The document discusses equations of straight lines. It covers determining the gradient and equation of a straight line, as well as representing a line in various forms such as y=mx+c, Ax + By + C = 0, and identifying lines from two points. Key concepts covered include finding the gradient from two points, the relationship between perpendicular lines, and deriving the equation of a line given values for the gradient and a point.

1. EQUATIONS OF
STRAIGHT LINES
(PERSAMAAN GARIS
LURUS)
Standard Competence : 1. Understanding an
algebraic form, relation, function, and linear
equation

2. Basic Competence : . Understanding an algebraic form,
relation, function, and linear equation
Standard Competence :
- 1.6 Determine the gradien, the equation, and the graph
of a straight line.

3. After learning this topic, the students are expected to be
able to :
 Identify Linear Equation in various form and
variables
 Draw the graph of Linear Equation in
cartessian coordinate
 Determine the gradient of Linear Equation in
various form
 Determine the Linear Equation passing
through two points

4. • Determine the Linear Equation passing through a
point and gradient
• Determine a point of itersection between two lines
in various positions
•And apply a linear equation concept in problem
solving
-

5. The Straight Line
y + ve gradient y
C C
x x
- ve gradient
All straight lines have an equation of the form y = mx + c
m = gradient y axis intercept

6. B ( x2 , y2 )
y2 − y1
A( x1 , y1 ) x2 − x1
Vertical Height
Gradient m =
Horizontal Distance
y2 − y1
=
x2 − x1

7. Undefined and zero gradient
Gradient is a measure of slope. If a line has zero gradient it has zero slope.
A line with zero slope is horizontal.
y
10 Consider two points on this graph.
8
y2 − y1
(−4, 4) 6
(6, 4) m=
4 x2 − x1
2 4−4
=
– 10 – 8 – 6 – 4 – 2 2 4 6 8 10 x 6+4
– 2
– 4 =0
– 6
– 8 The equation of the line is y = 4
– 10
All horizontal lines have an equation of the form y = c

8. y
Consider two points on this graph.
10
(4,8) y2 − y1
8
m=
6 x2 − x1
4
8+6
2 =
4−4
– 10 – 8 – 6 – 4 – 2 2 4 6 8 10 x
– 2 = ∞ (undefined)
– 4
– 6 (4, −6)
– 8
The equation of the line is x = 4
– 10
All vertical lines have an equation of the form x=a

9. y
From the diagram we can see that
∆y
tan θ = =m
∆x
θ
∆y
∆x
θ
m = tan θ x
Note that θ is the angle the line makes with the positive direction of
the x axis.

10. Collinearity
Two lines can either be:
C B C
B
B
D
A
A C A
At an angle Parallel and Distinct Parallel and form a straight line
Points that lie on the same straight line are said to be collinear.
To prove points are collinear:
1. Show that two pairs of points have the same gradient. (parallel)
2. If the pairs of points have a point in common they MUST be
collinear.

11. 1. Prove that the points P(-6 , -5), Q(0 , -3) and R(12 , 1) are collinear.
−3 + 5 1+ 3
m pq = mqr =
0+6 12 − 0
2 4
= =
6 12
1 1
= =
3 3
As the gradients of PQ and QR are equal PQ is parallel to QR.
Since Q is a point in common to PQ and QR, the points P, Q and R
are collinear.
Page 3 Exercise 1B

12. Perpendicular Lines
y
( a, b)
x
If we rotate the line through 900.

13. Perpendicular Lines
y
( a, b)
x
If we rotate the line through 900.

14. Perpendicular Lines
y
( a, b)
x
If we rotate the line through 900.

15. Perpendicular Lines
y
b−0 a−0
B(−b, a ) mOA = mOB =
a−0 −b − 0
b a
A ( a, b) = =−
a b
b a
O x mOA × mOB = × − = −1
a b
This is true for all perpendicular lines.
If two lines with gradients m1 and m2 are perpendicular then m1m2 = −1
Conversely, if m1m2 = −1 then the lines with gradients m1 and m2
are perpendicular.

16. 1. If P is the point (2,-3) and Q is the point (-1,6), find the gradient
of the line perpendicular to PQ.
6+3 9
mPQ = = − = −3
−1 − 2 3
To find the gradient of the line perpendicular to PQ we require the
negative reciprocal of –3.
a b
Remember: × − = −1
b a
3 1
Since − 3 = − The negative reciprocal would be
1 3
1
The gradient of the line perpendicular to PQ is .
3

17. 2. Triangle RST has coordinates R(1,2), S(3,7) and T(6,0). Show that
the triangle is right angled at R.
T 7−2 0−2
mRS = mRT =
3 −1 6 −1
5 2
= =−
2 5
5 2
R S mRS × mRT = × − = −1
2 5
Since mRS × mRT = −1, RS is perpendicular to RT.
Hence the triangle is right angled at R.

18. Equation of a Straight Line
All straight lines have an equation of the form y = mx + c
y P(x,y) is any point on the line except A.
P ( x, y ) For every position P the gradient of AP is
m y−c y−c
m= =
x−0 x
A(0,C)
x y − c = mx
y = mx + c

19. 1. What is the equation of the line with gradient 2 passing through
the point (0,-5)?
m = 2 and c = −5
∴ the equation of the line is y = 2 x − 5
2. Find the gradient and the y intercept of the line with equation
4x + 3y = 2
Rearranging gives: 3 y = −4 x + 2
4 2
⇒ y =− x+
3 3
4 2
Gradient is - and the y axis intercept is .
3 3

20. 3. Show that the point (2,7) lies on the line y = 4 x − 1.
When x = 2, y = 4 × 2 − 1 = 7
Because (2,7) satisfies the equation y = 4x – 1, the point must lie on the
line.

21. General Equation of a Straight Line
Ax + By + C = 0 is used as an alternative to y = mx + c
Ax + By + C = 0 is the GENERAL EQUATION of a straight line.
1. Rearrange y = 2 x + 5 into the form Ax + By + C = 0
and identify the values of A, B and C.
2x − y + 5 = 0 A = 2, B = −1, C = 5
4x
2. Rearrange y = − into the form Ax + By + C = 0
3
and identify the values of A, B and C.
3 y = −4 x ⇒ 4 x + 3 y = 0 A = 4, B = 3, C = 0

22. 3. Rearrange x = 7 into the form Ax + By + C = 0
and identify the values of A, B and C.
x−7 = 0 A = 1, B = 0, C = −7

23. Finding the equation of a Straight
Line
To find the equation of a straight we need
•A Gradient
•A Point on the line
The equation of a straight line with gradient m passing through (a,b) is
y − b = m( x − a )

24. P(x,y) is any point on the line except A.
y
P ( x, y ) For every position P the gradient of AP
m y −b m y −b
m= ⇒ =
x−a 1 x−a
A(a,b)
x
y − b = m( x − a )

25. 1
1. Find the equation of the straight line passing through (5, − 2) with gradient .
2
1
P (5, −2), m=
2
Equating with y − b = m( x − a)
1
y − (−2) = ( x − 5)
2
1 5
y+2= x−
2 2
1 9
y = x− or 2 y = x − 9
2 2

26. 2. Find the equation of the line passing through P(-2,0) and Q(1,6).
6−0 6−0
mPQ = =2 mPQ = =2
1+ 2 1+ 2
Using point P Using point Q
y − 0 = 2( x + 2) y − 6 = 2( x − 1)
y = 2x + 4 y − 6 = 2x − 2
y = 2x + 4
But what if we used point Q?
Regardless of the point you use the equation of the straight line
will ALWAYS be the same as both points lie on the line.

27. Lines in a Triangle
1. The Perpendicular Bisector.
A perpendicular bisector will bisect a line at 900 at the mid point.
The point of intersection
is called the
Circumcentre.

28. 1. A is the point (1,3) and B is the point (5,-7). Find the
equation of the perpendicular bisector of AB.
To find the equation of any straight line we need a point and a gradient.
 1 + 5 3 + (−7) 
Mid-point of AB =  , ÷ = ( 3, −2 )
 2 2 
−7 − 3 −10 5
mAB = = =−
5 −1 4 2
2
m⊥ = ( since m1m2 = −1)
5
2 2 16
y + 2 = ( x − 3) ⇒ y = x −
5 5 5

29. 2. The Altitude.
An altitude of a triangle is a line from a vertex perpendicular to
the opposite side. A triangle has 3 altitudes.
The point of intersection
is called the
Orthocentre.

30. 3. The Median of a Triangle.
The median of a triangle is a line from a vertex to the mid point
of the opposite side. A triangle has 3 medians.
The point of intersection
is called the centroid
1
2
A further point of information regarding the centroid.
The centroid is a point of TRISECTION of the medians. It divides
each median in the ratio 2:1.

31. 1. F, G and H are the points (1,0), (-4,3) and (0,-1) respectively. FJ
is a median of triangle FGH and HR is an altitude. Find the
coordinates of the point of intersection D, of FJ and HR.
(Draw a sketch – It HELPS!!)
MEDIAN  −4 + 0 3 − 1 
÷ = ( −2,1)
G J ,
 2 2 
1− 0 1
mFJ = =−
J −2 − 1 3
1 1 1
y − 0 = − ( x − 1) ⇒ y = − x +
3 3 3
F H

32. ALTITUDE
G
H(0,-1)
3−0 3 5
mFG = =− ∴mHR =
R J −4 − 1 5 3
D
5 5
y +1 = x ⇒ y = x −1
F H 3 3
1 1 5
The point D occurs when; − x + = x − 1
3 3 3
4
= 2x
3
2 5 2 1
x= ⇒ y = × −1 =
3 3 3 9
2 1
∴D , ÷
3 9