The Straight Line

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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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1

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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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3

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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