5 = -11x + 33
11x + y - 28 = 0
9. Perpendicular Bisectors
1. A is the point ( -2, 1 ) and B is the point ( 4, 7 ). Find the equation of the perpendicular bisector of AB.
x1 = -2 x2 = 4 y1 = 1 y2 = 7
Midpoints = ( ( x1 + x2 )/2 , ( y1 + y2 )/2 )
= ( ( -2 + 4 )/2 , ( 1 + 7 )/2 )
= ( 1, 4 )
Gradient (AB) = ( y2 - y1 ) / ( x2 - x1 )
= ( 7 - 1 ) / ( 4 - (-2) )
= 6/6
= 1
Perpendicular Gradient = -1
m = -1 a = 1 b = 4
y - b = m ( x - a )
y - 4 = -1 ( x - 1 )
y - 4 = -x + 1
x + y - 4 = 0
10. Intersection of Lines
1. Find the point of intersection of the lines 2x - y + 11 = 0 and x + 2y - 7 = 0.
2x - y = -11 .... x2
so
4x - 2y = -22 ADD x + 2y = 7
5x = -15
x = -3
Substiture x = -3 into x + 2y = 7
-3 + 2y = 7
2y = 7 + 3
2y = 10
y = 4
The point of intersection is ( -3, 5 )
2. Triangle PQR has vertices P ( 8, 3 ) , Q ( -1, 6 ) and R ( 2, -3 )
a.) Find the equation of altitude QS.
x1 = 8 x2 = 2 y1 = 3 y2 = -3
Gradient (PQ) = ( y2 - y1 ) / ( x2 - x1 )
= ( (-3) -3 ) / ( 2 - 8 )
= -6 / -6
= 1
Perpendicular gradient = -1
m = -1 a = -1 b = 6
y - b = m ( x - a )
y - 6 = -1 ( x + 1 )
y - b = - x - 1
x + y - 5 = 0
b.) Find the equation of median RT.
T = ( 8 + (-1) )/2 , ( 3 + 6 )/2
= ( 7/2 , 9/2 )
Gradient (RT) = ( (-3) - 9/2 ) / ( 2 - 7/2 )
= (-15/2) / ( - 3/2)
= -15 / -3
= 5
m = 5 a = 2 b = -3
y - b = m ( x - a )
y - (-3) = m ( x - 2)
5x - y - 13 = 0
c.) Hence find the coordinates of M.
x += y = 5 ADD 5x - y = 13
6x = 18
x = 3
Substitute x = 3 into x + y = 5
3 + y = 5
y = 2
Point of intersection (M) is ( 3, 2 )
1. Calculate the distance between the points ( -7, -3 ) and ( 16, -3 )
( -7, -3 ) and ( 16, -3 )
x1 = -7 x2 = 16 y1 = -3 y2 = -3
Distance (d) = x2 - x1
= 16 - ( -7 )
= 16 + 7
= 23
2. A is the point ( -2, 4 ) and B ( 3, 1). Calculate the length of the line AB.
( -2, 4 ) and ( 3, 1 )
x1 = -2 x2 = 3 y1 = 4 y2 = 1
Distance (d) = √ ( x2 - x1 )² + ( y2 - y1)²
= √ ( 1 - 4 )² + ( 3 - (-2) )²
= √ ( 9 + 25)
= √ 34
= 5.83
3. Calculate the distance between the points ( 1/2, - 15/4 ) and ( -1, -1 )
Distance (d) = √ ( x2 - x1 )² + ( y2 - y1)²
= √ ( (-1) - 1/2 )² + ( (-1) - (-15/4) )²
= √ (-3/2)² + (11/4)²
= √ 9/4 + 121/16
= √ 36/16 + 121/16
= √ 157/16
= (√157) / 4
2. The Midpoint Formula
1. Calculate the midpoint of the points ( 1, -4 ) and ( 7, 8 )
x1 = 1 x2 = 7 y1 = -4 y2 = 8
Midpoint (m) = ( (x1 + x2)/2 , (y1 + y2)/2 )
= ( ( 1 + 7 )/2 , ( (-4) + 8 )/2 )
= ( 4, 2 )
2. In the diagram below, A ( 9, -2 ) lies on the circumference of the circle with centre C ( 17, 12 ), and the line AB is a diameter of the circle. Find the coordinates of B.
Let ( x, y ) be B.
Midpoint C, ( 17, 12 ) = ( ( x + 9 )/2 , ( y + (-2) )/2 )
( x + 9 ) / 2 = 17 ( y - 2 ) / 2 = 12
x + 9 = 17 x 2 y - 2 = 12 x 2
x + 9 = 34 y - 2 = 24
x = 34 - 9 y = 24 + 2
x = 25 y = 26
Therefore the Midpoint C is ( 25, 36 )
3. Gradients
1. Calculate the gradient of the straight line shown in the diagram below.
m = tan32° = 0.62
2. Find the angle that the line joining P ( -2, -2 ) and Q ( 1, 7 ) makes with the positive direction of the x-axis.
Gradient (Mpq) = ( 7 - (-2) ) / ( 1 - (-2) )
= 9 / 3
= 3
tanX = 3 X = tan -¹ (3) = 71.6
The angle X = 71.6°
3. Find the size of the angle X shown in the diagram below.
Tan x° = gradient(m) = 5
x° = tan-¹ (5)
x° = 78.7
X = 90° - 78.7
X = 11.3°
4. Collinearity
1. Show that the points P ( -6, -1) , Q ( 0, 2 ) and R ( 8, 6 ) are collinear.
Gradient (Mpq) = ( 2 - (-1) ) / ( 0 - (-6) )
= 3 / 6
= 1 / 2
Gradient (Mqr) = ( 6 - 2 ) / ( 8 - 0 )
= 4 / 8
= 1 / 2
Mpq = Mqr so P, Q and R are Collinear.
2. The points A ( 1, -1 ) , B ( -1, k ) and C ( 5, 7 ) are collinear. Find the value of k.
Mab = Mbc
( k - (-1) ) / ( (-1) - 1 ) = ( 7 - k ) / ( 5 - (-1) )
( k + 1 ) / (-2) = ( 7 - k ) / 6
-2 ( 7 - k ) = 6 ( k + 1 )
-14 + 2k = 6k + 6
-20 = 4k
k = -5
5. Gradients of Perpendicular Lines
1. Given that T is the point ( 1, -2 ) and S i s ( -4, 5), find the gradient of a line perpendicular to ST.
Gradient (ST) = ( 5 - (-2) ) / ( (-4) - 1 )
= - ( 7 / 5 )
Perpendicular Gradient (ST) = 5 / 7
2. Triangle MOP has vertices M ( -3, 9 ) , O ( 0, 0 ) and P ( 12, 4 ). Show that the triangle is right-angled.
Gradient (MO) = ( 9 - 0 ) / ( (-3) - 0 )
= -3
Gradient (OP) = ( 4 - 0 ) / ( 12 - 0 )
= 1 / 3
Since the Gradient (MO) x Gradient (OP) = -1 then Gradient (MO) is perpendicular to Gradient (OP) so the triangle MOP is right-angled.
6. The Equation of a Straight Line
1. Find the equation of the line with gradient 1/3 passing through the point ( 3, -4 ).
m = 1/3 a = 3 b = -4
y - b = m ( x - a )
y - (-4) = 1/3 ( x - 3 )
y + 4 = 1/3 ( x - 3 )
3y + 12 = 1 ( x - 3 )
3y = x - 3 - 12
3y = x - 15
x - 3y - 15 = 0
2. Find the equation of the line passing through A ( 3, 2 ) and B ( -2, 1 )
x1 = 3 x2 = - 2 y1 = 2 y2 = 1
Gradient (AB) = ( y2 - y1 ) / ( x2 - x1 )
= ( 1 - 2 ) / ( (-2) - 3 )
= -1 / -5
= 1 / 5
m = 1/5 a = 3 b = 2
y - b = m ( x - a )
y - 2 = 1/5 ( x - 3 )
5y - 10 = 1 ( x - 3 )
5y = x - 3 + 10
x - 5y + 7 = 0
3. Find the equation of the line passing through ( -3/5, 4 ) and ( -3/5, 5 ).
x1 = -3/5 x2 = -3/5 y1 = 4 y2 =5
Gradient (M) = ( y2 - y1 ) / ( x2 - x1 )
= ( 5 - 4 ) / ( (-3/5) - (-3/5) )
= 1 / 0
= undefined
so equation is x = -3/5
4. Find the gradient of the line with equation 3x + 2y + 4 = 0
3x + 2y + 4 = 0
For y = mx + c
y = (-3/2x) - 4/2
m = -3/2
5. The line through points A ( 3, -3 ) and B has equation 5x - y - 18 = 0. Find the equation of the line through A which is perpendicular to AB.
5x - y - 18 = 0
y = 5x - 18
m = 5
perpendicular m = -1/5
A ( 3, -3 ) a = 3 b = -3 m = -1/5
y - b = m ( x - a )
y - (-3) = -1/5 ( x - 3 )
5y + 15 = -1 ( x - 3 )
5y + 15 = -x + 3
x + 5y + 12 = 0
7. Medians
1. Triangle ABC has vertices A ( 4, -9 ) , B ( 10, 2 ) and C ( 4, -4 ). Find the equation of the median from A.
Midpoint BC = ( ( x1 + x2 ) / 2 ) + ( ( y1 + y2 ) / 2 )
= ( ( 10 + 4 ) / 2 ) + ( ( 2 + (-4) ) / 2 )
= ( 14 / 2 ) + ( (-2) / 2 )
= ( 7, -1 )
Gradient (AM) = ( (-1) - (-9) ) / ( 7 - 4 )
= 8 / 3
m = 8/3 a = 4 b = -9
y - b = m ( x - a )
y - (-9) = 8/3 ( x - 4 )
3y + 27 = 8x - 32
8x - 3y - 59 = 0
8. Altitudes
1. Triangle ABC has vertices A ( 3, -5 ) , B ( 4, 3 ) and C ( -7, 2 ). Find the equation of the altitude from A.
B ( 4, 3 ) and C ( -7, 2 )
x1 = 4 x2 = -7 y1 = 3 y2 = 2
Gradient (BC) = ( y2 - y1 ) / ( x2 - x1 )
= ( 2 - 3 ) / ( (-7) - 4 )
= (-1) / (-11)
= 1/11
Gradient perpendicular = - 11
m = -11 a = 3 b = -5
y - b = m ( x - a )
y - (-5) = -11 ( x - 3 )
y + 5 = -11x + 33
11x + y - 28 = 0
9. Perpendicular Bisectors
1. A is the point ( -2, 1 ) and B is the point ( 4, 7 ). Find the equation of the perpendicular bisector of AB.
x1 = -2 x2 = 4 y1 = 1 y2 = 7
Midpoints = ( ( x1 + x2 )/2 , ( y1 + y2 )/2 )
= ( ( -2 + 4 )/2 , ( 1 + 7 )/2 )
= ( 1, 4 )
Gradient (AB) = ( y2 - y1 ) / ( x2 - x1 )
= ( 7 - 1 ) / ( 4 - (-2) )
= 6/6
= 1
Perpendicular Gradient = -1
m = -1 a = 1 b = 4
y - b = m ( x - a )
y - 4 = -1 ( x - 1 )
y - 4 = -x + 1
x + y - 4 = 0
10. Intersection of Lines
1. Find the point of intersection of the lines 2x - y + 11 = 0 and x + 2y - 7 = 0.
2x - y = -11 .... x2
so
4x - 2y = -22 ADD x + 2y = 7
5x = -15
x = -3
Substiture x = -3 into x + 2y = 7
-3 + 2y = 7
2y = 7 + 3
2y = 10
y = 4
The point of intersection is ( -3, 5 )
2. Triangle PQR has vertices P ( 8, 3 ) , Q ( -1, 6 ) and R ( 2, -3 )
a.) Find the equation of altitude QS.
x1 = 8 x2 = 2 y1 = 3 y2 = -3
Gradient (PQ) = ( y2 - y1 ) / ( x2 - x1 )
= ( (-3) -3 ) / ( 2 - 8 )
= -6 / -6
= 1
Perpendicular gradient = -1
m = -1 a = -1 b = 6
y - b = m ( x - a )
y - 6 = -1 ( x + 1 )
y - b = - x - 1
x + y - 5 = 0
b.) Find the equation of median RT.
T = ( 8 + (-1) )/2 , ( 3 + 6 )/2
= ( 7/2 , 9/2 )
Gradient (RT) = ( (-3) - 9/2 ) / ( 2 - 7/2 )
= (-15/2) / ( - 3/2)
= -15 / -3
= 5
m = 5 a = 2 b = -3
y - b = m ( x - a )
y - (-3) = m ( x - 2)
5x - y - 13 = 0
c.) Hence find the coordinates of M.
x += y = 5 ADD 5x - y = 13
6x = 18
x = 3
Substitute x = 3 into x + y = 5
3 + y = 5
y = 2
Point of intersection (M) is ( 3, 2 )
Hey
Hey,
Quite of a few of you guys have been asking me and CJ about our notes from last year so we thought it'd be much easier if we just set up a blog for you guys to see them online. So here you are! Maybe we'll all have a better chance of passing our exam this way haha.
Quite of a few of you guys have been asking me and CJ about our notes from last year so we thought it'd be much easier if we just set up a blog for you guys to see them online. So here you are! Maybe we'll all have a better chance of passing our exam this way haha.
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