This document provides information about coordinate geometry, including finding the distance between two points, the midpoint and division of a line segment, area of polygons, and equations of straight lines. It gives formulas and examples for calculating the distance between points using the Pythagorean theorem, finding the midpoint and points dividing a line segment in a given ratio, and computing the area of triangles and quadrilaterals. It also explains how to determine the gradient, x-intercept, and y-intercept of a straight line and write the equation of a straight line in general and gradient forms. Exercises are provided to apply these concepts.
Matematik tambahan spm tingkatan 4 geometri koordinat {add maths form 4 coord...Hafidz Sa
Nota padat Bab 6 Geometri Koordinat Matematik Tambahan Tingkatan 4 SPM
Slide Chapter 6 Coordinate Geometry Additional Mathematics Form 4
Topik Bab 6: Geometri Koordinat
Jarak di Antara Dua Titik
Pembahagian Tembereng Garis
Luas Poligon
Persamaan Garis Lurus
Garis Lurus Selari dan Garis Lurus Serenjang
Persamaan Lokus yang Melibatkan Jarak Antara Dua Titik
Matematik tambahan spm tingkatan 4 geometri koordinat {add maths form 4 coord...Hafidz Sa
Nota padat Bab 6 Geometri Koordinat Matematik Tambahan Tingkatan 4 SPM
Slide Chapter 6 Coordinate Geometry Additional Mathematics Form 4
Topik Bab 6: Geometri Koordinat
Jarak di Antara Dua Titik
Pembahagian Tembereng Garis
Luas Poligon
Persamaan Garis Lurus
Garis Lurus Selari dan Garis Lurus Serenjang
Persamaan Lokus yang Melibatkan Jarak Antara Dua Titik
The following presentation is a part of the level 4 module -- Digital Logic and Signal Principles. This resources is a part of the 2009/2010 Engineering (foundation degree, BEng and HN) courses from University of Wales Newport (course codes H101, H691, H620, HH37 and 001H). This resource is a part of the core modules for the full time 1st year undergraduate programme.
The BEng & Foundation Degrees and HNC/D in Engineering are designed to meet the needs of employers by placing the emphasis on the theoretical, practical and vocational aspects of engineering within the workplace and beyond. Engineering is becoming more high profile, and therefore more in demand as a skill set, in today’s high-tech world. This course has been designed to provide you with knowledge, skills and practical experience encountered in everyday engineering environments.
Module 5 (Part 3-Revised)-Functions and Relations.pdfGaleJean
The Cavite Mutiny of 1872 was an uprising of military personnel of Fort San Felipe, the Spanish arsenal in Cavite, Philippines on January 20, 1872, about 200 Filipino military personnel of Fort San Felipe Arsenal in Cavite, Philippines, staged a mutiny which in a way led to the Philippine Revolution in 1896. The 1872 Cavite Mutiny was precipitated by the removal of long-standing personal benefits to the workers such as tax (tribute) and forced labor exemptions on order from the Governor General Rafael de Izquierdo. Many scholars believe that the Cavite Mutiny of 1872 was the beginning of Filipino nationalism that would eventually lead to the Philippine Revolution of 1896.
The student is able to (I can):
• Find the midpoint of two given points.
• Find the coordinates of an endpoint given one endpoint
and a midpoint.
• Find the distance between two points.
Dear students get fully solved assignments
Send your semester & Specialization name to our mail id :
“ help.mbaassignments@gmail.com ”
or
Call us at : 08263069601
GDG Cloud Southlake #33: Boule & Rebala: Effective AppSec in SDLC using Deplo...James Anderson
Effective Application Security in Software Delivery lifecycle using Deployment Firewall and DBOM
The modern software delivery process (or the CI/CD process) includes many tools, distributed teams, open-source code, and cloud platforms. Constant focus on speed to release software to market, along with the traditional slow and manual security checks has caused gaps in continuous security as an important piece in the software supply chain. Today organizations feel more susceptible to external and internal cyber threats due to the vast attack surface in their applications supply chain and the lack of end-to-end governance and risk management.
The software team must secure its software delivery process to avoid vulnerability and security breaches. This needs to be achieved with existing tool chains and without extensive rework of the delivery processes. This talk will present strategies and techniques for providing visibility into the true risk of the existing vulnerabilities, preventing the introduction of security issues in the software, resolving vulnerabilities in production environments quickly, and capturing the deployment bill of materials (DBOM).
Speakers:
Bob Boule
Robert Boule is a technology enthusiast with PASSION for technology and making things work along with a knack for helping others understand how things work. He comes with around 20 years of solution engineering experience in application security, software continuous delivery, and SaaS platforms. He is known for his dynamic presentations in CI/CD and application security integrated in software delivery lifecycle.
Gopinath Rebala
Gopinath Rebala is the CTO of OpsMx, where he has overall responsibility for the machine learning and data processing architectures for Secure Software Delivery. Gopi also has a strong connection with our customers, leading design and architecture for strategic implementations. Gopi is a frequent speaker and well-known leader in continuous delivery and integrating security into software delivery.
UiPath Test Automation using UiPath Test Suite series, part 3DianaGray10
Welcome to UiPath Test Automation using UiPath Test Suite series part 3. In this session, we will cover desktop automation along with UI automation.
Topics covered:
UI automation Introduction,
UI automation Sample
Desktop automation flow
Pradeep Chinnala, Senior Consultant Automation Developer @WonderBotz and UiPath MVP
Deepak Rai, Automation Practice Lead, Boundaryless Group and UiPath MVP
Connector Corner: Automate dynamic content and events by pushing a buttonDianaGray10
Here is something new! In our next Connector Corner webinar, we will demonstrate how you can use a single workflow to:
Create a campaign using Mailchimp with merge tags/fields
Send an interactive Slack channel message (using buttons)
Have the message received by managers and peers along with a test email for review
But there’s more:
In a second workflow supporting the same use case, you’ll see:
Your campaign sent to target colleagues for approval
If the “Approve” button is clicked, a Jira/Zendesk ticket is created for the marketing design team
But—if the “Reject” button is pushed, colleagues will be alerted via Slack message
Join us to learn more about this new, human-in-the-loop capability, brought to you by Integration Service connectors.
And...
Speakers:
Akshay Agnihotri, Product Manager
Charlie Greenberg, Host
The Art of the Pitch: WordPress Relationships and SalesLaura Byrne
Clients don’t know what they don’t know. What web solutions are right for them? How does WordPress come into the picture? How do you make sure you understand scope and timeline? What do you do if sometime changes?
All these questions and more will be explored as we talk about matching clients’ needs with what your agency offers without pulling teeth or pulling your hair out. Practical tips, and strategies for successful relationship building that leads to closing the deal.
GraphRAG is All You need? LLM & Knowledge GraphGuy Korland
Guy Korland, CEO and Co-founder of FalkorDB, will review two articles on the integration of language models with knowledge graphs.
1. Unifying Large Language Models and Knowledge Graphs: A Roadmap.
https://arxiv.org/abs/2306.08302
2. Microsoft Research's GraphRAG paper and a review paper on various uses of knowledge graphs:
https://www.microsoft.com/en-us/research/blog/graphrag-unlocking-llm-discovery-on-narrative-private-data/
Essentials of Automations: Optimizing FME Workflows with ParametersSafe Software
Are you looking to streamline your workflows and boost your projects’ efficiency? Do you find yourself searching for ways to add flexibility and control over your FME workflows? If so, you’re in the right place.
Join us for an insightful dive into the world of FME parameters, a critical element in optimizing workflow efficiency. This webinar marks the beginning of our three-part “Essentials of Automation” series. This first webinar is designed to equip you with the knowledge and skills to utilize parameters effectively: enhancing the flexibility, maintainability, and user control of your FME projects.
Here’s what you’ll gain:
- Essentials of FME Parameters: Understand the pivotal role of parameters, including Reader/Writer, Transformer, User, and FME Flow categories. Discover how they are the key to unlocking automation and optimization within your workflows.
- Practical Applications in FME Form: Delve into key user parameter types including choice, connections, and file URLs. Allow users to control how a workflow runs, making your workflows more reusable. Learn to import values and deliver the best user experience for your workflows while enhancing accuracy.
- Optimization Strategies in FME Flow: Explore the creation and strategic deployment of parameters in FME Flow, including the use of deployment and geometry parameters, to maximize workflow efficiency.
- Pro Tips for Success: Gain insights on parameterizing connections and leveraging new features like Conditional Visibility for clarity and simplicity.
We’ll wrap up with a glimpse into future webinars, followed by a Q&A session to address your specific questions surrounding this topic.
Don’t miss this opportunity to elevate your FME expertise and drive your projects to new heights of efficiency.
Builder.ai Founder Sachin Dev Duggal's Strategic Approach to Create an Innova...Ramesh Iyer
In today's fast-changing business world, Companies that adapt and embrace new ideas often need help to keep up with the competition. However, fostering a culture of innovation takes much work. It takes vision, leadership and willingness to take risks in the right proportion. Sachin Dev Duggal, co-founder of Builder.ai, has perfected the art of this balance, creating a company culture where creativity and growth are nurtured at each stage.
State of ICS and IoT Cyber Threat Landscape Report 2024 previewPrayukth K V
The IoT and OT threat landscape report has been prepared by the Threat Research Team at Sectrio using data from Sectrio, cyber threat intelligence farming facilities spread across over 85 cities around the world. In addition, Sectrio also runs AI-based advanced threat and payload engagement facilities that serve as sinks to attract and engage sophisticated threat actors, and newer malware including new variants and latent threats that are at an earlier stage of development.
The latest edition of the OT/ICS and IoT security Threat Landscape Report 2024 also covers:
State of global ICS asset and network exposure
Sectoral targets and attacks as well as the cost of ransom
Global APT activity, AI usage, actor and tactic profiles, and implications
Rise in volumes of AI-powered cyberattacks
Major cyber events in 2024
Malware and malicious payload trends
Cyberattack types and targets
Vulnerability exploit attempts on CVEs
Attacks on counties – USA
Expansion of bot farms – how, where, and why
In-depth analysis of the cyber threat landscape across North America, South America, Europe, APAC, and the Middle East
Why are attacks on smart factories rising?
Cyber risk predictions
Axis of attacks – Europe
Systemic attacks in the Middle East
Download the full report from here:
https://sectrio.com/resources/ot-threat-landscape-reports/sectrio-releases-ot-ics-and-iot-security-threat-landscape-report-2024/
Software Delivery At the Speed of AI: Inflectra Invests In AI-Powered QualityInflectra
In this insightful webinar, Inflectra explores how artificial intelligence (AI) is transforming software development and testing. Discover how AI-powered tools are revolutionizing every stage of the software development lifecycle (SDLC), from design and prototyping to testing, deployment, and monitoring.
Learn about:
• The Future of Testing: How AI is shifting testing towards verification, analysis, and higher-level skills, while reducing repetitive tasks.
• Test Automation: How AI-powered test case generation, optimization, and self-healing tests are making testing more efficient and effective.
• Visual Testing: Explore the emerging capabilities of AI in visual testing and how it's set to revolutionize UI verification.
• Inflectra's AI Solutions: See demonstrations of Inflectra's cutting-edge AI tools like the ChatGPT plugin and Azure Open AI platform, designed to streamline your testing process.
Whether you're a developer, tester, or QA professional, this webinar will give you valuable insights into how AI is shaping the future of software delivery.
Mission to Decommission: Importance of Decommissioning Products to Increase E...
Chapter 6 coordinate geometry
1. Additional Mathematics Module Form 4
Chapter 6- Coordinate Geometry SMK Agama Arau, Perlis
Page | 60
CHAPTER 6- COORDINATE GEOMETRY
6.1 DISTANCE BETWEEN TWO POINTS
y
y2 Q (x2, y2)
y2- y1
y1 P(x1, y1)
x2- x1
x
To find the distance or length of PQ, use the concept of Pythagoras’ Theorem.
2
12
2
12
2
)()( yyxxPQ −+−=
Therefore:
where D is distance.
Example 1:
The distance point A(6, 3t) and point B(12, -t) is 10 units. Find the possible values of t.
Solution:
2
4
6416
1636100
)4(6100
)3()612(10
2
2
2
22
22
±=
=
=
+=
−+=
−−+−=
t
t
t
t
t
tt
Example 2:
Point A(h, 2h) and point B(h -3 , 2h + 1) are two points which are equidistant from the origin. Find the
value of h.
Solution:
2222
)012()03()02()0( −++−−=−+− hhhh
2
12
2
12 )()( yyxxD −+−=
How to obtain the formula?
Square the both sides
2. Additional Mathematics Module Form 4
Chapter 6- Coordinate Geometry SMK Agama Arau, Perlis
Page | 61
5
102
10255
144964
)12()3()2(
22
2222
2222
=
=
+−=
++++−=+
++−=+
h
h
hhh
hhhhhh
hhhh
EXERCISE 6.1
1. Find the perimeter of triangle ABC with points A(2, 1), B(4, 5) and C(-2, 8).
2. The points (-3, -5) lies on the circumference of a circle with centre (2, 1). Calculate the radius of the
circle.
3. Given that the distance between points A(1,3) and B(7, k) is 10 units . Find the possible value of k.
4. Given point A( 1, 7) and B(p, 2) and the distance between the points A and B is 13 units. Find the value
of p.
6.2 DIVISION OF A LINE SEGMENT
Line segment is a line that has distant.
6.2.1 Mid-point
The formula to find mid-points is just the same as we have learned in Form Two that is:
Example:
Given C(2, 5) is the mid-point of the point B(h, 3) and point D(-4, k). Find the values of h and k.
Solution:
Use the formula,
8
44
2
4
2
=
=−
+−
=
h
h
h
7
103
2
3
5
=
=+
+
=
k
k
k
EXERCISE 6.2.1
1. The coordinates of A and B are (m, 5) and (6, n) respectively. Find the values of m and n if the mid-
point of the points is (4, 10).
2. The coordinates of M and N are (4, 2) and (6, 5) respectively. Find the mid-point of these two points.
3. The coordinates of X and Y are (-1, b) and (a, 7) respectively. Find the values of a and b if the mid-
point of the points is (1, 2).
)
2
,
2
(),( 2121 yyxx
yx
++
=
3. Additional Mathematics Module Form 4
Chapter 6- Coordinate Geometry SMK Agama Arau, Perlis
Page | 62
6.2.2 Point which divides a line segment in the ratio m: n
Q (x2, y2)
n
y2- y y2- y
X(x, y)
m y- y1
y- y1
P(x1, y1) x- x1 x2- x
If the line PQ moves downwards, it will reach the horizontal line.
From that we know that:
nm
mxnx
x
mxnxnmx
mxnxmxnx
mxmxnxnx
xxmxxn
n
m
xx
xx
+
+
=
+=+
+=+
−=−
−=−
=
−
−
21
21
21
21
21
2
1
)(
)()(
If the line PQ moves upwards, it will reach the vertical line
nm
myny
y
mynynmy
mynymyny
mymynyny
yymyyn
n
m
yy
yy
+
+
=
+=+
+=+
−=−
−=−
=
−
−
21
21
21
21
21
2
1
)(
)()(
Hence, the formula to find the point that divides the line segment in ratio m : n is
When nm = , the point will be the midpoint of the line segment.
+
+
+
+
=
nm
myny
nm
mxnx
yx 2121
,),(
4. Additional Mathematics Module Form 4
Chapter 6- Coordinate Geometry SMK Agama Arau, Perlis
Page | 63
Example:
Find the coordinates of point P that divides the straight line that joins E(-6, 10) and F(4, -5) in the ratio
2: 3
E(-6, 10)
Solution:
m: n = 2 : 3 2
P(x, y)
3
F(4, -5)
EXERCISE 6.2.2
1. Point R divides the line segment joining J(-1, -7) and Q(10 ,7) internally in the ration PR: RQ = 1: 3. Find
the coordinate of R.
2. The points P(t, 2t), Q(2a, b) and R(4a, 3b) are on a straight line. Q divides PR internally in the ratio 1: 4.
Show that ab 6= .
3. Given points A(k, 5), B(0, 3) and C(5, 4). Find the possible values of k if the length of AB is twice the
length of BC.
6.3 AREA OF POLYGONS
6.3.1 Area Of Triangle
C(x3, y3)
B(x2, y2)
A(x1, y1)
K L M
The area of ABC∆ = Area of trapezium ACLK + Area of trapezium BCLM - Area of trapezium ABMK
o
o
o
+
−+
+
+−
=
32
)5)(2()10)(3(
,
32
)4)(2()6)(3(
),( yxP
+
+
+
+
=
nm
myny
nm
mxnx
yx 2121
,),(
−+−
=
5
1030
,
5
818
),( yxP
( )4,2−=
How to obtain the formula?
5. Additional Mathematics Module Form 4
Chapter 6- Coordinate Geometry SMK Agama Arau, Perlis
Page | 64
))((
2
1
))((
2
1
))((
2
1
122132321313 xxyyxxyyxxyy −+−−++−+=
[ ])()(
2
1
)(
2
1
)[
2
1
)([
2
1
312312133221
312312133221
211112223323322231111333
211112223323322231111333
yxyxyxyxyxyx
yxyxyxyxyxyx
yxyxyxyxyxyxyxyxyxyxyxyx
yxyxyxyxyxyxyxyxyxyxyxyx
++−++=
−−−++=
++−−−−++−−+=
−−+−−−++−−+=
Area of ∆ ABC =
1
1
3
3
2
2
1
1
2
1
y
x
y
x
y
x
y
x
=
[ ])()(
2
1
312312133221 yxyxyxyxyxyx ++−++=
The formula of the area of triangle is
6.3.2 Area of Quadrilateral
The formula of the area of quadrilateral is
( ) ( )4134231214433221
2
1
yxyxyxyxyxyxyxyx +++−+++=
Example 1:
Find the value of m if the point P(m, 2), Q(4, -3) and R(-2, 5) lie on a straight line.
Solution:
Method 1- Using Concept Area Of Triangle
0
25
2
3
4
22
1
=
−
−
mm
1
1
3
3
2
2
1
1
2
1
y
x
y
x
y
x
y
x
The area of a straight line is
zero.
1
1
4
4
3
3
2
2
1
1
2
1
y
x
y
x
y
x
y
x
y
x
Simplify
6. Additional Mathematics Module Form 4
Chapter 6- Coordinate Geometry SMK Agama Arau, Perlis
Page | 65
028
0514163
0)514(163
0)568()4203(
=+−
=−−+−
=+−+−
=++−−+−
m
mm
mm
mm
028
028
=+−
=+−
m
m
4
1
28
=
=
m
m
Method 2
Using the concept of gradient of straight line
6
8
)2(4
53
−
=
−−
−−
=
QR
QR
m
m
3
4
−=
4
1
14
15164
3
4
4
)3(2
=
=
−=−
−=
−
−−
=
m
m
m
m
mm QRPQ
Example 2:
Find the possible values of k if the area of triangle with vertices A (9, 2), B(4, 12) and C(k, 6) is 30 unit2
Solution:
30
2
9
612
4
2
9
2
1
=
k
601070
60)54128()224108(
=−
=++−++
k
kk
Point P, Q and R lie on the
same line, so QRPQ mm =
7. Additional Mathematics Module Form 4
Chapter 6- Coordinate Geometry SMK Agama Arau, Perlis
Page | 66
(i) 601070 −=− k
1
1010
=
=
k
k
(ii) 601070 −=− k
13
13010
=
=
k
k
Example 3:
Find the area of quadrilateral KLMN given K (1, 3), L(-1, 2) and M(-4, -3) and N(6, -9).
Solution:
3
1
9
6
3
4
2
1
3
1
2
1
−−
−−
=
2
5.48
97
2
1
3853
2
1
)38(59
2
1
)91883()183632(
2
1
unit=
=
+=
−−=
−−−−−+++=
EXERCISE 6.3
1. Find the possible values of p if the area of triangle with the vertices D(p, -p), E(1, 0) and F(-3, 6) is 10
unit2
.
2. Find the area of the rhombus PQRS if the coordinates of the points P, Q and R are (6, 4). (8, 7) and
(-6, 3) respectively.
3. The points G(4, -2), F(1, 1) and H(-2, p) lie on a straight line. Find the value of p.
4. Find the area of the triangle PQR if the coordinates of the vertices are:
(a) P(1, 3), Q (4, 2) and R (7, 0)
(b) P(-2, 5), Q (7, -2) and R (-3, 1)
(c) P(-1, -5),Q (1, -4) and R(-1, -3)
The modulus sign will always result a
positive number. If k1070 − would
result 60, there will be two values of k. If
the value of k is 1, the value in the
modulus is 60. If the value of k is 13, the
value in the modulus is -60 and at last
becomes 60 because the modulus sign
will always result a positive number.
1
1
4
4
3
3
2
2
1
1
2
1
y
x
y
x
y
x
y
x
y
x
8. Additional Mathematics Module Form 4
Chapter 6- Coordinate Geometry SMK Agama Arau, Perlis
Page | 67
6.4 EQUATION OF A STRAIGHT LINE
6.4.1 x-intercept and y-Intercept of a straight line
y
b
x
0 a
1. The line intersects with the x-axis at a and the line intersects with the y-axis at b
2. a is called x-intercept and b is called y-intercept.
EXERCISE 6.4.1
1. Determine the x-intercept for the following straight lines:
(a) 1
32
=+
yx
(b) 0943 =−− yx
(c) 66 −= xy
2. Determine the y-intercept for the following straight lines:
(a) 1
51
=+
yx
(b) 01475 =−− yx
(c) 653 −= xy
6.4.2 The Gradient of a Straight Line
(i) (ii) (iii) (iv)
The gradient is positive the gradient is negative the gradient is undefined the gradient is zero
9. Additional Mathematics Module Form 4
Chapter 6- Coordinate Geometry SMK Agama Arau, Perlis
Page | 68
Finding the gradient (m) of a straight line
1-
Example:
y
A(0,5)
x
0 B(0, 4)
12
12
xx
yy
m
−
−
=
40
05
−
−
=ABm
4
5
−=
2-
y
(0, b)
x
0 (a, 0)
a
b
m
−
−
=
0
0
12
12
xx
yy
m
−
−
=
10. Additional Mathematics Module Form 4
Chapter 6- Coordinate Geometry SMK Agama Arau, Perlis
Page | 69
a
b
−=
Hence,
Example:
y
A(0,5)
x
0 B(0, 4)
erceptx
ercepty
m
int
int
−
−
=
4
5
−=ABm
3-
Example: y
A(0,5)
5
x
0 4 B(0, 4)
θ
erceptx
ercepty
m
int
int
−
−
−=
θtan=m
11. Additional Mathematics Module Form 4
Chapter 6- Coordinate Geometry SMK Agama Arau, Perlis
Page | 70
θtan=ABm
4
5
−=ABm
EXERCISE 6.4.2
Find the gradient of the following points.
(a) A (2, 3) and B (4, 5)
(b) M (-3, 1) and N (4, -2)
(c) P (2, 3) and Q (4, 3)
(d) C (2, 5) and D (2, 8)
(e) F(0, 6) and G( 3, 0)
6.4.3 The Equation of a Straight Line
1-General Form
The equation of general form is 0=++ cbyax
Example:
Given the equation of a straight line is 542 += xy . Change the equation into the general form.
Solution:
542 += xy
0524 =+− yx
2-Gradient Form
The equation of gradient form is cmxy += where m is the gradient and c is y-intercept
Example:
Given the equation of a straight line is 542 += xy . Determine the gradient and the y-intercept of
the straight line.
Solution:
542 += xy
2
5
2 += xy
Hence, the gradient of the straight line is 2 while and the y-intercept of the straight line is
2
5
.
12. Additional Mathematics Module Form 4
Chapter 6- Coordinate Geometry SMK Agama Arau, Perlis
Page | 71
3-Intercept form
The equation of intercept form is 1=+
b
y
a
x
where a is x-intercept and b is y-intercept.
Example:
Given the equation of a straight line is 632 += xy . Convert the equation into the intercept form.
Hence, state the x-intercept and y-intercept of the straight line.
Solution:
632 += xy
1
32
1
32
1
6
2
6
3
623
=+
−
=+
−
=+
−
=+−
yx
yx
yx
yx
Hence, the x-intercept of the straight line is -2 and y-intercept of the straight line is 3.
Example:
Find the equation of the straight line which has a gradient of -3 and passes through the mid-point of the
line joining A (1, 4) and B(7, -2).
Solution:
Mid-point of AB
−++
=
2
)2(4
,
2
17
( )1,4=
The equation of the straight line which has a gradient of -3 and passes through (4, -1) is
xy
x
y
3121
3
4
1
−=−
−=
−
−
0133 =−+ yx or 133 +−= xy
EXERCISE 6.4.3
1. Write each of the following equations to intercept form. Hence, state the gradient of the straight line.
(a) 042 =−+ yx
(b) 63 =− yx
(c) 234 =+ yx
(d) 132 =+ yx
2. Find the equation of the straight line which has a gradient of -2 and passes through point B (7, -2).
Use the general point (x, y) and
specific point (4, -1) to find the
gradient of the line and it is equal
to the given gradient.
13. Additional Mathematics Module Form 4
Chapter 6- Coordinate Geometry SMK Agama Arau, Perlis
Page | 72
6.4.4 The Point of intersection of two Straight lines
1. When two lines intersect, the point of intersection lies on both lines.
2. This means the coordinates of the point satisfy both the equations of the lines.
3. Therefore, we need to solve the equation simultaneously in order to determine the point of
intersection.
Example 1:
The straight line which has a gradient of 2 and passes through the point (4, -1) intersects with the
straight line 6−=+ yx at the point P. Find the coordinates of the point P.
Solution:
First of all, we have to find the equation of the straight line.
92
821
2
4
)1(
=−
−=+
=
−
−−
yx
xy
x
y
6−=+ yx
+ ,
1
33
=
=
x
x
Substitute 1=x into ,
7
9)1(2
−=
=−
y
y
Hence the coordinates of P is (1, -7)
Example 2:
The straight line 04 =−+ yx and 01132 =−+ yx intersect at point A. Find the equation of the
straight line which passes through the point A and point B (5, 2).
Solution:
0822
04
=−+
=−+
yx
yx
01132 =−+ yx
- ,
3
03
=
=−
y
y
1
2
1 2
1
1
2
2 1
14. Additional Mathematics Module Form 4
Chapter 6- Coordinate Geometry SMK Agama Arau, Perlis
Page | 73
Substitute 3=y into ,
1
22
08)3(22
=
=
=−+
x
x
x
Hence the coordinates of A is (1, 3)
15
32
−
−
=ABm
4
1
−=ABm
The equation of the straight line that passes through point A and B is
xy
x
y
−=−
−=
−
−
1124
4
1
1
3
0134 =−+ yx or 134 =+ yx or
4
13
4
1
+−= xy
EXERCISE 6.4.4
1. Find the coordinates of the point of intersection of the line 01134 =−+ yx and 01762 =+− yx .
2. Find the points of intersection of the following pairs of straight lines.
(a) 025 =−− yx (b) 42 += xy
052 =+− yx 5+= xy
3. Find the equation of that is parallel to the line 52 += xy and passing through the point of
intersection of lines 092 =−− yx and 22 =+ yx .
6.5 PARALLEL AND PERPENDICULAR LINES
Parallel Lines
1. When two lines are parallel, they have the same gradient.
B Q
A P
2. If line AB and line PQ are parallel, so PQAB mm = .
1
We can write the equation in any
form. Either the general form or
intercept form or gradient form.
15. Additional Mathematics Module Form 4
Chapter 6- Coordinate Geometry SMK Agama Arau, Perlis
Page | 74
Example:
The straight line AB passes through the point (6, 3) is parallel to the straight line PQ. Given point P (0, 2)
and point Q(4, 0). Find the equation of the straight line AB.
Solution:
First of all, we have to find the gradient of straight line PQ
4
2
−=PQm
2
1
−=
PQAB mm =
2
1
−=ABm
The equation of the straight line that passes through point (6, 3) is
xy
x
y
−=−
−=
−
−
662
2
1
6
3
0122 =−+ yx or 122 =+ yx or 6
2
1
+−= xy
Perpendicular Lines
y
x
1. Given that line AB and BC are perpendicular to each other.
2. We already know θtan=m .
θtan1 =m αtan2 =m
AB
BC
=
BC
AB
−=
B
C
A
θ α
1m
2m
We can write the equation in any
form. Either the general form or
intercept form or gradient form.
16. Additional Mathematics Module Form 4
Chapter 6- Coordinate Geometry SMK Agama Arau, Perlis
Page | 75
3. When
BC
AB
AB
BC
mm −×=× 21
Hence, if two lines are perpendicular to each other, then the product of their gradient is 1− .
Example:
Given the straight line 9−= txy and 32 += xy is perpendicular to each other. Find the value of t.
Solution:
32 += xy
21 =m
9−= txy
tm =2
121 −=× mm
2
1
12
12
−=
−=
−=×
t
t
t
EXERCISE 6.5
1. The equation of the straight line PQ is 0786 =+− xx . Each of the following straight line is parallel to
PQ. Find the value of t in each case.
(a) 064 =−+ ytx
(b) 8
2
+= x
t
y
(c) 012 =−− tyx
2. Find the equation of the straight line which passes through point B (2, -5) and perpendicular to the
straight line 13 +−= xy .
3. PQRS is a rhombus with P (0, 5) and the equation of QS is 12 += xy . Find the equation of diagonal of
PR.
4. Find the value of h if the straight line 02 =+− hxy is perpendicular to the straight line
035 =++ xy .
5. Given that the equation of the line PQ is 1532 += xy and point Q lies on the y-axis. Point R is (4, 1)
lies on line QR. Find the equation of QR if the line PQ and QR are perpendicular to each of other.
Use the concept
121 −=× mm
17. Additional Mathematics Module Form 4
Chapter 6- Coordinate Geometry SMK Agama Arau, Perlis
Page | 76
6.6 LOCUS OF A MOVING POINT
Locus represents the path followed by a moving point with the reference to one or more fixed points,
satisfying certain conditions.
6.6.1 Equation of Locus
Equidistant from a fixed point
y
P (x, y)
3 unit
A (1, 1)
x
The equation of locus is
0722
91212
9)1()1(
3)1()1(
22
22
22
22
=−−−+
=+−++−
=−+−
=−+−
yxyx
yyxx
yx
yx
Equidistant from two fixed points
Find the equation of the locus of a moving point P such that its distance from the point A (1, 2) and point
B (3, 4) are equal.
B (3, 4)
A (1, 2)
Equation of locus is actually
involving the distance between
two points. So we have to use
the formula of the distance
between two points to find the
equation of locus. There is no
specific formula to find the
formula to find the equation of
locus.
In this case, P is the moving point
such that its distance is always 3
unit from point A.
05
02044
2586542
168964412
)4()3()2()1(
)4()3()2()1(
2222
2222
2222
2222
=−+
=−+
+−−+=+−−+
+−++−=+−++−
−+−=−+−
−+−=−+−
=
yx
yx
yxyxyxyx
yyxxyyxx
yxyx
yxyx
BPAP
locus
),( yxP
Square the both sides
Square the both sides
18. Additional Mathematics Module Form 4
Chapter 6- Coordinate Geometry SMK Agama Arau, Perlis
Page | 77
Constant ratio between two fixed points
Find the equation of locus of a moving point R such that its distance from the point E (4, 3) and the point
F (1, 5) is in the ration 3: 1.
0209841088
961682259099189
96168)251012(9
)3()4()5()1(3
3
1
3
22
2222
2222
2222
=+−−+
+−++−=+−++−
+−++−=+−++−
−+−=−+−
=
=
yxyx
yyxxyyxx
yyxxyyxx
yxyx
RERF
RF
RE
EXERCISE 6.6
1. Given the point A (0, 3) and the point B (1, 4).Find the equation of locus of a moving point Q such that
AQ= 2QB.
2. Given A (5, -2) and B (2, 1) are two fixed points. Point Q moves such that the ratio of AQ: QB = 2: 1.
Show that the equation of the locus of point Q is 034222
=−−−+ yxyx .
3. P is a moving point such that its distances from the points A(2, 5) and B(0, 3) is in the ratio of 2: 1. Find
the equation of locus P.
4. N is a locus which moves in such a way that NP=NQ. Given that P and Q are coordinates (-3, 6) and
(6,- 4) respectively, find the equation of locus N.
5. Show that the equation of the locus of a point that moves in such way that is distance from a fixed
point (3, -1) is 6 units, is by 0262622
=−+−+ yxyx .
CHAPTER REVIEW EXERCISE
1. Given the equation of straight lines AB and CD are 1
6
=+
k
yx
and 0432 =−+ yx respectively, find
the value of k if AB is perpendicular to CD.
2. The coordinates of the point A and B are (-2, 3) and (7, -3) respectively. Find
(a) the coordinates of C given that AB: BC = 1: 2.
(b) the equation of the straight line that passes through B and is perpendicular to AB.
3. ABCD is a parallelogram with coordinates A (-2, 3), B (3,4), C (2, -1) and D (h, k).
(a) Find the value of h and k.
(b) Find the equations of the diagonals AC and BD.
(c) State the angle between the diagonals AC and BD.
(d) Find the area of the parallelogram ABCD.
Square the both sides
19. Additional Mathematics Module Form 4
Chapter 6- Coordinate Geometry SMK Agama Arau, Perlis
Page | 78
4. P, Q and R are three points on a straight line. The coordinates of P and R are (-2, 3) and (3, 5)
respectively. Point Q lies on the y-axis. Find
(a) the ratio PQ: QR
(b) the coordinates of point Q
5. H is a point which moves such that its distance from point P (1, -2) and Q (-3, 4) is always equal. Show
that the equation of the locus H is given by the equation 0532 =+− yx .
6. Find the equation of straight line that passes through point P( 1 -2) and parallel to 4x – 2y = 8.
7. In diagram below, °=∠ 90PRS .
Find
(a) the coordinates of R
(b) the ratio of PQ: QR
(c) the equation of RS.
8. Diagram below shows a triangle BCD. The point A lies on the straight line BD.
Find
(a) the value of k
(b) the equation of CD, giving your answer in general form.
)9,6(−P
S
Q
R
3
x
y
0
)7,1(−C
)8,13(A
)2,5(B
),17( kD
x
0
y