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ight

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es

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Straight Lines

SERIES TOPIC

K 5

What do the values of m and c in the equation y mx c= + represent?

Why can the gradient of a straight line be found from interceptintercept

mxy--

=

How are the gradients of parallel and perpendicular lines related?

STRAIGHT LINES

Try to answer these questions before working through this unit.

Answer these questions, after working through the chapter.

Every straight line is related to a linear equation y mx c= + and vice versa. This relationship allows us to find the equation from a line, or a line from an equation. There are also interesting properties of parallel or perpendicular lines.

But now I think:

What do I know now that I didn’t know before?

I used to think:

What do the values of m and c in the equation y mx c= + represent?

Why can the gradient of a straight line be found from interceptintercept

mxy--

= ?

How are the gradients of parallel and perpendicular lines related?

2 100% Straight Lines


Straight Lines

SERIES TOPIC

K 5

Basics

Every straight line has a gradient, which measures how steep the line is. The greater the gradient, the steeper the line. The symbol of gradient is m. The line leans different ways depending on whether the gradient (or 'slope') is positive or negative.

Remembering Gradient

Negative gradient (m is negative)

Negative gradient (m is largely negative)

Negative gradient ( 0)m 1

i is called the angle of inclination. It is the angle between the line and positive x-axis.

Method 2

Find the gradient from the angle made by the line.

Gradient is defined as:

Positive gradient (m is positive)

Positive gradient (m is largely positive)

Positive gradient ( )m 02

horizontal runvertical risem =

y

xadjacent

opposite

The gradient of any line can be found in two ways:

Method 1

Pick any two points ,x y1 1^ h and ,x y2 2^ h.

,x y1 1^ h

y

x x2 1-

,x y2 2^ h

x

yy

21

-

horizontal runvertical risem

mx xy y

2 1

2 1`

=

=--

horizontal runvertical rise

adjacentopposite

tan

m

m

m

`

` i

=

=

=

i



Straight Lines

SERIES TOPIC

K 5

Basics

Every straight line corresponds to an algebraic equation. This can be written like:

What is the gradient and y-intercept of the straight line corresponding to 2 3y x=- + ?

Write the equation of this line in gradient-intercept form:

The equation 2 3y x=- + is in gradient-intercept form.

The gradient is 2m =-The y-intercept is c 3=

"Gradient-Intercept" form depends on the gradient and the y-intercept. y is always the subject of the equation.

Gradient-Intercept Form

General Form of a Line

If all the terms are moved to the left side of the equation and the coefficient of x is NOT negative, then the equation is in general form (or standard form). Also, all the coefficients must be integers.

gradient-intercept form: Make y the subject general form: Move all the terms to one side

y mx c= +

Gradient

y-intercept

y-intercept

Must be integers

Not negative

gradient

Answer these questions involving gradient-intercept form

Write y x4 7 8= - + in a gradient-intercept form and b general form.

a

a b

b

Since 45ci = the gradient is

The y-intercept is 3c =-

tan

tan

m

45

1

i=

=

=

c

The equation of the line is y mx c

y x 3

= +

= -

y x47 2=- + 7 4 8 0x y+ - =

ax by c 0+ + =

y

-2 -1 1 2 3 4 5 6 7 8

321

-1-2-3-4-5-6

45cx



Straight Lines

SERIES TOPIC

K 5

BasicsQuestions

1. Match the gradients to the graphs below them:

2. Find the gradient of these lines:

y

y

y

y

y

x

x

x

x

x

a m 2=

y

x

b 1m =-

y

x

c 5m =-

y

x

d m 4=

a

c

b

d

-5 -4 -3 -2 -1 0 1

-5 -4 -3 -2 -1 0 1-1 0 1 2 3 4 5

5

4

3

2

1

-1

-2

-3

-4

5

4

3

2

1

-1

-2

-3

-4

5

4

3

2

1

-1

-2

-3

-4

-1 0 1 2 3 4 5

5

4

3

2

1

-1

-2

-3

-4

3,3^ h

,2 1^ h,2 0-^ h

,0 4-^ h,0 3^ h

,0 3-^ h

4,3-^ h

1, 2-^ h



Straight Lines

SERIES TOPIC

K 5

Questions Basics

3. Find the gradient of a line which:

4. What angle does the line make with the positive x-axis if it has a gradient of 3? (Answer to nearest degree)

5. If a line has a gradient of 3m = and a y-intercept of c 1= - , write the corresponding equation in:

6. If a line has a y-intercept of c 5= and gradient m 6= - , write the corresponding equation in:

Goes through the points and(1,3) ( 1, 1)- - .

Goes through the points and( 2,8) (3, 7)- - .

Gradient-intercept form.

Gradient-intercept form.

General form.

General form.

a

c

a

a

b

b

Has angle of inclination 0i = c.

An angle of inclination 63.434i = c (nearest unit).

b

d



Straight Lines

SERIES TOPIC

K 5

BasicsQuestions

7. Points A, B and C are collinear (on the same line) if the gradient of AB is equal to the gradient of BC. Are the following points collinear?

a

b

, and( 1, 5) (2,7) ( 3, 13) .A B C- - - -

and( 2,4), (0,3) (3, 1) .A B C- -



Straight Lines

SERIES TOPIC

K 5

Knowing More

Find the intercepts of these lines:

y

x-5 -4 -3 -2 -1 1 2 3 4 5

Intercepts are the points where a line cuts the axes. Here are some examples:

The intercepts of a line are used to graph a line from an equation. The intercepts can be found from the equation like this:

The x-intercept is 4

The y-intercept is 8-

The x-intercept is 3-

The y-intercept is 9

As a shortcut, the y-intercept of a line in gradient-intercept form, y mx c= + , is the value of c.

This page shows how to find the intercepts from an equation. The next page shows how to find the equation from intercepts.

Intercepts

Find the intercepts of a straight line with these equations:

a

a b

b

y

x

To find the x-intercept, make y 0= and solve for x

To find the x-intercept, make y 0= and solve for x

Rewrite in gradient-intercept form

To find the y-intercept, make 0x = and solve for y

To find the y-intercept, make 0x = and solve for y

The x-intercept is 2-The y-intercept is 3

The x-intercept is 4The y-intercept is 2-

y x2 8= - 3 9 0x y

y x3 9

- + =

= +

0 2 8

2 8

4

x

x

x`

= -

=

=

y 2 0 8

8

= -

=-

^ h y 3 0 9

9

= +

=

^ h

x

x

x

0 3 9

3 9

3`

= +

=-

=-

x-intercepty-intercept

x-intercept

y-intercept

-5 -4 -3 -2 -1 1 2 3 4 5

5

4

3

2

1

-1

-2

-3

-4

5

4

3

2

1

-1

-2

-3

-4



Straight Lines

SERIES TOPIC

K 5

Knowing More

The equation of a line y mx c= + simply required values for m and c. These values can be found from the intercepts.

This can be converted to general form if necessary:

The intercepts can be used to find the equation of a line. Here is an example:

Finding an Equation from Intercepts

-intercept-intercept

-intercept

mx

y

c y

=-

=

Find the equation of a line with these intercepts

Find the equations of the following line in gradient-intercept form:

a

a

b

b

x-intercept 3- , y-intercept 12 x-intercept 2, y-intercept 5


mxy

25

25=- =- =-` j

-interceptc y 5= =

5y x25` =- +


mxy

312 4=- =--

=` j

-interceptc y 12= =

4 12y x` = +

x y4 12 0- + = x y

x y

x y

25 5 0

25 5 0

5 2 10 0

`

`

- - + =

+ - =

+ - =

Remember, in general form the coefficient of x must be positive and all the coefficients must be integers

• Find the intercepts from the graph.

The x-intercept is 4. The y-intercept is 2-

• Use the intercepts to find the gradient.

• Use the intercepts to find the value for c.

c = y-intercept = 2-

the equation of the line is 2y x21` = - .

y

x

x-intercept

y-intercept

-5 -4 -3 -2 -1 0 1 2 3 4 5

5

4

3

2

1

-1

-2

-3

-4


mxy

42

21=- =- - =` j



Straight Lines

SERIES TOPIC

K 5

Knowing More

The intercepts can also be used to draw the line from an equation. This is done in two steps.

If the line is in general form, then convert it to gradient-intercept form first.

Convert the equation to gradient-intercept form:

Step 2: Draw a line through the intercepts.

Step 2: Draw a line through the intercepts.Step 1: Find the intercepts.



Step 1: Find the intercepts.

The x-intercept is 2-

The y-intercept is 6-

Using an Equation to Draw a Line

Draw the straight line corresponding to the equation y x3 6= - -

Draw the straight line corresponding to the equation 4 2 8 0x y+ - =

To find the x-intercept, make y = 0 and solve for x

To find the y-intercept, make x = 0 and solve for y

x

x

x

0 3 6

3 6

2`

=- -

=-

=-

y 3 0 6

6

=- -

=-

^ h

4 2 8 0x y

y x2 4`

+ - =

=- +

x

x

x

0 2 4

2 4

2`

=- +

=

=

y 2 0 4

4

=- +

=

^ h

y-intercept

y

x-5 -4 -3 -2 -1 1 2 3 4 5

3

2

1

-1

-2

-3

-4

-5

-6

x-intercept

y

x-5 -4 -3 -2 -1 1 2 3 4 5

5

4

3

2

1

-1

-2

-3

-4

y-intercept

x-intercept

To find the x-intercept, make y = 0 and solve for x

To find the y-intercept, make x = 0 and solve for y



Straight Lines

SERIES TOPIC

K 5

Knowing MoreQuestions

1. Find the intercepts of these lines:

a

c

e

g

b

d

f

h

3 6y x=- +

x y5 2 10 0- + =

y x41 1=- +

y x23 3=- -

4 8y x=- -

x y3 7 21 0+ + =

x y3 5 15 0- - =

x y5 6 30 0+ - =



Straight Lines

SERIES TOPIC

K 5

Questions Knowing More

2. Find the equation of a line which has these intercepts, in gradient-intercept form:

3. Find the equation of a line which has these intercepts, in general form:

a

c

a

c

b

d

b

d

-intercept is -intercept is1, 3.x y


-intercept is -intercept is6, 4.x y-




-intercept is -intercept is4, 5.x y- -

-intercept is -intercept is3, 15.x y- -



Straight Lines

SERIES TOPIC

K 5

Knowing MoreQuestions

4. Answer the following questions about this straight line:

5. Write the equation of the following line in general form.

a

d

b

e

c

What are the intercepts of this line?

Write the equation of this line in gradient-intercept form.

Will the gradient be positive or negative?

Write the equation of this line in general form.

Find the gradient of the line.

y

x-5 -4 -3 -2 -1 0 1 2 3 4 5

5

4

3

2

1

-1

-2

-3

-4

y

x-5 -4 -3 -2 -1 0 1 2 3 4 5

5

4

3

2

1

-1

-2

-3

-4



Straight Lines

SERIES TOPIC

K 5

Questions Knowing More

6. Use the axes below to draw straight lines from the following equations:

a

b

c

y x 2= -

y x4 4- =

x y7 4 28 0- - =

y

x-5 -4 -3 -2 -1 0 1 2 3 4 5

5

4

3

2

1

-1

-2

-3

-4

y

x-5 -4 -3 -2 -1 0 1 2 3 4 5

5

4

3

2

1

-1

-2

-3

-4

y

x-5 -4 -3 -2 -1 0 1 2 3 4 5

1

-1

-2

-3

-4

-5

-6

-7

-8



Straight Lines

SERIES TOPIC

K 5

Using Our Knowledge

In the equation y mx c= + , only values for m and c need to be found for the equation of the line.

If a line passes through the two points and, ,x y x y1 1 2 2^ ^h h , then:

Use the formula for gradient to find m.

• The line goes through points and, ,A B2 5 3 5- -^ ^h h.

• Use the formula for gradient to find m.

• Substitute point A (or B) into the equation 2y x c=- + .

Substitute a point into the above equation and solve for c. Substituting ,A 1 1-^ hpoint gives:

The equation of the line is y x4 5= - .

Sometimes the graph needs to be used to find the points.

• The value for m (the gradient) can be found using the formula:

• The value for c is found by substituting either of the points into the equation of the straight line.

Finding an Equation from any Two Points

ormx xy y

mx xy y

2 1

2 1

1 2

1 2=

--

=--

Find the equation of the line which passes through and, ,A B1 1 2 3-^ ^h h

Find the equation of the following line:

x1

x1

y1

y1

x2

x2

y2

y2

mx xy y

m

m

y x c

2 1

3 1

4

4

2 1

2 1

`

`

=--

=-

- -

=

= +

^ h

x y

c

c

1 4 1

5`

- = +

=-

^ h

If B was chosen as ,x y1 1^ h and A was chosen as ,x y2 2^ h, then m would still be the same.

Either A or B can be substituted here. The answer will always work out.

y

x

,A 2 5-^ h

,B 3 5-^ h

2m

x xy y

m

35 5

2

2 1

2 1

`

=--

=- -- -

=-

^ h

c

c

5 2 2

1`

=- - +

=

^ h



Straight Lines

SERIES TOPIC

K 5

Using Our Knowledge

In the equation y mx c= + , only values for m and c need to be found for the equation of the line. If the gradient of the line is given, then there is no need to find m. All that is required is to find c.

To find c, substitute the point into the equation.

Here is an example using both methods

From the question 3m =- .

Substitute point ,A 3 4-^ h into this equation.

For the line passing through AB:

For the line passing through CD:

From the question 2m =- .

Substitute point ,C 5 13-^ h into this equation.

• The line goes through points and, ,A B4 22 3 6- -^ ^h h.

• Use the formula for gradient to find m.

• Substitute point B (or A) into the equation y x c4= + .

Finding an Equation from a Point and Gradient

Find the equation of a line which goes through the point ,A 3 4-^ h and has a gradient of 3-

Find the equations of the lines passing through AB and CD below, if CD has a gradient of 2-

x

x

y

y

The equation of the line is 3 5.y x` =- +

The equation of the line is 2 3y x` =- +

The line passing through has equation 4 6.AB y x` = -

3y x c` =- +

c

c

4 3 3

5`

- =- +

=

^ h

y

x

,C 5 13-^ h

,A 4 22- -^ h

,B 3 6^ h

D

2y x c` =- +

c

c

13 2 5

3`

=- - +

=

^ h

x1 y1 x2 y2

4mx xy y

3 4

6 22

2 1

2 1=

--

=- -

- -=^

^hh

c

c

6 4 3

6`

= +

=-

^ h



Straight Lines

SERIES TOPIC

K 5

Using Our KnowledgeQuestions

1. Find the equations of lines which pass through these points:

a

c

b

d

and, , .A B2 1 3 2^ ^h h

and, , .P Q4 11 2 13- -^ ^h h

and, , .C D2 1 4 5- -^ ^h h

and, , .A B3 16 1 10- -^ ^h h



Straight Lines

SERIES TOPIC

K 5

Questions Using Our Knowledge

2. Find the equations of these lines. Write the equation in gradient-intercept form and general form.

a

c

b

d

A line passing through with, 4.A m1 6- - =^ h

A line passing through with, .C m6 132=^ h

A line with a gradient of passing through1 , .B 3 6- -^ h

A line with a gradient of andpassing

through , .D52

15 9

-

-^ h



Straight Lines

SERIES TOPIC

K 5

Using Our KnowledgeQuestions

3. Use the graph on the right to answer these questions below:

y

x

,A 2 10-^ h

,D 6 3^ h

,C 2 5-^ h

M

B

a

b

c

d

What is the equation of line AB if it has gradient 4m =- ?

What is the equation of the line passing through points C and D?

What are the coordinates of M, the midpoint of C and D?

What would the equation of a line passing through AM be?



Straight Lines

SERIES TOPIC

K 5

Questions Using Our Knowledge

4. If the intercepts of a straight line are known, then we know it passes through the points:

a

b

Find the gradient of the line passing through these two points.

(x-intercept, 0) and (0, y-intercept)

Where have you used this before?



Straight Lines

SERIES TOPIC

K 5

Thinking More

yparallel lines have the same gradient

x

Parallel lines make the same angle with the x-axis.

In the diagram, the lines are parallel. This means 1 2i i= .

Since the gradient of a line is tanm i= , this means that parallel lines have the same gradient.

Sometimes the fact that lines are parallel can be used to find points on the line.

Let m1be the gradient of AB and let m2 be the gradient of CD. Parallel lines have equal gradients so m m1 2= .

So D has coordinates 4, 12D - -^ h.

Line 1 and Line 2 are parallel since they have the same gradient m = 3.

Rewrite each line in gradient-intercept form:

• Line 1: y x3 1- =

• Line 2: y x2 2 6+ =

• Line 3: y x6 4 0- + =

• Line 1: 3 1y x= +

• Line 2: y x3 1= -

• Line 3: y x6 4= -

Parallel Lines

Which of the following lines are parallel?

Four points have the coordinates and2, 11 , 3,9 , 2, 4 , 12A B C D x- - - - -^ ^ ^ ^h h h h.Solve for x so that AB is parallel to CD.

m2x

x

x

x

3 2

9 11

2

12 4

428

2 2

4

`

`

- -

- -=

- -

- - -

=+-

+ =-

=-

^^

^^

hh

hh

1i 2i

tan tan1 2` i i=

m1



Straight Lines

SERIES TOPIC

K 5

Thinking More

Lines are perpendicular if they cut each other at 90c:

Let's say Line 1 has a gradient of m1 and Line 2 has a gradient of m2 . If they are perpendicular then:

Remember to make sure that the equations are in gradient-intercept form before checking for gradients.

or

Straight lines will be perpendicular if the product of their gradients is 1- (except for vertical and horizontal lines as 0 1#3 !- ). For example:

• A line with a gradient of 5 will be perpendicular to a line with a gradient of 51- , since 5 1

51#- =- .

• A line with a gradient of 21 will be perpendicular to a line with a gradient of 2- , since 2 1

21 #- =- .

• A line with a gradient of 1- will be perpendicular to a line with a gradient of 1, since 1 1 1#- =- .

Perpendicular Lines Non-Perpendicular Lines

• Line 1: y x2 3= -

• Line 2: y x21 4= +

• Line 3: y x21 5=- -

Perpendicular Lines

Which of the lines below are perpendicular to each other?

Line gradient Line gradient

Not perpendicular

1 2

221

1

#

#

`

=

=


Perpendicular

1 3

221

1

#

#

`

= -

=-


Not perpendicular

2 3

21

21

41

#

#

`

= -

=-

1m m1 2# =-

1mm1

2 =-c m The perpendicular gradient is the negative reciprocal



Straight Lines

SERIES TOPIC

K 5

Thinking More

The gradient can be used to find parallel lines. Remember, only values for m and c are needed for the equation of a line.

If the equation of the line is not in gradient-intercept form, remember to change it first.

The gradient can be used to find perpendicular lines too, in the same way.

This line is in general form, so first convert to gradient-intercept form: 3 1y x=- -

Substitute ,6 1-^ h into y x c31= +

3y x31` = + is the equation of the line perpendicular to x y3 1 0+ + = that passes through ,6 1-^ h.

Finding Parallel Lines

Finding Perpendicular Lines

Substitute 4,2^ h into y x c2= +

• Since the line is parallel to y x2 3= + , it must have the same gradient.

• The gradient of the line is m 2= . the equation of the line is 2 .y x c` = +

• To find c, substitute the given point into the equation:

Find the equation of a line parallel to y x2 3= + that goes through the point 4,2^ h.

Find the equation of a line perpendicular to x y3 1 0+ + = that goes through the point ( , )6 1-

2 2

6

c

c

4`

`

= +

=-

^ h

is the equation of the line parallel to that passes through2 6 2 3 , .y x y x 4 2` = - = + ^ h

c

c

131 6

3

`

`

= - +

=

^ h

m31

31=-

-=` j

y x c31= +

• Since the line is perpendicular to 3 1y x=- - :

• The gradient of the line is m31= . the equation of this line is` :

• To find c, substitute the given point into the equation:



Straight Lines

SERIES TOPIC

K 5

Questions Thinking More

1. What is the difference between parallel lines and perpendicular lines?

2. Match the lines on the left to the lines parallel to them on the right.

3. Match the lines on the left to the lines perpendicular to them on the right.

3y x=-

y x41 3= +

1y x4 0= +

y x21=

x y2 3 0- - =

y x41 1=- +

2 8 6y x+ =-

y x61 4=- -

x y7 2 0- - =

2y x=- +

y x2 1= +

x y3 2 0+ - =

4 4y x=- +

x y4 1 0- - =

2 3y x=- +

y x 7= -

2 6y x=- +

y x4 3= +

y x6 9= +

x y4 1 0+ - =



Straight Lines

SERIES TOPIC

K 5

Thinking MoreQuestions

4. The two lines below are parallel.

a

b

Find the value of y in B.

Write the equation of each line in gradient-intercept and general form.

y

x

,A 1 5-^ h

,B y3-^ h

,C 2 2^ h

,D 1 3-^ h



Straight Lines

SERIES TOPIC

K 5


5. The two lines below are perpendicular.

a

b

Find the value of x in C.

Write the equation of each line in gradient-intercept and general form.

y

x

1,3A^ h

,B 3 1- -^ h,C x 1-^ h



Straight Lines

SERIES TOPIC

K 5


6. Find the following parallel lines in general form:

a

b

A line parallel to y x3 4= - passing through the point 3,14^ h.

A line parallel to x y4 7 0+ - = passing through the point ,4 17-^ h.



Straight Lines

SERIES TOPIC

K 5


7. Find the following perpendicular lines in general form:

a

b

A line perpendicular to y x21= passing through the point ,8 12-^ h.

A line perpendicular to x y3 2 0+ - = passing through the point ,9 2-^ h.



Straight Lines

SERIES TOPIC

K 5


8. Is 2y x41= - perpendicular to the line joining ,A 3 13-^ h and ,B 2 7-^ h?

9. The line ax by c 0+ + = is perpendicular to one of the following lines and parallel to the other:

Which line is it perpendicular to, and which line is it parallel to?

• bx ay 0- =

• 3yba x=- +



Straight Lines

SERIES TOPIC

K 5

Answers

Basics: Basics:

Knowing More:

1.

1.

3.

4.

5.

6.

7.

2.

y

x

a m 2=

b 1m =-

c 5m =-

d m 4=

y

x

y

x

y

x

m 2=

m 2=

m 1=-a b

c d m 2=-

a

c

b

d

m 2= m 0=

2m =m 3=-

to nearest degree72ci = ^ h

a b

b

b

y x6 5=- +

y x3 1= - x y3 1 0- - =

x y6 5 0+ - =a

a Since both AB and BC have the same gradient, Points A, B and C are collinear.

Since both AB and BC have different gradients, Points A, B and C cannot be collinear.

a

c

b

d


The x-intercept is -2





The y-intercept is -3


e

g

f

h











Straight Lines

SERIES TOPIC

K 5

Answers

Knowing More: Knowing More:

Using Our Knowledge:

2.

3.

4.

5.

6.

6.

1.

2.

a

c

a

b

d

b

y x3 3=- +

y x21 2= +

3 4 12 0x y+ - =

x y7 2 14 0- + =

5 15y x=- -

y x2 14=- +

c

d

x y2 3 12 0- + =

x y5 4 20 0+ + =

2m =

a

d

b

e

c



Positive

y x2 4= -

2 4 0x y- - =

x y5 3 15 0- + =

a

b

y

x-5 -4 -3 -2 -1 0 1 2 3 4 5

5

4

3

2

1

-1

-2

-3

-4

y

x-5 -4 -3 -2 -1 0 1 2 3 4 5

5

4

3

2

1

-1

-2

-3

-4

c

y

x-5 -4 -3 -2 -1 0 1 2 3 4 5

1

-1

-2

-3

-4

-5

-6

-7

-8

a

c

b

d

y x 1= - y x2 3=- +

y x4 5= - 3 7y x=- +

a

c

b

d

4 2 0x y- - =

x y2 3 9 0- - =

x y2 5 15 0+ + =

3 0x y+ - =

y x 2= -

y x4 4- =

x y7 4 28 0- - =



Straight Lines

SERIES TOPIC

K 5

Answers

Using Our Knowledge:

Thinking More:

Thinking More:

3.

1.

2.

3.

4.

6.

5.

4.

a

b 2 9 0x y- - =

4 2 0x y+ - =

c

d

,4 1-^ h

11 6 38 0x y+ - =

a

b

interceptintercept

mxy--

=-

This is the formula used to find the gradient when you are working out the equation of a line and you have been given the two intercepts.

3y x=-

1y x4 0= +

x y2 3 0- - =

2 8 6y x+ =-

x y7 2 0- - =

y x2 1= +

x y3 2 0+ - =

4 4y x=- +

x y4 1 0- - =

2 3y x=- +

Parallel lines have the same gradient.Or, more formally, two lines with gradients m1 and m2 are parallel if m m1 2=

The gradients of perpendicular lines multiply with one another to give -1. Or, more formally, two lines with gradients m1 and m2 are perpendicular if m m 11 2# =-

y x41 3= +

y x21=

y x41 1=- +

y x61 4=- -

2y x=- +

y x 7= -

2 6y x=- +

y x4 3= +

y x6 9= +

x y4 1 0+ - =

a

b

y 5=-

Line AB:

Line AB:

Line CD:

Line AC:

gradient-intercept: y x5 10= +

gradient-intercept: y x 2= +

gradient-intercept: 5 8y x= -

gradient-intercept: 4y x=- +

general form: 5 10 0x y- + =

general form: 2 0x y- + =

general form: 5 8 0x y- - =

general form: 4 0x y+ - =

x 5=a

b

a

b

3 5 0x y- + =

4 1 0x y+ + =



Straight Lines

SERIES TOPIC

K 5

Answers

Thinking More:

8.

9.

2y x41= - is perpendicular to line AB

ax by c 0+ + = is parallel to yba x 3=- +

and perpendicular to bx ay 0- =

7. a 2 4 0x y+ - =

b 15 0x y3- + =

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