Ch14

LEY_bk9_c14_finalpp Page 437 Wednesday, January 12, 2005 11:47 AM

Chapter 14 Straight Lines and Regions This chapter deals with the use and application of various standard forms of the equation of the straight line and the graphing of regions on the number plane. After completing this chapter you should be able to: ✓ find the equation of a straight line ✓ rearrange equations of straight lines into various forms ✓ sketch straight lines given their intercepts ✓ demonstrate that two lines are perpendicular if the product of their gradients is –1 ✓ find the equation of a line parallel or perpendicular to a given line ✓ graph a variety of regions of the number plane involving straight lines.

Syllabus reference PAS5.3.3 WM: S5.3.1–S5.3.5

LEY_bk9_c14_finalpp Page 438 Wednesday, January 12, 2005 11:47 AM

438

Straight Lines and Regions (Chapter 14) Syllabus reference PAS5.3.3

Diagnostic Test 1

2

3

4

The equation of the line passing through (–2, 3) with gradient 2 is: A y = 2x + 3

B y = 2x – 7

C y = 2x – 2

D y = 2x + 7

The equation of the line passing through K (–3, 2) and L (4, –5) is:

7

When written in gradient–intercept form the equation 3x – 2y + 8 = 0 is: A 2y = 3x + 4 3 C y = --- x + 4 2

8

B 3x – 2y = –8 3 D y = --- x + 8 2

The x and y intercepts of 3x – 4y – 12 = 0 are:

A y = –x – 1

B y=x+5

A 4 and –3

B 3 and –4

C y = –3x – 7

D y = –x + 1

C –4 and 3

D –3 and 4

The equation of the line cutting the y-axis at 4 and the x-axis at –5 is: 4 --- x + 4 A y= 4 B y = – --- x + 4 5 5 5 5 C y = --- x + 4 D y = – --- x + 4 4 4

9

10

The equation of this line is:

The line perpendicular to 3x – 5y + 7 = 0 is: A 3x + 5y – 2 = 0 B 3x – 5y – 7 = 0 5 C –3x – 5y + 7 = 0D y = – --- x + 2 3

y 3 2

11

1 –3 –2 –1

The line parallel to y = –2x + 1 is: --- x + 1 --- x + 1 A y= 1 B y= –1 2 2 C y = –2x + 3 D y = –2x + 1

x 1

–1

2

3

–2

--- x – 1 A y= 2 3

3 B y = --- x – 1 2

C y = –x + 3

D y = 2x – 1

12

The triangle ABC with vertices A (–1, 0), B (1, 4) and C (5, 2) is: A scalene

B isosceles

C equilateral

D right angled

The region defined by y < 3 is: A y 4

5

6

The value of a if (a, –2) lies on the line y = 3x – 2 is:

3

A 0

B 1

1

C 2

D 3

--- x – 2 In general form y = 1 --- is: 2 3 A 6y – 3x + 4 = 0 B 3x – 6y – 4 = 0 --- x – y – 2 --- = 0 C 3x + 2y – 2 = 0 D 1 2 3

2

–3 –2 2 –1 1

x –1 –2 –3

1

2

3 4

LEY_bk9_c14_finalpp Page 439 Wednesday, January 12, 2005 11:47 AM

Straight Lines and Regions (Chapter 14) Syllabus reference PAS5.3.3

B

14 y

The region defined by 3x – y < 9 is: A

4

2 y

3

x

2

–4 –3 –2 2 –1 1

1 –3 –2 2 –1 1

–1

x –1

1

2

2

3 4

–2 –3

3 4

–4

–2 –3

C

1

–5 5 –6 –7

y

–8 –9

4 3 2 1 –3 –2 2 –1 1

B

x –1

1

2

2 y

3 4

x –4 –3 –2 2 –1 1

–2 –3

3

–5 5 –6

2

–7

1

–8 –9

x 1

2

3 4

–2 –3

13

3 4

–4

y 4

–1

2

–2 –3

D

–3 –2 2 –1 1

–1

1

C

2 y

This region is defined by:

x –4 –3 –2 2 –1 1

y 4

–2 –3

3 2

–4

1 –4 –3 3 –2 2 –1 1

–1

x –1

1

2

3 4

–7

–2 –3

A x ≤ 3 and y < 2

–5 5 –6 –8 –9

B x > –3 and y < 2

C x < –3 and y ≤ 2 D x ≥ –3 and y > 2

1

2

3 4

439

LEY_bk9_c14_finalpp Page 440 Wednesday, January 12, 2005 11:47 AM

440

Straight Lines and Regions (Chapter 14) Syllabus reference PAS5.3.3

D

If you have any difficulty with these questions, refer to the examples and questions in the sections listed in the table. Question

1–5

6–8

9, 10

11

12, 13

14

Section

A and B

C

D

E

F

G

A. EQUATIONS OF LINES The equation of a line is the simplest relationship that connects the x and y values for every point on the line.

For example, this line contains infinitely many points.

y

3

Some of these are (–1, 0), (0, 1), (1, 2), (2, 3), (3, 4), (4, 5).

2 1 –3 –2 –1

x –1 –2 –3

1

2

3 4

Notice that the y-coordinate is always 1 more than the x-coordinate. This is summarised by writing y = x + 1. Thus y = x + 1 is the equation of all points on this line.

–4 y

Similarly, this line contains the points (5, –1), (4, 0), (3, 1), (2, 2), (1, 3), (0, 4), (–1, 5) and so on.

4

Notice that the sum of the x and y coordinates is always 4.

2

3 1

i.e. x + y = 4 Thus the equation of the line is x + y = 4.

–2 –1

x –1 –2

1

2

3 4

LEY_bk9_c14_finalpp Page 441 Wednesday, January 12, 2005 11:47 AM

Straight Lines and Regions (Chapter 14) Syllabus reference PAS5.3.3

The equation of a line The equation of a line can be determined if we know: • the gradient of the line, and • the coordinates of a point on the line.

If a straight line has gradient m and passes through the point with coordinates (x1, y1), y–y x – x1

then its equation is y – y1 = m(x – x1) or -----------1 = m.

Proof:

Suppose P (x, y) is any point on the line with gradient m.

y P(x2,y2) Q(x1,y1) x

Equating the slopes gives y – y1 ------------ = m x – x1 or y – y1 = m(x – x1).

Example 1 Find the equation of the line through (–1, 2) having a gradient of 4. The equation of the line is y – y1 = m(x – x1) where (x1, y1) = (–1, 2) and gradient m = 4. y – 2 = 4(x – (–1)) y – 2 = 4(x + 1) y – 2 = 4x + 4 y = 4x + 6

Exercise 14A 1

Find the equation of the line through: a (2, –3) with gradient 3

b (–4, 2) with gradient –2

c (7, 4) with gradient – 1--2-

d (2, –5) with gradient 5

e (3, 7) with gradient

1 --3

g (3, 4) with gradient –1 1--2i

( – 1--2

,

1 1--2-

) with gradient 4

f

(–5, –2) with gradient – 1--2-

h (0, 0) with gradient 6 j

(5, 4) with gradient 0

441

LEY_bk9_c14_finalpp Page 442 Wednesday, January 12, 2005 11:47 AM

442

Straight Lines and Regions (Chapter 14) Syllabus reference PAS5.3.3

Example 2 Find the equation of the line that passes through the points A(–1, 5) and B(2, –3). First we find the gradient of AB using y2 – y1 m = --------------y2 – x1

[use (–1, 5) for (x1, y1) and (2, –3) for (x2, y2)]

–3–5 ∴ m = -------------------2 – ( –1 ) 8 ∴ m = – --3 Second, find the equation of the line using y – y1 = m ( x – x1 ) 8 ∴ y – 5 = – --- ( x – ( – 1 ) ) 3 8 ∴ y – 5 = – --- ( x + 1 ) 3 8 8 ∴ y = – --- x – --- + 5 3 3 8 7 ∴ y = – --- x + --3 3

2

1 8 or y = – --- x + 2 --3 3

Find the equation of the line passing through the following pairs of points. a A(2, 3) and B(4, 7) b A(0, 2) and B(–2, 4) c A(–1, –3) and B(5, –5) d A(6, 3) and B(4, 1) e A(5, –2) and B(2, –5) f P(0, 0) and Q(3, 5) g P(–3, 5) and Q(1, –2) h L(–3, –2) and M(0, 4) i X(2, 2) and Y(–3, –1) j X(0, 6) and Y(–4, 0)

Example 3 Find the equation of the line that cuts the y-axis at 3 and the x-axis at –2. The y-axis is cut when x = 0, i.e. (0, 3) is one point. The x-axis is cut when y = 0, i.e. (–2, 0) is the other point.

B(2, –3) could have been used as (x1, y1 ), giving the same equation.

LEY_bk9_c14_finalpp Page 443 Wednesday, January 12, 2005 11:47 AM

Straight Lines and Regions (Chapter 14) Syllabus reference PAS5.3.3

y2 – y1 Gradient, m = --------------x2 – x1

[using (x1, y1) = (0, 3) and (x2, y2) = (–2, 0)]

0–3 = ----------------–2–0 –3 = -----–2 3 = --2 But, y – y1 = m(x – x1) 3 ∴ y – 3 = --- (x – 0) 2 3 ∴ y – 3 = --- x 2 3 i.e. y = --- x + 3 2

Find the equation of the line a cutting the y-axis at –4 and the x-axis at 2 b cutting the y-axis at 7 and the x-axis at –2 c cutting the y-axis at –5 and the x-axis at –3 d with y-intercept –3 and x-intercept 2

3

B. THE GRADIENT–INTERCEPT FORM If a straight line has gradient m, and y-intercept b, then it has equation y = mx + b.

Proof: Since the y-intercept is b, the point (0, b) lies on the line.

y

Thus the equation is y – b = m(x – 0) i.e. y = mx + b b

x

Example 1 Find the equation of the line with gradient –3 and y-intercept 2. Since m = –3 and b = 2, the equation is y = –3x + 2.

443

LEY_bk9_c14_finalpp Page 444 Wednesday, January 12, 2005 11:47 AM

444

Straight Lines and Regions (Chapter 14) Syllabus reference PAS5.3.3

Exercise 14B 1

Find the equation of the line with: a gradient 2 and y-intercept 7

b

c gradient –3 and y-intercept –1 2 e gradient – --- and y-intercept 6 3

gradient 4 and y-intercept –6 1 gradient – --- and y-intercept 2 2 gradient 1 and y-intercept 0

d f

Example 2 Find the equation of the line below. (0, 2) and (4, 3) lie on the line 3–2 1 ∴ gradient is = ------------ = --4–0 4 1 i.e. m = --- and b = 2 4 1 ∴ equation is y = --- x + 2 4

y 4 3

(4, 3)

2 1

x 1

2

2

3

4

Find the equations of the following lines. a b

c y

y

y

4

4

4 3

3

(2, 3)

2

3

(3, 3)

2

2

1

1

1

x

x 1

2

3

4

1

–1

2

3

x –1

4

1

2

3

4

–1

d

e

f

y 3

2

4

2

1

3

1

x x

(–2, 3) 2

1

–4 –3 –2 –1

1

x –2 –1

y

y

5

1

2

3

4

2

3

–3 –2 –1

1

2

3

4

5

–1

–1

–2

–2

–3

(4, –3)

LEY_bk9_c14_finalpp Page 445 Wednesday, January 12, 2005 11:47 AM

Straight Lines and Regions (Chapter 14) Syllabus reference PAS5.3.3

Find the equation connecting the variables given. a b

3

M

c G

y

4

3

4

3

2

3

2

slope 13

1 1

2

3

x 1

–2 –1

t –2 –1

2

1 2

3

4

1

5

s

–1

4

–1

2

3

4

–1

–2

d

1

e M

F 5

2 1

4

g 1

–1

2

3

4

3

6

5

–1

2

–2

1 –1

(10, 2) x2 1

2

3

4

5

6

7

8

9

10

f p

√n 1

2

3

4

5

6

–1 –2 –3

(6, 3)

–4

Check the variable on each axis.

–5 –6

An alternative method to find the equation of a line is to use y = mx + b.

Example 3 Use y = mx + b to find the equation of the line passing through (–3, 5) with gradient 3. y = mx + b i.e. y = 3x + b We substitute x ∴5 ∴5 ∴ 14 i.e. y

= = = = =

– 3 and y = 5 3 ( –3 ) + b –9+b b 3x + 14

[as the gradient m = 3] [as (–3, 5) lies on the line]

445

LEY_bk9_c14_finalpp Page 446 Wednesday, January 12, 2005 11:47 AM

446

Straight Lines and Regions (Chapter 14) Syllabus reference PAS5.3.3

4

Find the equation of the line through a (3, –4) with gradient 2 c (8, –2) with gradient –3 e (3, 0) with gradient 5 g (0, 5) with gradient – 1--3i (9, 9) with gradient 1--3-

b d f h j

(5, –2) with gradient 1 (–4, –4) with gradient –2 (6, 8) with gradient 1--2(0, 0) with gradient 8 (6, 2) with gradient 0

Example 4 Find a given that (a, –1) lies on the line y = 2x + 3. Substitute x = a and y = –1 into the equation y = 2x + 3. ∴ –1 = 2(a) + 3 ∴ –1 = 2a + 3 ∴ –4 = 2a ∴ a = –2

5

Find a given that each point below lies on the line with the given equation. a (a, 3) y = 2x – 1 b (a, –2) y=4–x c (4, a) y= x+3 d (–2, a) y = 1 – 3x

Example 5 Draw the graph of the line with equation y = –2x + 1. Method 1 Table of values: x

–2

0

2

y

5

1

–3

Method 2 –2 y-intercept is 1 and gradient is -----1 ∴ start at 1 on y-axis, x-step of 1, then y-step of –2.

y

y y= –2x +1

5

5

4

4

3

3

2

2 1

1 –4 –3 –2 –1

–2

x –1 –2 –3

1

2

3 4

–4 –3 –2 –1

–1 –2 –3

1

x 2

3 4

LEY_bk9_c14_finalpp Page 447 Wednesday, January 12, 2005 11:47 AM

Straight Lines and Regions (Chapter 14) Syllabus reference PAS5.3.3

6

Draw the graph of the lines with the following equations. 1 a y = --- x + 2 b y = 2x + 1 2 1 d y = –3x + 2 e y = – --- x 2 3 2 g y = --- x h y = --- x + 2 2 3

c y = –x + 3 f y = –2x – 2 3 i y = – --- x – 1 4

C. GENERAL FORM OF A STRAIGHT LINE EQUATION The general form of the equation of a straight line is Ax + By + C = 0 where A, B and C are all integers and A ≥ 0.

Example 1 Write, in general form, the equations a y = 3x – 1

b

a

b

y = 3x – 1 0 = 3x – y – 1 (subtracting y) (swapping sides) 3x – y – 1 = 0

2 y = – --- x + 4 3 2 y = – --- x + 4 3 2 0 = – --- x – y + 4 3 0 = – 2x – 3y + 12 – 2x – 3y + 12 = 0 2x + 3y – 12 = 0

(subtracting y) (multiplying by 3) (swapping sides) (multiplying by –1 so that A ≥ 0)

Exercise 14C 1

Write the following equations in general form. a y = 2x + 1

b y = 5x – 2

d y = –2x – 5

e y = –3x + 4

1 g y = – --- x – 5 2 2 1 j --- y = --- x + 1 3 4

2 h y = – --- x – 3 3 1 3 k y = --- x – --2 4

c y = 2x + 5 1 f y = --- x + 2 2 3 2 i y = – --- x – --4 3 1 1 l --- y = x + --5 2

447

LEY_bk9_c14_finalpp Page 448 Wednesday, January 12, 2005 11:47 AM

448

Straight Lines and Regions (Chapter 14) Syllabus reference PAS5.3.3

Example 2 Rewrite the equation 4x – 3y – 12 = 0 in the form y = mx + b and hence find the gradient and y-intercept. 4x – 3y – 12 = 0 4x – 12 = 3y (adding 3y to both sides) 3y = 4x – 12 (swapping sides) 4 (dividing by 3) y = --- x – 4 3 4 ∴ gradient is --- m = and y-intercept is –4. 3

2

Rewrite the following equations in y = mx + b form and hence find the gradient and y-intercept. a x + 2y – 4 = 0 b 3x + 2y – 24 = 0 c 2x – y + 4 = 0 d 4x – 2y – 6 = 0 e 5x + 2y + 10 = 0 f 3x + 2y – 8 = 0 g 4x – y – 6 = 0 h 3x – 2y + 17 = 0 i 8x – 2y – 7 = 0

3

Find b if each point below lies on the line with the given equation. a (2, b) x + 2y = –4 b (–1, b) 3x – 4y = 6 c (b, 4) 5x + 2y = 1 d (b, –3) 4x – y = 8

Example 3 Find the x- and y-intercepts of the line with equation 4x – 3y – 12 = 0. x-intercept when y = 0

y-intercept when x = 0

4x – 3(0) – 12 = 0

4(0) – 3y – 12 = 0

4x – 12 = 0

–3y – 12 = 0

4x = 12

–3y = 12

x =3 i.e. x-intercept is 3 4

y = –4 i.e. y-intercept is –4

Find the x- and y-intercepts of the following lines. a x + 2y – 8 = 0 b 3x + 2y – 12 = 0 d 3x – 2y – 36 = 0 e 5x + 2y + 20 = 0 g 4x – y – 5 = 0 h 3x – 2y + 15 = 0

c 2x – y + 6 = 0 f 3x + 2y – 5 = 0 i 9x – 2y – 5 = 0

LEY_bk9_c14_finalpp Page 449 Wednesday, January 12, 2005 11:47 AM

Straight Lines and Regions (Chapter 14) Syllabus reference PAS5.3.3

Example 4 Draw the graph of the line with equation 5x – 3y – 15 = 0. Find the x-and y-intercepts If x = 0

5(0) – 3y – 15 = 0

If y = 0

5x – 3(0) – 15 = 0

–3y – 15 = 0

5x – 15 = 0

–3y = 15

5x = 15

y = –5 ∴ y-intercept is –5

x=3 ∴ x-intercept is 3

Use a third point as a check. Try x = 1.

y

5(1) – 3y – 15 = 0

3

5 – 3y – 15 = 0

2 1

–3y – 10 = 0 –3y = 10 10 y = – -----3 10 ∴ (1, – ------ ) is on the line. 3

–4 –3 –2 –1

x –1

1

2

3 4

–2 –3 –4 –5

5

Draw the graph of the line with equation a x + 2y – 8 = 0 b 3x – y – 6 = 0 d 4x + 3y – 8 = 0 e x+y–5=0 g 3x – 4y – 12 = 0 h 5x + 2y + 10 = 0

c 2x – 3y – 4 = 0 f x–y+5=0 i x – 2y = 0

Investigation 1 WM: Reasoning, Communicating, Applying Strategies

Graphs of lines 1

a On the same number plane draw the straight lines with equations y = 2x, y = 2x + 1, y = 2x – 3, y = 3x + 1 b What do you notice about these lines?

☞

449

LEY_bk9_c14_finalpp Page 450 Wednesday, January 12, 2005 11:47 AM

450

Straight Lines and Regions (Chapter 14) Syllabus reference PAS5.3.3

2

a On the same number plane draw the straight lines with equations y = –3x + 1, 3x + y – 2 = 0, 6x + 2y + 3 = 0. b What do you notice about these three lines? c Rewrite the second two lines in y = mx + b form and find their gradients.

3

Copy and complete: Straight lines are _____ if their _____ are equal.

4

a On the same number plane draw the straight lines with equations 1 y = 2x + 1 and y = – --- x + 2. 2 b What do you notice about these lines?

5

a On the same number plane draw the straight lines with equations y = 3x – 1 and x + 3y – 6 = 0. b What do you notice about these lines? c Rewrite x + 3y – 6 = 0 in y = mx + b form and find the gradient.

6

Copy and complete: Straight lines are ____ if the product of their gradient is ____.

D. PARALLEL AND PERPENDICULAR LINES From Investigation 1: If two straight lines have gradients m1 and m2 then: •

the lines are parallel if m1 = m2

•

the lines are perpendicular if m1 × m2 = –1.

Example 1 a Find the equation of the line parallel to the line y = –5x – 7 passing through the point (2, 3). b Find the equation of the line parallel to the line 3x – 6y + 8 = 0 passing through the point (–1, –2).

Use a graphics calculator.

LEY_bk9_c14_finalpp Page 451 Wednesday, January 12, 2005 11:47 AM

Straight Lines and Regions (Chapter 14) Syllabus reference PAS5.3.3

a The gradient is –5, since the lines are parallel. i.e. equation is

y ∴3 ∴3 ∴ 13 y

= = = = =

– 5x + b – 5 ( 2 ) + b (substituting (2, 3)) – 10 + b b – 5x + 13 is the equation of the line.

b 3x – 6y + 8 = 0 must be rearranged into y = mx + b form to find the gradient. 3x – 6y + 8 = 0 ∴ 3x + 8 = 6y 3x 8 ∴ ------ + --- = y 6 6 1 4 ∴ y = --- x + --2 3

1 So gradient = --- . 2 1 Use y – y1 = m(x – x1) with m = --- and (–1, –2) for (x1, y1) 2 1 ∴ y – (–2) = --- (x –(–1)) 2 1 1 1 ∴ y + 2 = --- (x + 1) = --- x + --2 2 2 1 1 y = --- x – 1 --2- is the equation of the line. 2

Exercise 14D 1

2

Which of the following pairs of lines are parallel? a y = 3x + 1, y = 3x – 5 b y = 2x – 1, c y = 5x + 3, y = 3x + 5 d y = 4x – 3, e y = 2x – 5, 2x – y + 4 = 0 f y = –x – 5, g 4x – 3y + 5 = 0, 3x + 4y + 2 = 0 h 2x + 3y – 2 = 0,

y = 2x y = 4 – 3x x – 2y + 3 = 0 2x + 3y – 5 = 0

Find the equation of the line: a parallel to the line y = 2x – 5 passing through the point (1, 4) b parallel to the line y = –7x – 2 passing through the point (–5, –2) 1 c passing through the origin parallel to y = – --- x + 5 2 d parallel to the line 5x – 7y + 3 = 0 passing through the point (2, –3) e passing through (–1, –3), parallel to the line passing through the points (1, 5) and (3, 6) f

with y-intercept –2, that is parallel to the line segment joining (–7, 5) and the origin

451

LEY_bk9_c14_finalpp Page 452 Wednesday, January 12, 2005 11:47 AM

452

Straight Lines and Regions (Chapter 14) Syllabus reference PAS5.3.3

Example 2 a Is the line y = 3x – 5 perpendicular to the line 2x + 6y + 9 = 0? b What is the gradient of the line perpendicular to the line 5x – 2y + 4 = 0? c Find the equation of the line passing through (6, –3) perpendicular to the line 2 y = – --- x + 4. 3 a y = –3x – 5 has gradient m1 = 3 We rearrange 2x + 6y + 9 = 0 to find its gradient. 6y = – 2x – 9 – 2x 9 ∴ y = --------- – --6 6 1 1 3 ∴ y = – --- x – ---, and so had gradient m2 = – --3 3 2 1 Now m1 × m2 = 3 × – --- = –1 ∴ the lines are perpendicular. 3 b Rearrange 5x – 2y + 4 = 0 into y = mx + b form. 5x – 2y + 4 = 0 ∴ 5x + 4 = 2y 5x 4 ∴ ------ + --- = y 2 2 5 y = --- x + 2 2 2 ∴ the gradient of the perpendicular line is – --5 2 2 c y = – --- x + 4 has gradient m1 = – --3 3 3 ∴ the gradient of the perpendicular line is --2 3 ∴ y = --- x + b and so we substitute (6, –3) 2 3 ∴ –3 = --- (6) + b 2

5 2 (since --- × – --- = –1) 2 5 2 3 (since – --- × --- = –1) 3 2

∴ –3 = 9 + b –12 = b 3 ∴ y = --- x – 12 is the equation. 2 3

Which of the following pairs of lines are perpendicular? a y = 2x – 5, y = 2x b y = –3x + 5, c y = 4x + 7,

y = –4x + 3

e 3x – 4y + 5 = 0, 4x + 3y – 2 = 0

1 y = --- x – 2 3 2 5 d y = --- x – 1, y = – --- x – 1 5 2 f 7x + 5y + 3 = 0, 5y – 7x = 0

LEY_bk9_c14_finalpp Page 453 Wednesday, January 12, 2005 11:47 AM

Straight Lines and Regions (Chapter 14) Syllabus reference PAS5.3.3

4

5

Find the gradients of the lines perpendicular to the given lines. 3 a y = – --- x + 2 b y = 7x – 2 5 4 d y = – --- x – 7 e 3x – 2y + 1 = 0 5

3 c y = --- x – 5 2 f

5x – 7y + 7 = 0

Find the equation of the line: a perpendicular to y = 5x – 2 passing through (–3, 2) 1 b perpendicular to y = – --- x + 7 passing through (0, 5) 4 3 1 c passing through the origin perpendicular to y = --- x + --4 3 d passing through (–2, 5) perpendicular to 3x – 4y + 12 = 0 e with y-intercept 7 perpendicular to 5x + 2y – 7 = 0 f

passing through (–2, –5), perpendicular to the line segment joining (2, 3) and (5, –3)

g passing through the origin, perpendicular to the line segment joining (3, 0) and (0, –5) 6

1 a Find the equations of five lines that are perpendicular to y = --- x – 3. 2 b Find the equations of five lines that are parallel to y = 3x – 2. c Write the equation of all lines parallel to 3x – 5y + 6 = 0. d Write the equation of all lines perpendicular to 5x – 3y + 7 = 0.

7

a Find the equations of three lines that are perpendicular to 3x – 4y + 8 = 0. b Write the equations of these lines in general form. c What do you notice?

8

a Find the equations of three lines that are parallel to 2x – 7y + 3 = 0. b Write the equations of these lines in general form. c What do you notice?

9

Write the equation of all lines: a parallel to 7x – 5y + 6 = 0

b perpendicular to 7x – 5y + 6 = 0

Investigation 2 WM: Communicating, Reasoning

Transformation of lines 1

a On the same number plane, sketch the graphs of y = 3x and y = 3x + 1. b Which transformation, rotation, reflection or translation, would be used to superimpose the graph of y = 3x over that of y = 3x + 1? c Describe a transformation that would superimpose the graph of y = 3x over the graph of y = 3x – 5.

453

LEY_bk9_c14_finalpp Page 454 Wednesday, January 12, 2005 11:47 AM

454

Straight Lines and Regions (Chapter 14) Syllabus reference PAS5.3.3

2

3

a On the same number plane, sketch the graphs of y = 2x and y = –2x. b Describe a transformation that would superimpose the graph of y = 2x on the graph of y = –2x. 1 a On the same number plane, sketch the graphs of y = 3x and y = – --- x. 3 b Describe a transformation that would superimpose the graph of y = 3x over the graph of 1 y = – --- x. 3

4

Describe a transformation that would superimpose the graph of y = 4x over the graph: 1 a y = –4x b y = 4x – 7 c y = – --- x 4

5

Describe the transformations needed to map: 1 1 a y = 3x to y = –3x to y = 5 – 3x b y = –4x to y = --- x to y = --- x – 2 4 4 1 1 1 c y = --- x to y = – --- x to y = 5 – --- x d y = 3x – 5 to y = 3x to y = –3x 2 2 2 Use a graphics calculator here.

E. FURTHER COORDINATE GEOMETRY This section uses some coordinate geometry aspects from Chapter 7 in conjunction with aspects of this chapter. Remember x1 + x2 y1 + y2 Midpoint formula:  ---------------, ---------------   2 2  Distance formula: d =

2

( x2 – x1 ) + ( y2 – y1 )

2

Exercise 14E 1

Line segments AB and CD bisect each other at T. Find the coordinates of C.

A(5, 3)

C T

B(–3, –1)

D(2, 0)

LEY_bk9_c14_finalpp Page 455 Wednesday, January 12, 2005 11:47 AM

Straight Lines and Regions (Chapter 14) Syllabus reference PAS5.3.3

2

A(3, 2)

D

B(8, 4)

ABCD is a parallelogram. Use the fact that the diagonals of the parallelogram bisect each other to find the coordinates of D.

C(6, –1)

3

Triangle ABC has A(–1, 4), B(2, –1) and C(5, 2) as vertices. Find the length of the line segment from A to the midpoint of BC.

4

For the points P(2, 3), Q(0, 0), R(7, 4), S(a, –1), find a if: a PQ is parallel to RS b PR is parallel to QS c PQ is perpendicular to RS d PR is perpendicular to QS

5

The triangle ABC has vertices A(–2, 0), B(2, 1) and C(1, –3). a Find the length of the sides AB, BC and AC using the distance formula. b Classify ∆ABC as scalene, isosceles or equilateral. Give a reason.

6

The triangle PQR has vertices P(1, 0), Q(3, 1) and R(7, 3). a Find the lengths of each of the sides. b Classify ∆PQR as scalene, isosceles or equilateral. Give a reason.

7

Classify ∆LMN with vertices L(–2, –1), M(0, 3) and N(4, 1) as scalene, isosceles or equilateral. Give a reason.

8

The triangle XYZ has vertices X(1, 2), Y(2, 5) and Z(4, 1). a Find the length of each of the sides XY, YZ and XZ. b Use Pythagoras’ rule to decide if ∆XYZ is right angled. Give a reason. c Find the gradient of the sides XY, YZ and XZ. d Use the gradients to decide if ∆XYZ is right angled. Give a reason.

9 10

Is the triangle with the vertices D(–2, –1), E(1, –1) and F(–2, 3) right angled? Give a reason. The triangle PQR has vertices P(–2, 5), Q(3, –1) and R(–4, –7). a Find the coordinates of S, the midpoint of PQ. b Find the coordinates of T, the midpoint of PR. c Show that the length of QR is twice the length of ST. d Show that ST is parallel to QR.

11

The quadrilateral PQRS has vertices P(2, 4), Q(5, 1), R(–1, –2) and S(–4, 1). a Prove that PQRS is a parallelogram by showing that: i the opposite sides are equal ii the opposite sides are parallel iii the diagonals bisect each other iv one pair of opposite sides are equal and parallel.

☞

455

LEY_bk9_c14_finalpp Page 456 Wednesday, January 12, 2005 11:47 AM

456

Straight Lines and Regions (Chapter 14) Syllabus reference PAS5.3.3

b Is PQRS a rectangle? Explain. c Is PQRS a rhombus? Explain. 12

ABC is a triangle with vertices A(–4, 4), B(3, 2) and C(–2, –3). a Show that ∆ABC is isosceles with AC = AB. b Find the midpoint D of the base BC. c Find the length of AD. d Prove that AD is perpendicular to BC.

13

Triangle XYZ has vertices X(–1, 3), Y(–4, 1) and Z(3, –3). a Show that ∆XYZ is right angled. Do this in two ways. b Find the equation of the perpendicular bisector of XY. c Find the equation of the side YZ. d Show that the point of intersection of the perpendicular bisector of XY and the side YZ is (– 1--2- , –1). e Show that the perpendicular bisector of XZ also passes through the point (– 1--2- , –1). f

14

Explain why (– 1--2- , –1) is the centre of a circle with X, Y and Z on its circumference. What names are given to this point and the circle?

The triangle STU has vertices S(–2, 5), T(2, 3) and U(–4, –1). a Find the equation of the perpendicular bisectors of each side. b Hence, find the circumcentre of ∆STU.

15

The quadrilateral WXYZ has vertices W(–2, –3), X(–4, 4), Y(3, 2) and Z(2, –4). The points A,B, C, D are the midpoints of each of the sides WX, XY, YZ and ZW. a Find the coordinates of A, B, C and D. b What type of quadrilateral is ABCD?

16

Find the area of the triangle enclosed by the lines y = 0, y = 2x and x + y = 6.

F. INEQUALITY GRAPHS USING LINES PARALLEL TO THE AXES y

We know that the solution to the inequality x ≥ 1 can be drawn on the number line as shown.

4 3

x=1

2 –3 –2 –1

0

1 2

3 4

The graph of x = 1 on the number plane is a line parallel to the y-axis passing through x = 1 as shown.

1 –4 –3 –2 –1

x –1 –2 –3 –4

1 2

3 4

5

LEY_bk9_c14_finalpp Page 457 Wednesday, January 12, 2005 11:47 AM

457

Straight Lines and Regions (Chapter 14) Syllabus reference PAS5.3.3

Combining these two graphs gives the graph of x ≥ 1 on the number plane. It includes all x-values to the right of x = 1.

4

This is shown by shading the half-plane to the right of the line x = 1.

3

y

2 1 This line is the boundary of the region.

–4 –3 –2 2 –1

x –1

1 2

3 4

5

–2 –3 –4

The graph of the inequality x ≥ 1 is shown opposite.

y 4

The dotted line indicates that x = 1 is not included, which is similar to the open circle on number lines.

3 2 1 –4 –3 –2 2 –1

x –1

1 2

3 4 5

–2 –3 –4

Example 1 Sketch the regions defined by: a x≥2

b

y<3

☞

LEY_bk9_c14_finalpp Page 458 Wednesday, January 12, 2005 11:47 AM

458

Straight Lines and Regions (Chapter 14) Syllabus reference PAS5.3.3

a x≥2

b

y<3

y

y

4

4

3

3

2

2 1

1 –4 –3 –2 2 –1

x

x –1

1

2

3 4

–4 4 –3 –2 2 –1 1

5

–1

–2 –3

–2

–4

–4 4

1

2

3 4

–3

Line is parallel to the y-axis. The line is solid and included in the shaded area

Line is parallel to the x-axis. Since y < 3 the shading is below the line which is not solid.

Exercise 14F 1

Sketch the following regions. a x≥2 d y≤4 g y > –5 1 j x < –1 --4

b e h k

x ≤ –1 x>3 y<7 3 y ≥ --2

c f i l

y ≥ –2 x<1 x ≥ –4 y<1

Example 2 Sketch the region defined by the intersection of: a x < 3 and y ≥ 1 b x ≥ 1 and y ≤ 2 y

a Sketch both graphs on the same number plane.

4 3 y≥1 –4 –3 –2 2 –1

The solution is the region that is shaded by both graphs as shown opposite.

2 1

x –1 –2 –3 –4

1

2

3 4

x<3

LEY_bk9_c14_finalpp Page 459 Wednesday, January 12, 2005 11:47 AM

Straight Lines and Regions (Chapter 14) Syllabus reference PAS5.3.3

Note: If one or more graphs are dotted lines then the point of intersection is not included.

4

y

3 An open circle shows that the point (3, 1) is not included.

2

y≥1

1

–4 –3 –2 2 –1

x –1

1

–2 –3

2

3 4

x<3

–4

b Sketch both graphs on the same number plane. 4

y x≥1

3 y≤2

2 1

–4 –3 –2 2 –1

x –1

1

2

3 4

–2 –3 –4

The solution is the region that is shaded by both graphs.

4

y x≥1

3

Note: The point (1, 2) is included because both graphs are solid lines.

y≤2 –4 –3 –2 2 –1

2 1

x –1

1

2

3 4

–2 –3 –4

2

3

Sketch the regions defined by the intersection of these regions. a x ≥ 2 and y ≤ 2 b x > 1 and y ≥ 3 d x > –3 and y < –1 e x ≥ 4 and y ≤ –5 Shade the region defined by the intersection of these regions. a x ≥ –2 and x < 4 b y > –3 and y ≤ 2 c x ≥ –4 and x < 1 d y ≥ 0 and y < 4

c x < –1 and y > 3 f x ≤ 2 and y > 4

459

LEY_bk9_c14_finalpp Page 460 Wednesday, January 12, 2005 11:47 AM

460

Straight Lines and Regions (Chapter 14) Syllabus reference PAS5.3.3

4

a Shade the region defined by x ≥ –3, y ≤ 2, x ≤ 1 and y ≥ –2. b Describe the region.

5

Write inequalities that would result in a region that is a rectangle 4 × 6 units on the number plane.

G. REGIONS IN THE NUMBER PLANE In the previous section all the lines were parallel to one of the axes. In this section regions involving any straight line are considered.

Example 1 Sketch the region defined by: a x+y≤3

b 2x – y > 6

a First, sketch the graph of the line x + y = 3. Here the x intercept is 3 and the y intercept is 3. 4

y

Draw an unbroken line because the inequality contains an equals sign.

3 2 1 –4 –3 –2 –1

x –1

1

2

3 4

–2 –3 –4

Second, decide which side of the line to shade. In the previous section this was easy, but here it may not be as clear. To decide where to shade, choose a point on one side of the line and test to see if it satisfies the inequality.

4

y

3

To decide where to shade, choose a point on one side of the line and test to see if it satisfies the inequality.

x

2

+ y ≤

1

3

For example, test (0, 0). x+y ≤3 0+0≤3 0 ≤ 3 is true. ∴ shade on the side of the line containing (0, 0).

–4 –3 –2 –1

–1 –2 –3 –4

1

2

3 4

x

LEY_bk9_c14_finalpp Page 461 Wednesday, January 12, 2005 11:47 AM

Straight Lines and Regions (Chapter 14) Syllabus reference PAS5.3.3

Always test (0, 0) when possible as it is easy to substitute.

b Step 1: 2x – y = 6 Sketch 2x – y = 6

Step 2: Test (0, 0) 2x – y > 6

x intercept when y = 0 is x = 3

2(0) – 0 > 6

y intercept when x = 0 is y = –6

0 > 6 is false ∴ shade the opposite side of the line to (0, 0).

y 4 3

x 2

>6

2

3 4 5

–2 –3

2x

–1

1

–y

3

1 –5 –4 –3 –2 –1

y

4

2

1 –5 –4 –3 –2 –1

x –1

–5

–2 –3

–6

–4

–4

1

2

3 4 5

–5

The line is dotted as the inequality does not contain an equals sign.

–6

Exercise 14G 1

2

Sketch these regions on the number plane. a x+y≤2 b x+y≥4 e x–y≥5 f x – y ≤ –3

c x+y>3 g y–x<3

d x + y < –2 h y–x>2

Sketch these regions on the number plane. a 2x – y ≤ 4 b 2x – 3y > 6 e y – 2x < –4 f 4x – y < 3

c 4x – y > –4 g 6x – 5y ≥ 15

d 3x + 5y ≤ 15 h 3x + 5y ≥ 15

461

LEY_bk9_c14_finalpp Page 462 Wednesday, January 12, 2005 11:47 AM

462

Straight Lines and Regions (Chapter 14) Syllabus reference PAS5.3.3

Example 2 Sketch the region defined by: b y ≤ 4x

a y > 3x a y > 3x

Step 1: Sketch y = 3x using a dotted line y

4

4

y = 3x

3

y = 3x

3 2

2

1

1 –4 –3 –2 –1

y

x

x –1

1

2

–4 –3 –2 –1

3 4

–1

–2 –3

–2 –3

–4

–4

1

2

3 4

Step 2: Choose a point on one side of the line, say (1, 0). Any point may be used.

(0, 0) cannot be used as the line passes through the origin.

Test (1, 0) y > 3x 0 > 3(1) 0 > 3 is false

∴ shade the opposite side of the line to (1, 0) b y ≤ 4x Step 1: Sketch y = 4x using an unbroken line. 4

y y = 4x

3

Step 2: Use the point (0, 2) to test.

2 1 –4 –3 –2 –1

x –1 –2 –3 –4

1

2

3 4

y ≤ 4x

4

2 ≤ 4(0)

3

2 ≤ 0 is false

2

∴ shade the opposite side of the line to –4 –3 –2 –1 (0, 2).

y

1

x –1 –2 –3 –4

1

2

3 4

LEY_bk9_c14_finalpp Page 463 Wednesday, January 12, 2005 11:47 AM

Straight Lines and Regions (Chapter 14) Syllabus reference PAS5.3.3

3

Sketch these regions on the number plane. a y ≥ 2x b y < 3x e y > 5x f y > –x

d y≥x h y ≤ –3x

c y<x g y ≤ –2x

Example 3 Write an equation to describe the region. a 4

b

y

6 5 y 4

3 2

3

1 –4 –3 –2 –1

(2, 6)

x –1

1

2

2

3 4

1 x

–2 –3

–4 –3 –2 –1

–4

–1

1

2

3 4

–2 –3 –4 –5 –6

a Step 1: Find the equation of the line. rise m = --------y intercept is 2. run 2 = + --3 2 ∴ equation is y = --- x + 2 3 ∴ 3y = 2x + 6

b Step 1: Find the equation of the line rise m = --------- y intercept is 0. run 6 = --2 =3 ∴ equation is y = 3x

3y – 2x = 6 Step 2: Substitute a point in the region, not on the line, e.g. (0, 0). 3y – 2x = 6 Test 3(0) – 2(0) = 6 0 = 6 is false

Step 2: Substitute a point in the region, e.g. (1, 0). y = 3x Test 0 = 3(1) 0 = 3 is false

Use an inequality to make the statement true, e.g. use <.

Use an inequality to make the statement true, e.g. use <.

∴ inequality is 3y – 2x < 6

∴ inequality is y < 3x

463

LEY_bk9_c14_finalpp Page 464 Wednesday, January 12, 2005 11:47 AM

464

Straight Lines and Regions (Chapter 14) Syllabus reference PAS5.3.3

4

Write equations to describe these regions. a 4

b

y

4

3

3

2

2

1 –4 –3 –2 –1

1

x –1

1

2

y

–4 –3 –2 2 –1

3 4

x –1

–2 –3

–2 –3

–4

–4

c

1

2

3 4

d 4

y

4

3

3

2

2

1 –4 –3 –2 –1

1

x –1

1

2

y

3 4

–5 5 –4 –3 –2 2 –1

x –1

–2 –3

–2 –3

–4

–4

1

2

3 4

5

–5

e

f y (–2, 2 8)

8

4

7

3

6

2

5

1

4

–5 5 –4 –3 –2 2 –1

(4, 2) x

3

–1

2

–2 –3

1 –4 –3 –2 –1

y

x –1 –2 –3 –4

1

2

3 4

–4 –5

1

2

3 4

5

LEY_bk9_c14_finalpp Page 465 Wednesday, January 12, 2005 11:47 AM

Straight Lines and Regions (Chapter 14) Syllabus reference PAS5.3.3

non-calculator activities 1

Holly spends $14.25. What is her change from $50?

2

Evaluate 8 – 2.63.

3

Convert 5.23 cm to mm.

4

Peter buys a pair of jeans marked at $75.00. If he received 10% discount, how much does he pay for the jeans?

5

Red and blue lollies are sold in the ratio 2 : 5. If there are 16 red lollies in a bag, how many are blue?

6

Solve 5 – 2x = 9.

7

Explain the difference between gross income and taxable income.

8

Factorise x2 – 11x + 24.

9

Find d in this triangle. 5 cm

13 cm

d cm

10

Al works in a shoe store and notes that the mode female shoe size sold is 7 1--2- . Explain the term mode.

11

What is the probability of rolling a 6 with a normal six-sided die?

12

1 – 1--What is the value of -----------51- ? 1 + --5-

13

My car uses 15 L of petrol per 100 km. Find the number of litres of petrol needed to travel 350 km.

14

Between which two consecutive whole numbers is the square root of 67.3?

15

If 153 × 4785 = 732 105, find the value of 732.105 ÷ 478.5.

465

LEY_bk9_c14_finalpp Page 466 Wednesday, January 12, 2005 11:47 AM

466

Straight Lines and Regions (Chapter 14) Syllabus reference PAS5.3.3

Language in Mathematics

Pierre de Fermat

(1601–1665)

Pierre de Fermat was born in Beaumont-de-Lomagne in France, near the border with Spain. He studied Latin and Greek literature, ancient science, mathematics and modern languages at the University of Toulouse, but his main purpose was to study law. In 1629 Fermat studied the work of Appollonius, a geometer of ancient Greece, and discovered for himself that loci or sets of points could be studied using coordinates and algebra. His work Introduction to Loci was not published for another fifty years, and together with La Geometrie by Descartes, formed the basis of Cartesian geometry. In 1631 Fermat received his degree in law, was later awarded the status of a minor nobleman, and in 1648 became King’s Councillor, Fermat was a man of great integrity who worked hard. He remained aloof from matters outside his own jurisdiction, and pursued his great interest in mathematics. He worked with Pascal on the theory of probability and the principles of permutations and combinations. He worked on a variety of equations and curves and the Archimedean spiral. In 1657 he wrote ‘Concerning the Comparison of Curved and Straight Lines’ which was published during his lifetime. Fermat died in 1665. He was acknowledged master of mathematics in France at the time, but his fame would have been greater if he had published more of his work while he was alive. He became known as the founder of the modem theory of numbers. In mid-1993, one of the most famous unsolved problems in mathematics, Fermat’s Last Theorem was solved by Andrew Wiles of Princeton University (USA). Wiles made the final breakthrough after 350 years of searching by many famous mathematicians (both amateur and professional). Wiles is a former student and collaborator of Australian Mathematician John Coates. Fermat’s Last Theorem is a simple assertion which he wrote in the margin of a mathematics book, but which he never proved, although he claimed he could. The theorem is: The equation x n + y n = z n, when the exponent n is greater than 2, has no solutions in positive integers. Wiles’ work establishes a whole new mathematical theory, proposed and developed over the last 60 years by the finest mathematical minds of the 20th century. 1

Answer these questions based on the biography of Pierre de Fermat. a How old was Fermat when he studied Appollonius?

LEY_bk9_c14_finalpp Page 467 Wednesday, January 12, 2005 11:47 AM

Straight Lines and Regions (Chapter 14) Syllabus reference PAS5.3.3

467

b When was his work Introduction to Loci published? What was it about? c Who solved Fermat’s last theorem? d Explain Fermat’s last theorem. Research this theorem and write a one-page report. 2

Write an explanation of each glossary term.

3

Use every third letter to find a sentence. DFTEDWVGOHULIOIKJNHGEFDSASAERRTGEBNPMIEKJRGTP QAEZSNWEDRFIDCCSHUKOLPLADURIOIDEFASTFVHBKENJPM HRKIOLODPOUTUCXYTTTORRFEETDDHFFEXFIVHRNPGOIRYT AEDDSVIHUEKANEDTASSCFIVGSFDNESEAEGAEAAETDRIFGV HNEMJOTGNAXEHUAJKNUPDZCTVBHMKELPYIOAUYRUIEIJPH GAIERETARELETLRTEASLCVIBHFNGTDSHFGEHNIMJRIKGUY RTRAEDDEDISWEADNASTCVSBNANNRMMESSERRQAAUSSAQWL

Glossary boundary gradient–intercept form linear relationship product translation

equation inequality parallel reflection y-intercept

CHECK YOUR SKILLS 1

2

3

4

✓

general form inequation perpendicular region

gradient line point sketch

The equation of the line passing through (–2, –6) with gradient 2 is: A y = 2x + 3 B y = 2x – 7 C y = 2x – 2

D y = 2x + 7

The equation of the line passing through A(–4, 5) and B(2, –13) is: A y = –x – 1 B y=x+5 C y = –3x – 7

D y = –x + 1

The equation of the line cutting the y axis at –4 and the x axis at +5 is: 4 4 5x A y = --- x – 4 B y = – --- x – 4 C y = ------ – 4 5 5 4

5x D y = – ------ – 4 4

The equation of this line is: A y = 3x – 3 B y = –3x – 3 C y = –x + 3 D y = 2x – 3

y 3 2 1 –3 –2 –1

x –1 –2 –3

1

2

3

LEY_bk9_c14_finalpp Page 468 Wednesday, January 12, 2005 11:47 AM

468

Straight Lines and Regions (Chapter 14) Syllabus reference PAS5.3.3

5

6

The value of a if (a, 1) lies on the line y = 3x – 2 is: A 0 B 1 C 2

D 3

1 In general form y = --- x – 2 A 6y – 3x + 8 = 0

D

4 --- is: 3 B 3x – 6y – 8 = 0

C 3x + 2y – 4 = 0

1 4 --- x – y – --- = 0 2 3

7

When written in gradient–intercept form, the equation 3x + 2y + 8 = 0 is: 3x 3x A 2y = 3x + 4 B 3x – 2y = –8 C y = – ------ + 8 D y = – ------ – 4 2 2

8

The x- and y-intercepts of 4x + 3y – 12 = 0 are: A 4 and –3 B 3 and 4

9

10

11

12

The line parallel to y = –3x + 1 is: 1 1 A y = --- x + 1 B y = – --- x + 1 3 3 The line perpendicular to 3x + 5y + 7 = 0 is: A 3x + 5y – 2 = 0 B 3x – 5y – 7 = 0

C –4 and 3

✓ D –3 and 4

C y = –3x + 3

D y = –x + 3

C 3x + 5y + 7 = 0

5x D y = ------ + 2 3

The triangle ABC with vertices A(2, 0), B(1, 4) and C(6, 3) is: A scalene B isosceles C equilateral The region defined by y ≤ 3 is: A 4

B

y

4

3

3

2

2

1 –4 –3 3 –2 2 –1 1

1

x –1

1

2

–4 –3 3 –2 2 –1 1

3 4

x

1

2

1

2

–1 –2 –3

–2 –3

C

D 4

y

4

3

3

2

2

1 –4 –3 3 –2 2 –1 1

y

–1 –2 –3

2

3 4

y

3 4

1

x 1

D right angled

–4 –3 3 –2 2 –1 1

x –1 –2 –3

3 4

LEY_bk9_c14_finalpp Page 469 Wednesday, January 12, 2005 11:47 AM

Straight Lines and Regions (Chapter 14) Syllabus reference PAS5.3.3

13

This region is defined by: A x ≤ –3 and y ≤ 2 B x > –3 and y ≤ 2 C x < –3 and y ≤ 3 D x ≥ –3 and y ≥ 2

4

469

y

3 2 1 –4 –3 3 –2 2 –1 1

x –1

1

2

3 4

–2 –3

14

The region defined by 3x – y ≤ 6 is: A

B

3

2

2

1 –4 –3 –2 –1 1

1

x –1

1

2

3 4

–4 –3 –2 –1 1

–2 –3 3

x

–1

1

2

3 4

1

2

3 4

–2 –3 3

–4

–4

–5

–5

–6

–6

–7

–7

C

D y

y

3

3

2

2

1 –4 –3 –2 –1 1

✓ y

y 3

1

x –1

1

2

3 4

–4 –3 –2 –1 1

–2 –3 3

x

–1 –2 –3 3

–4

–4

–5

–5

–6

–6

–7

–7 –8

If you have any difficulty with these questions, refer to the examples and questions in the sections listed in the table. Question

1–5

6–8

9, 10

11

12,13

14

Section

A and B

C

D

E

F

G

LEY_bk9_c14_finalpp Page 470 Wednesday, January 12, 2005 11:47 AM

470

Straight Lines and Regions (Chapter 14) Syllabus reference PAS5.3.3

REVIEW SET 14A 1

For the points A(3, –1) and B(–5, 0) find: a the gradient of AB b the equation of the line through A and B c the x- and y-intercepts of the line AB d the equation of the line perpendicular to AB passing through A e the equation of the line parallel to AB passing through the point (3, 0)

2

a Write 3y = 4x – 7 in general form. b Find the equation of the line passing through (–4, 7) and (2, 6) in general form. c Sketch the line 3x – 4y + 12 = 0.

3

a Find the fourth vertex of the parallelogram ABCD for A(–7, 11), B(6, 5) and C(3, 8). b Find the equation of the line from A(7, 2) to the midpoint of the line segment joining B(–6, 4) to C(3, –1).

4

a Find k if 2x + ky = 5 is perpendicular to x – 3y = 11. b A(–1, 3), B(2, 4) and C(t, –1) are collinear. Find t.

5

Write the equation of each inequality. a b 4

y

y

4

3

3

2

2 1

1 –4 –3 –2 –1

x

x –1

1

2

3 4

–6 6 –5 5 –4 –3 –2 –1

–1

–2 –3

–2 –3 3

–4

–4

1

2

3 4

LEY_bk9_c14_finalpp Page 471 Wednesday, January 12, 2005 11:47 AM

Straight Lines and Regions (Chapter 14) Syllabus reference PAS5.3.3

REVIEW SET 14B 1

Find the equation of the line passing through M(–4, 2) and N(3, 7).

2

Find the equation of the line passing through (–4, 5) perpendicular to 3x – 2y + 7 = 0.

3

2 Write y = – --- x – 4 in general form. 3

4

Find the equation of the perpendicular bisector of the join of (–2, 3) and (4, 5).

5

Sketch these regions: a y < –1

b

x–y>6

c 2x – 3y ≤ 12.

REVIEW SET 14C 1

For the points P(–6, 3) and Q(–1, –5) find: a the gradient of PQ b the equation of the line through P and Q c the x- and y-intercepts of the line PQ d the equation of the line perpendicular to PQ passing through Q e the equation of the line parallel to PQ passing through the origin

2

a Write 4x – 7y + 8 = 0 in gradient–intercept form. b Find the equation of the line passing through (4, –2) and the origin in general form. c Sketch the line 6x – 7y + 9 = 0.

3

a Determine z if 2x + 5y = 11 and zx – 3y = 9 are perpendicular. b A(–2, 4), B(3, 7) and D(–1, d) lie on the same straight line. Find the value of d. c Find the coordinates of D for parallelogram ABCD if A(–2, 3), B(1, 7) and C(5, 1) are vertices.

4

Sketch x – 3y = 7.

5

Sketch these regions: a x ≤ –1

b 2x – y > 4

471

LEY_bk9_c14_finalpp Page 472 Wednesday, January 12, 2005 11:47 AM

472

Straight Lines and Regions (Chapter 14) Syllabus reference PAS5.3.3

REVIEW SET 14D 1

a Find the equation of the line with gradient 3 passing through (–2, 4). b Find the equation of the line parallel to 3x – 5y = 8 passing through the origin.

2

3x Write y = – ------ + 2 in general form. 4

3

Find the area of the triangle formed when 2x – 3y = 6 cuts the coordinate axes.

4

Does (1, 4) lie on the line 3x – 5y + 2 = 0? Explain.

5

Sketch: a the intersection of y ≥ –2 and x < –1 b 3x – y > 6 c y ≤ –4x