Sample
Consider the figure and answer the questions that follow.
1. Write down the coordinates of A, B, C and D.
2. Calculate the gradients of AD and BC. What do you notice?
3. Calculate the gradients of AB and DC. What do you notice?
4. What type of figure is ABCD? Give a reason for your answer.
REMEMBER
The order of the letters by which we name a figure is important.
The name ABCD tells us that we move from point A to point B, from B to point C, from C to point D and then back to point A in order to form figure ABCD.
Solution
1. A(−1; 3), B(2; −1), C(8; 1), D(5; 5)
2. Gradient of AD = 5 3
5 ( 1) = 2 6 = 1 3
Gradient of BC = 1 ( 1) 8 2 = 2 6 = 1 3
The gradients (inclinations) are equal.
3. Gradient of AB = 1 3 2 − (−1) = 4 3
Gradient of DC = 1 5 8 − 5 = −4 3
The gradients (inclinations) are equal.
4. It is a parallelogram, because both pairs of opposite sides are parallel.
9.1 THE DISTANCE FORMULA
In the diagram below, two points are given: (−3; −1) and (2; 3). We want to find the distance between them.
In order to find the distance between the points (−3; −1) and (2; 3), we will add a point vertically below (2; 3) and horizontally to the right of (−3; −1). This new point is therefore (2; −1). If we connect the three points, they form a rightangled triangle with the 90° angle at the new point.
We use the theorem of Pythagoras to calculate the distance between (−3; −1) and (2; 3).
The length of the vertical side is (3 −(−1)) = 4 units and the length of the horizontal side is (2 − (−3)) = 5 units. (distance)² = (3 −(−1))² + (2 − (−3))² Pythagoras = 4² + 5²
= 16 + 25 = 41
∴ distance = √ 41
We can generalise this calculation and get a formula – a formula is simply the generalisation of a specific case.
REMEMBER
The formula used for calculating the distance between two points, A (�� 1; ��1) and B (��2; ��2), is:
the
Worked example 1
Worked example 2
The distance between A(��;4) and B(2; 1) is 5 units. Find the value of ��.
Solution
4)
−2)
The coordinates of A can therefore be (6; 4) or (−2; 4).
There are two answers, because there are two possible positions for A. This is shown in the sketch below.
4)
1)
4)
Important
Before you start solving a problem in analytical geometry, first draw a sketch showing the information given. Such a sketch helps you to use the information correctly and also to ensure that your answer is correct.
Exercise 9.1: The distance formula
1. Draw the points A(−1; 4) and B(7; −2) on a system of axes and calculate the distance from A to B.
A(–1; 4) = (����1; y1) B(7; –2) = (����2; y2) AB = √ (�� 2 �� 1 ) 2+ (�� 2 �� 1) 2 = √ (7 (−1)) 2+ (−2 4) 2 = √ (8) 2+ (−6) 2 = √ 100 = 10 units
2. Draw the points C(−4; 1) and D(3; −4) on a system of axes and calculate the distance from C to D correct to one decimal figure.
C(–4; 1) = (����1; y1)
Sample
D(3; –4) = (����2; y2) CD = √ (�� 2 �� 1 ) 2+ (�� 2 �� 1) 2 = √ (3 (−4)) 2+ (−4 1) 2 = √ (7) 2+ (−5) 2 = √ 74 = 8,60232… = 8,6 units
3. Calculate the lengths of the line segments in the following sketch. Leave your answers in the simplest surd form.
B(–1; 7)
F(–1; 1) E(3; 9)
–1)
–3)
(−1) 2 + (−2) 2
√ 5 units AB = √ (�� 2 �� 1 ) 2 + (�� 2 �� 1) 2
√ (4 (−1)) 2 + (−2 7) 2 = √ (5) 2+ (−9) 2 = √ 106 units
EF = √ (�� 2 �� 1 ) 2+ (�� 2 �� 1) 2 = √ (−1 3) 2 + (1 9) 2 = √ (−4) 2 + (−8) 2 = √ 80 units
A(4; –2)
4. The point B(b; −3) is 5 units from the origin.
Find the value of b.
Note the difference between B and b.
The capital letter (B) indicates the point, and the lower case letter (b) indicates a coordinate (in this case the ��-coordinate). √ (��2− ��1)2 + (��2− ��1)2 = 5
√ (�� − 0)2 + (−3 − 0)2 = 5
√ (��2 + 9) = 5
Sample��2 + 9 = 25
��2 = 16
��= ±4
There are therefore two possible positions for B.
5. Given: A(−7; m), B(−3; 4) and AB = 4√ 5 .
Calculate the value of m. Show your answer on a sketch.
A(–7; m) =(����1; y1)
6. A(−1; −1) is equally far from M(0; 2) and P(p; −2).
A (–7; m) =(����2; y2)
y
B(–3; 4) =(����2; y2) 4 √5 4 √5 AB = √ (�� 2 �� 1 ) 2 + (�� 2 �� 1) 2
∴ 4√ 5 = √ (−3 − (−7)) 2 + (4 − m) 2
∴ 80 = 16 + 16 − 8m + m 2
∴ 48 = m 2 − 8m
∴ m 2 − 8m − 48 = 0
∴ (m − 12)(m + 4) = 0
∴ m = 12 or m = −4
There are therefore two possible positions for A: A(−7; 12) and A'(−7; −4) Sample
Find the value of p and also show your answer on a sketch. ���� y M(0; 2) =(����2; y2) A(–1; –1) =(����1; y1) P (p; –2) = (����2; y2) = (–4; –2)
(p; –2) = (����2; y2) = (2; –2)
(p + 4)(p − 2) = 0 ∴ p = −4 or p = 2
There are therefore two possible positions for P. (See sketch.)
7. A circle with the origin as its centre is given. A(3; 4) and B(−4; q) are points on the circumference of the circle.
Calculate the value of q.
4)
q)
8. Given: P(2; −3), M(−2; 1) and N(��; −7).N is equally far fromP and M.
Draw a sketch to represent the information given. Then find the value of �� and show your answer on the sketch.
There are therefore two possible positions for B. (See sketch.) Sample
B(–4; q) =(–4; 3) B(–4; q) =(–4; –3) A(3; 4) O(0; 0) OA = OB Radiuses are equal
9. A(0; 0), B(p; q) and C(−q; p) are the vertices of a triangle. Prove by calculation that ∆ABC is isosceles and right-angled.
A(0; 0)
In order to prove that a triangle is isosceles, we prove that two sides are equal.
B(p; q)
C(–q; p)
In order to prove that a triangle is right-angled, we prove that one angle is 90° (i.e. that two sides are perpendicular to each other).
AC = AB ∆ABC is therefore right-angled and isosceles.
10. A(0; 0), B(√ ; 1) and C(√ 3 ; −1) are the vertices of a triangle.
Prove by calculation that ΔABC is equilateral.
B(√3 ; 1)
A(0; 0)
C(√3 ; –1)
AB = √ (�� 2 − �� 1) 2 + (�� 2 − �� 1) 2 = √ (√ 3 − 0) 2 + (1 − 0) 2 = √ 3 + 1 = √ 4 = 2 units
AC = √ (�� 2 − �� 1) 2 + (�� 2 − �� 1) 2 = √ (√ 3 − 0) 2 + (−1 − 0) 2 = √ 3 + 1 = √ 4 = 2 units
BC = √ (�� 2 − �� 1) 2 + (�� 2 − �� 1) 2
= √ (√ 3 − √ 3 ) 2 + (−1 − 1) 2
Sample
= √ 0 + 4 = √ = 2 units
∴ AB = AC = BC = 2 ΔABC is therefore equilateral.
11. A parallelogram with vertices A(−2; 2), B(3; 2), C(1; −2) and D(−4; p) is given.
Calculate the value of p. ���� y
A(–2; 2)
B(3; 2)
D(–4; p)
C(1; –2)
REMEMBER
The opposite sides of a parallelogram are equal and parallel.
mAB = mDC Opposite sides of parallelogram are parallel
2 2 3 (−2) = −2 p 1 (−4)
5(−2 p) = 0 Cross multiply
−10 − 5p = 0 ∴ −5p = 10 ∴ p = −2
12. A circle with M(−2; 1) as its centre and radius √ 13 units are given.
12.1 Prove that the circle passes through the points A(−5; 3) and B(1; 3).
√ 9 + 4
√ 13 units
3) B(1; 3)
M(–2; 1)
For the circle with centre M and radius √ 13 units to pass through points A and B, we must prove that AM = BM = √ 13 units, so that radiuses are equal.
∴ AM = BM = √ 13 units
The circle therefore passes through points A and B.
12.2 Calculate the circumference and area of this circle if the units are in centimetres. Round off your answers to the nearest integer.
Circumference of circle = 2��r = 2��× √ 13 = 23 cm to the nearest integer
Area of circle = ��r 2 = ��(√ 13 ) 2 = 40,84070... = 41 cm 2 to the nearest integer
9.2 THE MID-POINT OF A L INE SEGMENT
In the diagram below, C(��; ��) is the mid-point of line segment AB, with A(��1; ��1) and B(��2; ��2).
The ��-coordinate ofC is exactly halfway between ��1 and ��2 on the ��-axis.
The value of the ��-coordinate ofC is therefore ��1 + ��2 2 . This is the average value of ��1 and ��2.
In the same way, the value of �� is ��1 + ��2 2 . It follows that:
The formula for calculating the mid-point, C, of a line segment between two points, A(��1; ��1) and B(��2; ��2), is:
Worked example 3
–1) Sample
The word “mid-point” indicates that a line segment is bisected.
Find the mid-point of the line segment between A(2; 5) and B(−8; −1) and show your answer on a sketch. Solution
2)
5)
Worked example 4
If M(−2; 1) is the mid-point of the line segment between A(−6; 8) and B(a; b), find the coordinates of B.
A(–6; 8) M(–2; 1)
Exercise 9.2: The mid-point of a line segment
1. Consider the sketch below. Find the coordinates of the mid-point of each of the following line segments: AB, CD, EF and EH.
9) C(–4; 5)
d; –4)
2. Given: Rectangle ABCD with A(−6; 2), B(−5; −2), C(5; 2) and D(4; 6).
In addition, AE = ED.
2)
y D(4; 6) C(5; 2) B(–5; –2)
2.1 Prove that the mid-point of BC is Q(0; 0).
2.3 Prove that AB = EQ = DC. AB = √ (−6 − (−5))2 + (2 − (−2))2 = √ (−1)2 + (4)2 = √ 17
= (0; 0) Line segment BC therefore passes through the origin.
2.2 Now find the coordinates of E, the mid-point of AD. M
∴ AB = EQ = DC = √ 17
2.4 Is the following statement true?
“If the mid-points of a pair of opposite sides of a rectangle are connected, two rectangles are formed which are congruent.”
Give reasons for your answer.
Tip
In order to prove that two rectangles are congruent, you need to prove that the corresponding sides are equal and that the corresponding angles are equal.
According to the calculations done in 2.1 to 2.3, the following was found:
AB = EQ = DC
Opposite sides are therefore equal.
AE = ED = BQ = QC
Each side is therefore bisected.
The angles A, B, C and D = 90° Interior angles of rectangle ABCD = 90°each
ABQE and DCQE are therefore parallelograms (both pairs of opposite sides are equal) with 90° angles. They are therefore rectangles.
The two rectangles’ sides are equal; therefore they are identical. The rectangles are therefore congruent.
3. If M(−3; −2) is the mid-point of A(−8; −5) and B(p; q), calculate the values of p and q.
Consider the answer in the sketch below:
–2)
–5)
4. Given: M is the mid-point of AB and N is the mid-point of BC. Find the coordinates of N.
y C(5; 8)
A(–13; 5)
For B(p; q):
M(−6; 3) =
M(–6; 3)
N(r; s)
B(p; q)
∴ −12 = p − 13 ∴ 6 = q + 5
∴ p = 1 ∴ q = 1
∴ B(1; 1)
For N(r; s):
M BC = N(r; s)
M BC = ��
P(–1; 4)
5. Square PQSR is given. ���� y Q(4; 4)
R(a; b)
S(c; d)
5.1 What are the values of a, b, c and d in the coordinates of R and S? Indicate the coordinates of R and S on the sketch. ���� y Q(4; 4)
P(–1; 4)
R(a; b) = (–1; –1) 5 5 5 5
S(c; d) = (4; –1)
∴ N(3; 4 1
5.2 Now prove that the diagonals of a square bisect each other using the mid-point formula.
6.1 Redraw the sketch, do the necessary calculations and draw all three medians of the given triangle on the sketch.
6. Consider triangle ABC below.
Both diagonals’ mid-points are at the same point.
This means that the diagonals bisect each other.
–2)
8) C(9; –6)
= F(1; −4)
B(–7; –2) Sample
Mid-points of the three sides
6.2 Indicate the median point of the triangle on the sketch.
Tip
A median connects a vertex with the mid-point of the side opposite that vertex. The point where the three medians intersect is called the median point (or centroid) of the triangle.
8)
D(–2; 3)
E(6; 1)
F(1; –4)
C(9; –6)
Rhombus ABCD is given.
–8)
7.1 Prove that the point of intersection of the diagonals lies on the ��-axis.
The diagonals of a rhombus bisect each other. Therefore the point of intersection of the diagonals is also the mid-point of each diagonal.
7.2 Find the length of AC rounded off to two decimal places. AC = √ (3 − 1) 2 + (−8 − 8) 2 = √ (2) 2 + (−16) 2 = √ 4 + 256 = √ 260 = 16,1245...
16,12 units (rounded off to two decimal figures)
= D(2; 0)
The ��-coordinate is 0. Therefore the point of intersection of the diagonals lies on the ��-axis.
Self-evaluation
Use the following scale to determine how comfortable you are with each topic in the table that follows:
1. Help! I am not at all comfortable with the topic; I need help.
2. Alarm! Alarm! I am not comfortable with the topic; however, I merely need more time to go through the topic again.
3. OK! I am fairly comfortable with the topic, but do get stuck occasionally.
4. Sharp! I am comfortable with the topic.
5. Whoohoo, it's party time! I am completely comfortable with the topic, and can even answer more difficult questions on this.
Complete the table.
I understand how the distance formula is derived.
I can use the distance formula to determine the distance between two given points (i.e. the length of a line segment) in the Cartesian plane.
I can use the distance formula to solve problems in analytical geometry.
I understand how the formula for the midpoint of a line segment is derived.
I can use the formula for the midpoint of a line segment to determine the midpoint of a line segment in the Cartesian plane.
I can use the formula for the midpoint of a line segment to solve problems in analytical geometry.
9.3 THE GRADIENT (INCL INATION) OF A LINE
The gradient or inclination of a line refers to the steepness of the line.
Sample
Earlier on, we defined this as: the change in the ��-values between 2 points on a line the change in the ��-values between the same 2 points on the line
The gradient can also be defined as: the vertical difference between 2 points on a line the horizontal difference between the same 2 points on the line
REMEMBER
The formula for calculating the gradient of a line connecting the two points
and
is:
2 − ��1 means “the change in ��”. We can also write this as ∆�� (delta ��)
2 − ��1 means “the change in ��”. We can also write this as ∆�� (delta ��)
Note
The order in which the ��- and ��-coordinates are subtracted from each other is important:
Correct: m
Worked example 5
Find the inclination and mid-point of the line segment connecting A(4; 1) and B(−8; 4).
Inclination Mid-point m
H orizontal and vertical lines
A line which is parallel to the ��-axis is calleda horizontal line. The gradient of a horizontal line is zero, because there is no vertical change in the gradient of the line:
m = change in �� change in �� = 0 change in �� = 0
A line which is parallel to the y-axis is called a vertical line.
The gradient of a vertical line is undefined, because there is no horizontal change in the gradient of the line:
m = change in �� change in �� = change in �� 0 = undefined
Consider the sketch below:
Everywhere on this horizontal line, the y-value is 3.
Everywhere on this vertical line, the ����-value is������.
In the sketch we see:
• The ��-coordinates of points that lie alongside each other on a horizontal line are equal.
• The ��-coordinates of points that lie directly below each other on a vertical line are equal.
REMEMBER
The gradients of parallel lines are equal: m1 = m2
The gradients of perpendicular lines: m1 × m
−1
Exercise 9.3: The gradient (inclination) of a line
1. Consider the line segments given below:
y B(–14; 2)
C(–8; p)
A(10; 5)
E(–8; –4) F(0; –4)
G(10; 2)
H(10; 0)
D(q; –4)
1.1 Find all the unknown coordinates and indicate these on the sketch.
Sample
B(–14; 2) C(–8; 2) H(10; 0) G(10; 2)
5) D(10; –4) E(–8; –4) F(0; –4)
1.2 Now find the gradient of each line segment.
mAB = �� 2 − �� 1 �� 2 − �� 1 = 5 − 2 10 − (−14) = 3 24 = 1 8
mCD = �� 2 − �� 1 �� 2 − �� 1 = −4 − 2 10 − (−8) = −6 18 = −1 3
mEF = �� 2 − �� 1 �� 2 − �� 1 = −4 − (−4) 0 − (−8) = 0 8 = 0
∴ EF || ��-axis
mGH = �� 2 − �� 1 �� 2 − �� 1 = 2 − 0
10 − 10 = 2 0 = undefined
∴ GH || ��-axis
Sample
B(b; 3)
2. Consider the sketch given below: ���� y A(a; 5)
C(m; n)
E(1; e)
D(d; –4)
P(����; y)
F(k; h)
Q(r; s) 5 –1
Find the values of the coordinates indicated by the letters a to s.
A(a; 5) = A(−2; 5)
B(b; 3) = B(−2; 3)
C(m; n) = C(−2; −1)
D(d; −4) = D(−2; −4)
E(1; e) = E(1; −1)
F(k; h) = F(5; −1)
P(��; ��) = P(0; −4)
Q(r; s) = Q(5; −4)
AD ⊥ ��-axis
ABD ⊥ ��-axis
C lies on AD and CF ⊥ ��-axis
AD ⊥ ��-axis
EF ⊥ ��-axis
FQ ⊥ ��-axis and FC ⊥ ��-axis
P lies on ��-axis and alongside D
Q lies on FQ and DQ
3. In the sketch, AB ⊥ CD and PQ ⊥ RS. The points A(−14; 5), B(2; 1), R(5; −2) and S(−4; −6) are given. Find the gradients of CD and PQ.
B(2; 1) A(–14; 5)
4. Line segment AB and CD are given, with AB ∥ CD. The points A( 5; 1) and B( 3; a) form line segment AB and the points C( 4; 3) and D( 1; 3) form line segment CD.
Calculate the value of a.
AB = ��CD Given: AB ∥ CD
–2)
C
Given ∴ ��AB . ��CD = −1
1 − 5 2 (−14) . ��CD = −1
−4 16 . ��CD = −1
��CD = 16 4 = 4
5. If AB ⊥ CD with A(−5; 2), B(b; −1), C(−4; −3) and D(−1; 3), calculate the value of b.
D(–1; 3)
A(–5; 2)
6. Quadrilateral ABCD is given.
D(–7; –3) A(–5; 4) y
B(b; –1)
C(–4; –3)
mAB . mCD = −1
Given: AB ⊥ CD ∴ −1 − 2 b − (−5) × −3 − 3 −4 − (−1) = −1
∴ −3 b + 5 × −6 −3 = −1
∴ 18 −3b − 15 = −1
∴ 18 = 3b + 15
∴ b = 1
6.1 Prove that AD ∥ BC.
AD ∥ BC Sample
��AD = ��BC
–3)
B(5; 4)
C(3;
6.2 Prove that AD = BC.
AD = √ (−7 − (−5))2 + (−3 − 4)2 = √ (−2)2 + (−7)2 = √ 53
BC = √ (3 − (5))2 + (−3 − (4))2 = √ (−2)2 + (−7)2 = √ 53
∴ AD = BC
6.3 What type of quadrilateral is ABCD? Give a reason for your answer. A parallelogram, because one pair of opposite sides is equal and parallel.
6.4 Determine in which quadrant the point of intersection of the diagonals lies. Show your calculations.
The diagonals of a parallelogram bisect each other.
M AC(��; ��) = ( −5 + 3 2 ; 4 3 2 ) = (−1; 1 2 )
The point of intersection of the diagonals is therefore in the second quadrant.
7. Consider quadrilateral ABCD. Show by calculation that the quadrilateral is a trapezium.
y ����
D(4; 4)
C(1; –3)
B(–2; –1)
A(1; 6)
∴ ABCD is a trapezium, because one pair of opposite sides is parallel.
8. Consider the sketch and use analytical methods to prove that the diagonals of kite ABCD intersect perpendicularly.
Sample
A(–6; 5)
B(–10; 3)
D(–2; 3)
C(–6; –5) y ����
The ��-values ofA andC are the same; therefore AC || ��-axis. The y-values of B and D are the same; therefore BD || ��-axis.
AC ⊥ BD
9. ABCDEF is a regular hexagon. Do the necessary calculations to prove that all three pairs of opposite sides are parallel.
10. In the triangle below, N is the mid-point of AB and P is the mid-point of AC. Use analytical methods to prove that NP ∥ BC and NP
4)
A(–3; 5)
B(–6; 1)
C(–3; –3)
F(2; 5)
E(5; 1)
D(2; –3)
REMEMBER
According to the mid-point theorem (which was dealt with in Theme 8), the following applies: If a line segment is drawn from the mid-point of one side of a triangle to the midpoint of another side of the triangle, this line is parallel to the third side and equal to half of the third side.
If D is the mid-point of AB, for example, and E is the mid-point of AC, it follows that DE ∥ BC and DE = 1 2 BC.
–2)
C(6;
B(–6; –4)
Mid-point of AB:
N(��; ��) = ( −2 − 6 2 ; 4 − 4 2 ) = ( −8 2 ; 0 2 )
∴ N(−4; 0)
Mid-point of AC:
P(��; ��) = ( −2 + 6 2 ; 4 − 2 2 ) = ( 4 2 ; 2 2 )
∴ P(2; 1) ��PN = 1 − 0 2 − (−4) = 1 6 ��BC = −2 − (−4) 6 − (−6) = 2 12 = 1 6 ∴ ��PN = ��BC ∴ PN ∥ BC PN = √ (2 − (−4)) 2 + (1 − 0) 2 = √ (6) 2 + (1) 2 = √ 37 BC = √ (6 − (−6))2 + (−2 − (−4))2 = √ (12)2 + (2)2 = √ 144 + 4 = √ 148 = 2√ 37 ∴ PN = 1 2 BC
11. Sketch the points A(−5; 6), B(−5; −8) and C(6; −8). Determine which angle in triangle ABC is 90° using analytical methods. A(–5; 6) B(–5; –8) C(6; –8) y ���� Sample
The ��-values of A and B are the same
∴ AB ⊥ ��-axis
The ��-values of B and C are the same
∴ BC ⊥ ��-axis
∴ BC ⊥ AB
∴ ˆ = 90°
Using the gradient method:
mAB = −8 6 −5 (−5) = −14 0 This is undefined and is therefore impossible to calculate.
Try something else: Use the sketch.
9.4 POINT S ON A STRAIGHT LINE (COLLINEAR POINTS)
Points are collinear if they lie on the same straight line.
We can prove this using one of the following methods:
1. by proving that the gradients between any two pairs of the three points are equal
2. by proving that the distance between the two furthest points is equal to the sum of the distance between the three points.
Worked example 6
Prove that A(2; −1), B(5; 2) and C(7; 4) are collinear.
Sample
The three points will lie on the same straight line if mAB = mBC.
We are therefore going to prove that mAB = mBC (gradient method):
∴ mAB = mBC
∴ A, B and C are collinear.
Alternative follows on next page
OR:
We can also prove that AB + BC = AC (distance method):
AB = √ (5 − 2) 2 + (2 − (−1)) 2
= √ 9 + 9
= √ 18 = 3√ 2
= √ (7 − 5) 2 + (4 − 2) 2 = √ 4 + 4 = 2√ 2 AC = √ (7 − 2) 2 + (4 − (−1)) 2 = √ 25 + 25 = √ 50
= 5√ 2
∴ AB + BC = 3 √ 2 + 2 √ 2 = 5 √ = AC
Therefore A, B and C are collinear.
Worked example 7
Sample
The points A(−6; 2), B(2; −2) and C(−3; p) are collinear. Find the value of p.
Solution
If A, B and C lie on the same straight line, the gradient of AB is the same as the gradient of BC.
Exercise 9.4: P oints on a straight line (collinear points)
1. Determine whether the following points lie on the same straight line: A(−1; 2), B(2; 5) and C(0; 3). Use the gradient method.
mAB = 5 − 2 2 − (−1) = 3 3 = 1
mAC = 3 − 2 0 − (−1) = 1 A, B and C lie on the same line
2. Determine whether the following points lie on the same straight line: A(1; 5), B(2; 7) and C(3; 6). Use the distance method.
B(2; 7)
A(1; 5) ���� y
C(3; 6)
In order to lie on the same line, AB + BC must be equal to AC.
LHS: AB + BC = √ (1 − 2) 2 + (5 − 7) 2 + √ ( 3 − 2) 2 + (6 − 7) 2 = √ 1 + 4 + √ 1 + 1
= √ 5 + √ 2
RHS: AC = √ (3 − 1) 2 + (6 − 5) 2 = √ 4 + 1
= √ 5
∴ A, B and C do not lie on the same line.
3. Determine whether the following points lie on the same straight line: A(−1; −1), B(2; 8), C(3; 11) and D(1; 5).
Sample
Therefore A, B, C and D lie on the same straight line, i.e. they are collinear.
4. The following seven points are given.
G(4; 7) y
C(1; 4)
D(3; 6) B(2; 3)
A(1; 1) F(3; 1)
E(4; 2)
Find three points that would lie on the same straight line. The points were purposely not placed accurately. Do calculations to support your answer.
mAB = 3 − 1 2 − 1 = 2 1 = 2
mAD = 6 − 1
3 − 1 = 5 2
mAG = 7 − 1
4 − 1 = 6 3 = 2
A, B and G will therefore lie on the same straight line.
In the same way, it can be proven that C, D and G lie in a straight line.
5. If A(1; 1), B(2; 3) and C(p; 5) are collinear, calculate the value of p.
mAB = mAC
3 − 1
2 − 1 = 5 − 1 p − 1
2 1 = 4 p − 1
2p − 2 = 4
2p = 6 p = 3
Given: points are collinear
6. P(3; 2), Q(4; 1) and R(1; m) are collinear. Calculate the value of m.
Sample
mPQ = mPR
Given: points are collinear ∴ 1 − 2 4 − 3 = m − 2 1 − 3 ∴ −1 1 = m − 2 −2
∴ m − 2 = 2
∴ m = 4
7. M(1; 4), N(−3; �� ) and P(−5; −2) lie on the same straight line. Calculate the value of �� .
mMN = mMP
Given: points are collinear a − 4 −3 − 1 = −2 − 4 −5 − 1 a − 4 −4 = − 6 − 6
−6a + 24 = 24 −6a = 0 a = 0
9.5 THE EQUATION OF A STRAIGHT LINE
The following formula is used to find the equation of a line:
In the first formula, m is the gradient of the line and (��1; ��1) represents a point on the line. In the second formula, (��1; ��1) and (��2; ��2) are two points on the line.
Worked example 8
Find the equation of the line passing through the points A(1; 2) and B(−4; 3).
Exercise 9.5: The equation of a straight line
1. Find the equation of the straight line passing through the points P(−10; 5) and Q(0; 0).
��= 2�� Sample
2. Find the equation of the straight line passing through the points M(1; 2) and H(3; 6).
��−2 = 6 − 2 3 − 1 (�� − 1)
��−2=2(�� − 1)
3. The straight lines ��(��) = −3��+6 and ��(��) = m��+ c are given.
Find the equation of �� if�� and�� are parallel to each other and�� passes through the point (1; 1). The gradients of parallel lines are equal.
∴ m �� = m�� = −3
Now substitute m = −3 and (�� 1; �� 1) = (1; 1):
�� 1 = m(�� − �� 1)
∴ ��−1= 3(�� − 1)
∴ ��= −3�� + 3 + 1
∴ ��= −3�� + 4
4. Consider parallelogram ABCD in the sketch below and answer the questions that follow:
4.1
Calculate the coordinates of D.
mAD = mBC Opposite sides of a parallelogram are parallel
∴ 0 − �� −2 − �� = 1 − (−3) 2 − 1
∴ −�� −2 − �� = 4 1
∴ ��=4(−�� − 2) Cross multiply ∴ ��= −4�� − 8 ∴ ��= 4�� + 8 … ①
mAB = mDC Opposite sides of a parallelogram are parallel 1 − 0 2 − (−2) = �� − (−3) �� − 1 1 4 = �� + 3 �� − 1 4��+ 12=�� − 1 4��=�� − 13 … ②
Substitute 1 into 2:
B(2; 1)
A(–2; 0) y ����
C(1; –3)
D(��������y)
Sample
∴ 4(4�� + 8) = ��− 13
∴ 16��+ 32=�� − 13
∴ 15��= −45
∴ ��= −3
Substitute into 1:
∴ ��=4(−3) + 8 = −12 + 8 = −4
∴ D(−3; −4)
OR: Simply use the sketch and apply the gradient of AB to the gradient of DC. (The gradients are equal, because the opposite sides of a parallelogram are parallel.)
If you move one unit down from C and four units to the left, you get D(−3; −4).
4.2 Find the equation of AC.
Substitute A(−2; 0) and C(1; −3):
∴ ��−0= −3 − 0 1 − (−2) (�� − (−2))
∴ ��−0= 1(�� + 2) ∴ ��=
4.3 Prove that the quadrilateral’s diagonals intersect perpendicularly.
mAC . mBD
= (−1) 1 − (−4) 2 − (−3)
= (−1)( 5 5 )
= −1
4.4 What type of parallelogram is ABCD?
It is a rhombus, because the diagonals intersect each other perpendicularly.
Sample
The lines AC and BD are perpendicular to each other.
4.5 Find the equation of the line which passes through B and is parallel to AC.
mAC = −1 Already found in 4.2
∴ gradient of new line is also −1 Lines are parallel
Now substitute m = −1 and B(2; 1):
�� 1 = m(�� − �� 1)
��−1= −1(�� − 2) ∴ ��−1= −�� + 2
��= −�� + 2 + 1 ∴ ��= −�� + 3
4.6 Calculate the distance from the ��-intercept to the ��-intercept of the line which is referred to in question 4.5.
Self-evaluation
A(–2; 0)
B(2; 1) P Q y ����
C(1; –3)
D(–3; –4)
For P: This is the ��-intercept of ��= �� + 3
Let ��= 0
∴ ��=0+ 3 = 3 P(0; 3)
For Q: This is the ��-intercept of ��= �� + 3
Let ��= 0
∴ 0 = −��+ 3
��=3Q(3; 0)
Distance PQ = √ (3 − 0) 2 + (0 − 3) 2 = √ 18 = 3 √ 2 units
Use the following scale to determine how comfortable you are with each topic in the table that follows:
1. Help! I am not at all comfortable with the topic; I need help.
2. Alarm! Alarm! I am not comfortable with the topic; however, I merely need more time to go through the topic again.
3. OK! I am fairly comfortable with the topic, but do get stuck occasionally.
4. Sharp! I am comfortable with the topic.
5. Whoohoo, it's party time! I am completely comfortable with the topic, and can even answer more difficult questions on this.
Complete the table.
I can calculate the gradient of a line.
I can use the gradient to solve problems in analytical geometry, including problems involving
• parallel lines
• perpendicular lines
• collinear points
I can determine the equation of a line.
Sample
1.1
Summary of theme
Now that we have come to the end of this theme, learners should:
1. be able to find die distance between two points
2. be able to find the coordinates of the mid-point of a line segment
3. be able to find the gradient of the line segment between two points
4. be able to find the equation of a straight line
5. know and be able to use the conditions for parallel and perpendicular lines
6. be able to work with collinear points
7. be able to apply all this knowledge in mixed exercises.
End of theme exercise
1. AB is the diameter of the circle with A(−3; −5) and B(4; 1).
M(a; b) is the centre of the circle.
Answer the questions that follow and round off your answers to two decimal figures where necessary.
Find the coordinates of the centre of the circle.
1.2 Find the radius of the circle.
Radius = AM
1.3 Find the area of the circle.
2. Consider the following sketch, where the two lines �� and �� are parallel to each other.
Find the equation of the line ��(��) = m��+ c, which passes through the origin.
3. P(2; 3), Q(5; q), R(4; 2) and S(1; 1) are the vertices of parallelogram PQRS.
3.1 Sketch the parallelogram on a system of axes.
S(1; 1) P(2; 3)
The gradients of parallel lines are equal.
m�� = m �� = 4
Now substitute m = 4 and (�� 1; �� 1) = (0; 0):
��− �� 1 = m(��− �� 1)
∴ ��−0=4(��−0)
∴ ��= 4��
(m = 4 and c = 0)
Sample
Q(5; q) = (5; 4)
R(4; 2) ����
3.2 Find the value of q. Opposite sides of a parallelogram are parallel.
mPS = mQR ∴ 1 − 3 1 − 2 = 2 − q 4 − 5
∴ −2 −1 = 2 − q −1
∴ −2 + q = 2
∴ q = 4
3.3 Find the coordinates of A, the point of intersection of the diagonals of PQRS.
The point of intersection of the diagonals is also the mid-point of each diagonal.
3.4 Prove that the diagonals do not intersect perpendicularly.
4. The equation of the line ��(��) = 2��+8 is given. Find the equation of the straight line ��, which is perpendicular to �� and passes through the point (2; −3).
The product of the gradients is not −1. The diagonals are therefore not perpendicular to each other. Sample
5. Prove that if the mid-points of the sides of a parallelogram are connected, the figure which is formed will also be a parallelogram. Use parallelogram ABCD given in the sketch below.
First find the mid-point of each side:
E(��; ��) = ( 0 − 2 2 ; 3 − 2 2 ) = (−1; 1 2 )
G(��; ��) = ( 0 + 6 2 ; 3 + 3 2 ) = (3; 3)
H(��; ��) = ( 6 + 4 2 ; 3 − 2 2 ) = (5; 1 2 )
F(��; ��) = ( 4 − 2 2 ; − 2 − 2 2 ) = (1; −2)
Find the gradient of each side: mEG = 3 − 1 2
∴ EFGH is a parallelogram, because both pairs of opposite sides are parallel.
6. PQRS is a polygon with vertices P(0; 3), Q(4; 3), R(5; −1) and S(−1; −1).
6.1 Prove the following:
6.1.1 PS = QR PS = √ (−1 − (0))2 + (−1 − (3))2 = √ (−1)2 + (−4)2 = √ 17 QR = √ (5 4)2 + (−1 (3))2 = √ (1)2 + (−4)2 = √ 17
∴ PS = QR
Sample
6.1.2 PQ || RS ��PQ = 3 − 3 4 − 0 = 0 4 = 0 ��RS = −1 (−1) −1 − 5 = 0 −6 = 0 ∴ PQ ∥ RS
6.2 What type of polygon is PQRS?
∴ PS ∦ QR Gradients are not equal
∴ PQRS is an isosceles trapezium, because only one pair of sides is parallel, and the other pair of sides is equal.
6.3 Prove that the diagonals PR and QS do not bisect each other.
Mid-point of PR =
Mid-point of QS =
The diagonals PR and QS therefore do not bisect each other, because their mid-points are not the same.
7. Given: ΔPQR with vertices P(1; 3), Q(4; 1) and R(6; 4).
7.1 Sketch ΔPQR in the Cartesian plane.
6; 4)
4; 1) P(1; 3)
7.2 Prove that PQR is an isosceles triangle.
= √ (4 − 1)2 + (1 − 3)2 = √ (3)2 + (−2)2 = √ 13 RQ = √ (4 6)2 + (1 4)2
Therefore P, Q and S(7; −1) lie on the same straight line, because mPQ = mQS. Sample
PQ = RQ
PQR is an isosceles triangle.
7.3 Find the coordinates of M, the mid-point of PR.
Mid-point of PR =
7.4 Find the gradient of PQ. ��PQ = 1 − 3 4 − 1 = −2 3
7.5 Prove that P, Q and S(7; −1) lie on the same straight line. ��QS = −1 − 1 7 − 4 = −2 3 ��PQ = 1 − 3 4 − 1 = −2 3
