1. The document summarizes key concepts from a chapter on coordinate geometry including calculating the distance and midpoint between two points, finding the slope and equation of a line, calculating the area of a triangle, finding the angle between two lines, and dividing a line segment in a given ratio. 2. Formulas are provided for finding the slope, equation, and angle of inclination of a line, as well as the perpendicular distance from a point to a line and the coordinates of the point where two lines intersect. 3. Worked examples demonstrate using the formulas to calculate values like the slope and equation of a line given two points, the area of a triangle using coordinates, and dividing a line segment based on a given

1. 1
ACTIVE MATHS 4 BOOK 2
Learning Outcomes
In this chapter you have learned to:
• Calculate the distance between two points
• Calculate the slope of a line through two points and consolidate the
understanding of slopes
• Find the equation of a line, given the slope and a point
• Find the equation of a line, given two points
• Calculate the area of a triangle
• Recognise the fact that the relationships y = mx + c and y – y1 = m(x – x1) are linear
• Find the equation of lines parallel and perpendicular to a given line and through a given point
• Find the point of intersection of two lines
• Recognise the fact that ax + by + c = 0 represents a linear relationship
• Calculate the length of the perpendicular from (x1,y1) to ax + by + c = 0
• Calculate the angle 𝜃 between two lines with slopes of m1 and m2
• Divide a line segment in a given ratio m : n
Chapter 8 – Co-ordinate Geometry: The Line
08

2. 2
ACTIVE MATHS 4 BOOK 2
Distance between two points: (𝑥2−𝑥1)2 + (𝑦2−𝑦1)2
A(8,–6) and B(5,–2). Find |AB|. |AB| = (𝑥2−𝑥1)2 + (𝑦2−𝑦1)2 (8,–6) (5,–2)
𝑥1 𝑦1 𝑥2 𝑦2
= (5 − 8)2 + (−2 − (−6))2
= (−3)2 + (4)2
= 9 + 16 = 25 = 5
Midpoint of a line segment:
𝑥1 + 𝑥2
2
,
𝑦1 + 𝑦2
2
A(8,–6) and B(5,–2). Find the midpoint of [AB].
Midpoint of [AB] =
𝑥1 + 𝑥2
2
,
𝑦1 + 𝑦2
2
(8,–6) (5,–2)
𝑥1 𝑦1 𝑥2 𝑦2
=
8 + 5
2
,
−6 − 2
2
=
13
2
,
−8
2
= (6∙5,–4)
Revision of Formulae
08 Co-ordinate Geometry: The Line

3. 3
ACTIVE MATHS 4 BOOK 2
Slope of a line given two points:
𝑦2 − 𝑦1
𝑥2 − 𝑥1
A(8,–6) and B(5,–2). Find the slope of AB. Slope of AB =
𝑦2 − 𝑦1
𝑥2 − 𝑥1
(8,–6) (5,–2)
𝑥1 𝑦1 𝑥2 𝑦2
=
−2 − (−6)
5 − 8
=
4
−3
= −
4
3
Equation of AB:
A(8,–6) and B(5,–2). Find the equation of AB.
𝑦 − 𝑦1 = 𝑚(𝑥 − 𝑥1)
Equation of AB is 𝑦 − 𝑦1 = 𝑚(𝑥 − 𝑥1) (8,–6)
𝑥1 𝑦1
𝑚 = slope
𝑚 = −
4
3
𝑦 + 6 = −
4
3
(𝑥 − 8)
3𝑦 + 18 = −4(𝑥 − 8)
3𝑦 + 18 = −4𝑥 + 32
⇒ 4𝑥 + 3𝑦 − 14 = 0
NOTE: Two forms of an equation.
4𝑥 + 3𝑦 − 14 = 0
𝑦 = −
4
3
𝑥 +
14
3
Form 1: 𝑎𝑥 + 𝑏𝑦 + 𝑐 = 0
Form 2: 𝑦 = 𝑚𝑥 + 𝑐 𝑚 = 𝑠𝑙𝑜𝑝𝑒 𝑎𝑛𝑑 𝑐 𝑖𝑠 𝑡ℎ𝑒 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 𝑜𝑛 𝑡ℎ𝑒 𝑦 𝑎𝑥𝑖𝑠.
08 Co-ordinate Geometry: The Line

4. 4
ACTIVE MATHS 4 BOOK 2
Slope of AB, given its equation:
Point of intersection of a line with the x- and y-axes:
Point of intersection of two lines:
Slope of parallel and perpendicular lines:
Equation form Slope Example
𝑦 − 𝑦1 = 𝑚(𝑥 − 𝑥1) slope is 𝑚
𝑦 = 𝑚𝑥 + 𝑐 slope is 𝑚
𝑦 − 2 = 3 𝑥 − 4 Slope = 3
𝑎𝑥 + 𝑏𝑦 + 𝑐 = 0 slope is
−𝑎
𝑏
𝑦 = 3𝑥 − 10 Slope = 3
3𝑥 + 2𝑦 − 10 Slope =
−3
2
• When a line cuts the x-axis, the value of the y coordinate is 0.
• When a line cuts the y-axis, the value of the x coordinate is 0.
Use the method of solving simultaneous equations to find the point of intersection.
If two lines 𝑎 and 𝑏 have slopes 𝑚1 and m2, respectively, then:
• 𝑎||𝑏 ⇔ 𝑚1 = 𝑚2
• 𝑎 ⊥ 𝑏 ⇔ 𝑚1. 𝑚2 = −1
08 Co-ordinate Geometry: The Line

5. 5
ACTIVE MATHS 4 BOOK 2
Graphical Approach
A triangle with vertices (2,1), (3,–2) and (–1,–1).
The area of the triangle can be found using the following graphical approach:
-6 -5 -4 -3 -2 -1 1 2 3 4 5 6
-4
-3
-2
-1
1
2
3
4
(3,−2)
(2,1)
(− 1,−1)
A B
D C
P Q
R
3 1
3
4
2
1
Construct a rectangle ABCD around the triangle as shown, and find its area: 4 × 3 = 12 units2.
Subtract the areas of the triangles P (3 units2), Q (1∙5 units2) and R(2 units2). Total = 6∙5 units2.
Area of original triangle = 12 − 6∙5 = 5∙5 units2.
The Area of a Triangle
08 Co-ordinate Geometry: The Line

6. 6
ACTIVE MATHS 4 BOOK 2
-4 -3 -2 -1 1 2 3 4 5 6 7 8 9 10 11 12
-4
-3
-2
-1
1
2
3
4
5
6
7
8
9
10
11
12
x
y
Formula Area of triangle: (F and T: P18)
1
2
|𝑥1𝑦2 − 𝑥2𝑦1|
To use this formula, one vertex must be (0,0).
Find the area of the triangle with vertices (–3,4), (4,2) and (6,10).
Translate the triangle so that one vertex is at (0,0).
(− 3,4)
(4,2)
(6,10)
(0,0) (7,-2)
(9,6)
The vertices of the translated triangle are (0,0), (7,–2) and (9,6).
(𝑥1, 𝑦1) (𝑥2, 𝑦2)
𝐴𝑟𝑒𝑎 =
1
2
𝑥1𝑦2 − 𝑥2𝑦1 =
1
2
|(7)(6) − (9)(−2)| =
1
2
|60| = 30 𝑢𝑛𝑖𝑡𝑠2
.
08 Co-ordinate Geometry: The Line
Formula Approach

7. 7
ACTIVE MATHS 4 BOOK 2
The shortest distance from a point to a line is the perpendicular distance.
The distance from a point (x1, y1) to the line 𝑎𝑥 + 𝑏𝑦 + 𝑐 = 0 is given by:
Formula 𝑑 =
|𝑎𝑥1 + 𝑏𝑦1 + 𝑐|
𝑎2 + 𝑏2
(F and T: P19)
Find the distance from the point (3,1) to the line 4x + 3y + 10 = 0.
𝑎 = 4 𝑏 = 3 𝑐 = 10
4𝑥 + 3𝑦 + 10 = 0 (3,1)
(𝑥1, 𝑦1)
𝑑 =
|(4)(3) + (3)(1) + 10|
(4)2+(3)2
=
|12 + 3 + 10|
16 + 9
=
|25|
25
=
25
5
∴ 𝑑 = 5
The Perpendicular Distance from a Point to a Line
08 Co-ordinate Geometry: The Line

8. 8
ACTIVE MATHS 4 BOOK 2
The Angle of Inclination
The angle of inclination is the angle formed between a line and the positive side of the x-axis.
Definition
• The angle of inclination is always between 0° and 180°.
α
180° 0°
α
180° 0°
• It is always measured anticlockwise from the positive side of the x-axis.
• The slope m of any line is equal to the tangent of its angle of inclination then
m = tan α (where α = angle of inclination).
The Angle of Inclination and the Angle Between Two Lines
08 Co-ordinate Geometry: The Line

9. 9
ACTIVE MATHS 4 BOOK 2
The Angle Between Two Lines
Generally, find the acute angle first (tan 𝜃 is positive),and use 180° – 𝜃 to find the obtuseangle.
Ɵ
p
q
Formula
tan 𝜃 = ±
𝑚1 − 𝑚2
1 + 𝑚1𝑚2
(F and T: P19)
If two linesp and q haveslopes m1 and m2 respectively, and 𝜃 is the anglebetweenthem, then:
Find the measure of the angles between the two lines: 𝟐𝒙 + 𝒚 − 𝟔 = 𝟎 and 𝟑𝒙 − 𝟐𝒚 + 𝟐 = 𝟎.
2𝑥 + 𝑦 − 6 = 0 3𝑥 − 2𝑦 + 2 = 0
𝑆𝑙𝑜𝑝𝑒 𝑚1 = −
𝑎
𝑏
= −2 𝑆𝑙𝑜𝑝𝑒 𝑚2 = −
𝑎
𝑏
= −
3
−2
=
3
2
tan 𝜃 = ±
𝑚1 − 𝑚2
1 + 𝑚1𝑚2
= ±
(−2) −
3
2
1 + (−2)
3
2
= ±
7
−4
⇒ tan 𝜃 = ±
7
4
𝜃 = tan−1
7
4
(𝐴𝑐𝑢𝑡𝑒 𝑎𝑛𝑔𝑙𝑒)
⇒ 𝜃 = 60∙26°
𝑂𝑏𝑡𝑢𝑠𝑒 𝑎𝑛𝑔𝑙𝑒 = 180° − 60∙26°
⇒ 𝜃 = 119∙74°
08 Co-ordinate Geometry: The Line

10. 10
ACTIVE MATHS 4 BOOK 2
P Q
R
a b
R divides [PQ] internally in the ratio a : b.
Formula 𝑅 =
𝑏𝑥1 + 𝑎𝑥2
𝑏 + 𝑎
,
𝑏𝑦1 + 𝑎𝑦2
𝑏 + 𝑎
(F and T: P18)
X(−5,3) and Y(3,−1) are points. Find the co-ordinates of Z, which divides [XY] in the ratio 3 : 1.
a = 3 b = 1
(−5,3) X Y
Z (3,−1)
(𝒙𝟏, 𝒚𝟏)
(𝒙𝟐, 𝒚𝟐)
𝑍 =
(1)(−5) + (3)(3)
1 + 3
,
(1)(3) + (3)(−1)
1 + 3
𝑍 =
4
4
,
0
4
𝑍 = (1,0)
Dividing a Line Segment in a Given Ratio a : b
08 Co-ordinate Geometry: The Line