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SMK. KOTA KLIAS, BEAUFORT
MATHEMATICS FORM FOUR
CHAPTER 5 THE STRAIGHT LINE
Disediakan oleh:
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MOHD NAZAN BIN KAMARUL ZAMAN Chapter 5: The Straight line
v v
v
Diagram A
h Diagram C
h Diagram B
h
vertical
Gradient of a straight line = m = horizontal
dis tan ce dis tan ce
m=
y 2 − y1 x 2 − x1
Gradient = The ratio of vertical distance to horizontal distance y 2 − y1 = x −x 2 1 6
= 10
Find the gradient of AB B 4 cm A
12 cm
C
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5.1 The Gradient of a Straight Line Gradient (m) a) y
m=
A(1, 8)
B (7, 2)
y 2 − y1 x 2 − x1
Gradient (m) c)
y
2 −8 6 m= m =− 7 −1 6 B (-3, 7)
A(8, 4)
m=-1
x
b)
y A(6 , 4)
x y − y1 m= 2 x 2 − x1
d) y A(5, -1)
m= B (-5, 3)
x
5.2 Intercepts
m=
B (-2, -9)
x
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a)
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y
b)
y A(0, 5)
B (8, 0)
x
x B (-2, 0)
x – intercept = y – intercept =
x – intercept = y – intercept =
gradient =
gradient =
5.3 Equation of a Straight Line EQUATION OFASTRAIGHT LINE : y = mx + c
with gradient m, and which cuts the y-axis at (0,c) @ y-intercept Gradient (m) a)A(3, 4) and B(2, 1)
m=
y 2 − y1 x 2 − x1
4 −1 m= 3 −2
m=3
y-intercept (c)
y = mx + c
m = 3 and c = -5 Use point Aor point B y = mx + c Point A (3, 4) x = 3 , y y = 3x - 5 = 4 and m = 3 y = mx + c 4 = 3(3) + c 4=9+c c=4–9 c = -5
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b)A(2, -2) and B( 5, 9)
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m=
y 2 − y1 x 2 − x1
m=
m=
c)A( -5, 12) and B(1, - 6)
m=
y 2 − y1 x 2 − x1
m=
m=
Gradient (m) d)A(0, 0) and B(-4, 8)
m=
y 2 − y1 x 2 − x1
m=
m=
e)A(0, -12) and B( 6, 0)
m= m=
y 2 − y1 x 2 − x1
y-intercept (c)
y = mx + c
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or
y −int ercept m= x − int ercept
f)A( -5, 0) and B(0, -15)
m=
m=
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5.4 Parallel Line – find the equation of the straight line CD The straight linesAB and CD are parallel
Gradient (mAB = mCD)
mAB =mCD y − y1 mAB = 2 x 2 − x1
a) y
C(6, 2) B (4, 0)
x
m CD= 3
A(0, -12)
b)
y
0 −( −12 ) m= =3 4 −0
C (5, 7) A(0, 6)
x B (2, 0)
c) C(-8, 2)
y B (-4, 0)
x A(0, -2)
y-intercept (c)
m = 3 and point C( 6, 2 ) y = mx + c 2 = 3(6) + c 2 = 18 + c 2 – 18 = c C = - 16
y = mx + c
m = 3 and c = -16 y = mx + c y = 3x – 16
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In the diagram, PQRS is a parallelogram.
y 1
The gradient of RS is 3 . Find (a) the y-intercept of the straight line RS, (b) the equation of the straight line PQ.
2
(Ans : 3) 1
(Ans : y = 3 x −2)
S
In the diagram, the straight line PQ is parallel to the straight line ST. O is the origin.
y Q (3,
R (−5, 8) Given the gradient of RS is −2. Find (a) the equation of the straight line PQ, (b) the x-intercept of the straight line RS.
O
(Ans : y = −2x + 14) (Ans : −1)
P
O
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In the diagram, OPQR is a parallelogram and O is the origin.
y P (3, 6)
Find 3 (Ans : 5 )
(a) the gradient of the straight line PQ,
(Ans : y = 2x −7)
(b) the equation of the straight line QR.
4
Q
In the diagram, OPQR is a parallelogram and O is the origin.
O
y
Find (a) the equation of the straight line PQ, (b) the y-intercept of the straight line QR.
(Ans : y = 3x −15) (Ans : 20)
O
R (5, 3)
R (4