Straight Lines

-5

-4

-3

-2

-1

0

1

2

3

4

5

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

y = 2x + 3 x + 2y – 2 = 0

x

y

Revision Notes Get Started

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Properties of the straight lineYou should know … (x1, y1) and x2, y2) are points on a line …

€

m =y2 − y1

x2 − x1

€

d = x2 − x1( )2

+ y2 − y1( )2

€

m = tanθ

recognise the

term locus

ax + by + c = 0

y – y1 = m(x – x1)the equation of a straight line given two points

the equation of a straight line given one point and the gradient

the gradients of

parallel lines are equal

Lines are perpendicular if and only if m1m2 = –1 Find where lines intersect

the concurrency properties of lines associated with the triangle

A(–2,1)

B(10,6)

x

y

O

€

d = x2 − x1( )2

+ y2 − y1( )2

C

Thinking it out …

AC = 10 – (–2) = 12

BC = 6 – 1 = 5

By Pythagoras’ theorem

AB2 = 122 + 52

= 144 + 25 = 169

AB = √169 = 13

Using the formula …

AB = √[(10 – (–2))2 + (6 – 1)2]

= √(122 + 52)

AB = √169 = 13

Test Yourself?

The gradient, m

The gradient, m, is a measure of the slope of a line.

It is defined to be

€

m =Δy

Δx=

y2 − y1

x2 − x1

We can see in the diagram that this is tan and that

is also the angle the line makes with the x-axis.

A(x1,y1)

x

y

O

Change in x … x2 – x1

Change in y … y2 – y1

B(x2,y2)

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m = tanθ

A(–1, –2)

x

y

O

B(6, 12)

€

m =12 − (−2)

6 − (−1)

=14

7= 2

€

m = tanθ

tanθ = 2

θ =127°

The gradient

The angle

Test Yourself?

Where ‘’ stands for ‘change in …’

The equation of a line

Locus n, pl loci a set of points whose position satisfies some condition or equation.

The locus of the points(x, y) which satisfy the equation y = mx + c is a straight line with gradient m and y-intercept c.

The equation of the line can take different forms • y = mx + c … gradient m and y-intercept c.• ax + by + c = 0 … a and b cannot both be zero.

When a = 0 … we get an equation of the form y = k a line parallel to the x-axis

When b = 0 … we get an equation of the form x = k a line parallel to the y-axis

• y – y1 = m(x – x1) … where (x1, y1) lies on the line and m is its gradient.

The line which passes through the points (x1, y1) and

(x2, y2) has a gradient

and an equation y – y1 = m(x – x1)

€

m =y2 − y1

x2 − x1

1

2

3

1 2 3 x

y

O

y = 1/2x + 1

1

2

3

1 2 3 x

y

O

x – 4y + 2 = 0

y = 3

x = –1

The line which passes through the points (–1, –5) and (1, 1) has a gradient

Its equation is y – 1 = 3(x – 1) … or y – (–5) = 3(x – (–1))

Which both simplify to y = 3x – 2€

m =–5 −1

−1−1=

−6

−2= 3

Test Yourself?

Parallel and perpendicular

When two lines are parallel

their gradients are equal

… m1 = m2

When their gradients are equal, two lines are parallel

… or the same line.

When two lines are perpendicular

The product of their gradients is negative 1.

… m1m2 = – 1

When the product of their gradients is negative 1, two lines are perpendicular

To find the equation of the line parallel to y = 3x + 1 passing through the point (2, 4)

Parallel to y = 3x + 1 means its gradient is 3.

Passing through (2, 4) means its equation is:

y – 4 = 3(x – 2)

To find the equation of the line perpendicular to y = 3x + 1 passing through the point (2, 4)

Perpendicular to y = 3x + 1 means its gradient is –1/3.

Passing through (2, 4) means its equation is:

y – 4 = –1/3(x – 2)

Test Yourself?

Intersecting lines

When two lines intersect, find the point of intersection by solving their equations simultaneously.

If the equations are in the form y = mx + c

… equate the expressions for y:

m1x + c1 = m2x + c

… solve for x

… substitute and solve for y.

If the equations are in the form ax + by + c = 0

… use the elimination method

… scaling the equations

… subtract to eliminate one variable

… solve

… substitute and solve

Where do

y = 3x + 4 and y = 5x – 2 intersect?

3x + 4 = 5x – 2

2x = 6

x = 3

y = 3.3 + 4 = 13

intersect at (3, 13)

Where do

2x + 3y – 9 =0 and 3x + 5y – 14 = 0 intersect?

2x + 3y = 9 …

3x + 5y = 14 …

5: 10x + 15y = 45 …

3: 9x + 15y = 42 …

– : x = 3

y = 1 (by substitution in )

intersect at (3, 1)

Test Yourself?

lines associated with the triangle

C

M

B

A

M is the midpoint of BC.

A is the vertex opposite BC.

AM is called a median .

C

M

B

A

N

M is the midpoint of BC.

MN is perpendicular to BC.

… a perpendicular bisector .

C

G

B

A

G lies on BC.

AG is perpendicular to BC.

AG is an altitude .

C

P

B

A

P lies on BC.

AP bisects the angle at A

AP is an angle bisector .

More…More…

lines associated with the triangle

A triangle has 3 medians.

They are concurrent …

[they all pass through the same point]

The point of concurrency is called the centroid .

A triangle has 3 perpendicular bisectors

They are concurrent.

The point of concurrency is called the circumcentre .

C

M

B

A

C

M

B

A

N

A triangle has 3 altitudes

They are concurrent.

The point of concurrency is called the orthocentre .

C

G

B

A

A triangle has 3 angle

bisectors

They are concurrent.

The point of concurrency is called the incentre .

More…More…

C

P

B

A

lines associated with the triangle

The centroid cuts each median in the ratio 2:1

Its coordinates are:

The circumcentre is the centre of the circumcircle … the circle which passes through the three vertices of the triangle

C

M

B

A

C

M

B

A

N

The orthocentre, the centroid and the circumcentre all lie on the same line which is called the Euler line.

C

G

B

A

The incentre is the centre of the incircle … the circle which has the three sides as tangents

C

P

B

A

€

xa + xb + xc

3,ya + yb + yc

3

⎛

⎝ ⎜

⎞

⎠ ⎟

MN

Test Yourself?

Between Central Station and The Kingston Bridge, Argyll St in Glasgow is a straight line.

Relative to a suitable set of axes and using suitable units, the end-points are (–10, 30) and (73, 17).

Calculate the length of Argyll St correct to the nearest unit.

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Between Central Station and The Kingston Bridge, Argyll St in Glasgow is a straight line.

Relative to a suitable set of axes and using suitable units, the end-points are (–10, 30) and (73, 17).

Calculate the length of Argyll St correct to the nearest unit.

Argyll St = √[(–10 – 73)2 + (30 – 17)2]

=√[(–83)2 + 132]

= √(6889 + 169)

= √7058

= 84 ( to nearest whole unit)

x

y

O

A meteor streaks through the atmosphere.

Relative to a suitable set of axes, it passes

through the points (–1, 4) and (5, 1).

(a) What is the gradient of its flight path?

(b) At what angle does it cross the x-axis?

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x

y

O

A meteor streaks through the atmosphere.

Relative to a suitable set of axes, it passes

through the points (–1, 4) and (5, 1).

(a) What is the gradient of its flight path?

(b) At what angle does it cross the x-axis?

€

m =1− 4

5 − (−1)

=−3

6

= −1

2

(a)

(b)

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m = tanθ

tanθ = −1

2θ = −53°

The negative sign tells us the angle is measured clockwise and is 53˚ below the x-axis

53˚ x

At an archaelogical dig a series of post holes

seems to form a straight line.

Relative to a convenient set of axes and units,

there are holes at (–3, 5) and (6, 1).

(a) Find the equation of the line expressing it in the form ax + by + c = 0.

(b) Another post hole is discovered at (3, –2). Does this lie on the same line?

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At an archaelogical dig a series of post holes

seems to form a straight line.

Relative to a convenient set of axes and units,

there are holes at (–3, –5) and (6, 1).

(a) Find the equation of the line expressing it in the form ax + by + c = 0.

(b) Another post hole is discovered at (3, –2). Does this lie on the same line?

€

m =−5 −1

−3− 6=

−6

−9=

2

3

(a)

€

y −1 =2

3x − 6( )

⇒ 3y − 3 = 2x −12

⇒ 2x − 3y − 9 = 0

(b) Consider the point (3, –2).

When x = 3 on the line2.3 – 3y – 9 = 03y = 6 – 93y = –3y = –1The point (3, –1) is on the line(3, –2) is not.

x

y

O

D

C

B

A

A trapezium has vertices A(1, 0), B(3, 4), D(2, –3) and C. The coordinates of C are not given.

AB is parallel to DC.

(a) Find the equation of both lines.

(b) When CB and DA are produced, they intersect at right angles. Find the equations of both lines.

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x

y

O

D

C

B

A

A trapezium has vertices A(1, 0), B(3, 4), D(2, –3) and C. The coordinates of C are not given.

AB is parallel to DC.

(a) Find the equation of both lines.

(b) When CB and DA are produced, they intersect at right angles. Find the equations of both lines.

(a) The equation of AB:

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mCD = mAB = 2

y – 0 = 2(x – 1)

y = 2x – 2

The equation of DC:

CD is parallel to AB so …

€

mAB =4 − 0

3 −1= 2

y – (–3) = 2(x – 2)

y = 2x – 7

(b) The equation of DA:

€

mDA =−3− 0

2 −1= −3

y – 0 = –3(x – 1)

y = –3x + 3

The equation of CB:

CB is perpendicular to DA so …

€

mCD =−1

mAB

=1

3

€

y − 4 =1

3x − 3( )

revealrevealx

y

O

D

C

B

A

A trapezium has vertices A(1, 0), B(3, 4), D(2, –3) and C. The coordinates of C are not given.

It can be found that the equation of

DC is: 2x – y – 7 = 0

BC is: x –3y + 9 = 0

What are the coordinates of C?

x

y

O

D

C

B

A

A trapezium has vertices A(1, 0), B(3, 4), D(2, –3) and C. The coordinates of C are not given.

It can be found that the equation

of DC is: 2x – y – 7 = 0;

of BC is: x –3y + 9 = 0

What are the coordinates of C?

2x – y = 7 …

x – 3y = –9 …

3: 6x – 3y = 21 …

1: x – 3y = –9 …

– : 5x = 30

x = 6

y = 5 (by substitution in )

C is the point (6, 5)

Three businessmen living in Airdrie A(5, 0), Crieff C(11, 18) and Dunfermline D(13, 4) pick Stirling as a place to hold a meeting as it is equidistant from all three towns.

This means that Stirling is the circumcentre of the triangle ACD.

Find the coordinates of Stirling.

[All coordinates are with reference to a convenient set of axes and units.]

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Three businessmen living in Airdrie A(5, 0), Crieff C(11, 18) and Dunfermline D(13, 4) pick Stirling as a place to hold a meeting as it is equidistant from all three towns.

This means that Stirling is the circumcentre of the triangle ACD.

Find the coordinates of Stirling.

[All coordinates are with reference to a convenient set of axes and units.]

Perpendicular bisector of AD:

Midpoint AD = (9, 2)

So required equation is:

y – 2 = –2(x – 9)

y = –2x + 20

€

mAD =4 − 0

13 − 5=

1

2⇒ m⊥ = −2

Perpendicular bisector of DC:

€

mDC =18 − 4

11−13= −7⇒ m⊥ =

1

7

Midpoint DC = (12, 11)

So required equation is:

y – 11 = 1/7(x – 12)

Intersection of Perpendicular bisectors:

–2x + 20 – 11 = 1/7(x – 12) –14x + 140 – 77 = x – 1215x = 75 x = 5y = –2.5 + 20 y = 10 Stirling is at S(5, 10)

This revision list takes you through the

Learning outcomes required for Higher

Maths, Unit 1, The Straight line.

Practice makes Perfect.