basic skills in higher mathematics robert glen adviser in mathematics mathematics 1(h) outcome 1

Basic Skills in

Higher Mathematics

Robert GlenAdviser in Mathematics

Mathematics 1(H)Outcome 1

Mathematics 1(Higher)

Outcome 1 Use the properties of the straight line

Straight lines

PC Index

PC(a) Gradients and straight lines

PC(b) Gradients and angles

PC(c) Parallel and perpendicular

Click on the one you want

PC(a) Determine the equation of a straight line given two points on the line or one point and the gradient

Index Click on the section you want

1 What is gradient?

2 The gradient of a line

3 The equation of a line given its gradient and the intercept on the y - axis

4 The equation of a line given one point on the line and the gradient

5 The equation of a line given two points on the line

Section 1

1 What is gradient?

Mathematics 1(Higher) 1.1

The gradient (slope) of this roof is

1 What is gradient?

Gradient

Gradient 1

Check this:The steeperthe slope, the greater the gradient.

What is the gradient of this roof ?

1 What is gradient?

Click on the letter of the correct answer

1 What is gradient?

Sorry, wrong answer

Have another go!

Gradient = vertical horizontal

1 What is gradient?

Click on the letter of the correct answer

1 What is gradient?

Correct!

1 What is gradient?

End of Section 1

Section 2

Read all lines from left to right

Line AB is uphill from left to right

Line AB has a positive gradient mAB 0

Line PQ is downhill from left to right

Line PQ has a negative gradient mPQ 0

Line AB has a positive gradient mAB 0

BGradient =

change in y

change in x

(9, 6)

(0, 3)mAB =39

BGradient =

change in y

change in x

(9, 6)

(0, 3)mAB =39

Note: we could have measured the gradient like this

Gradient = change in y

change in x

mPQ =-6 9

(0, 7)

(9, 1)

change in x

mPQ =-6 9

Note: we could have measured the gradient like this

y (0, 7)

Q (9, 1)x

B (9, 6)

(0, 3)

change in x

9 - 0 6 - 3 9 - 0

= 3913

change in x

= 9 - 0

1 - 7 1 - 7 9 - 0-6 923

(0, 7)

Q (9, 1)x

2 The gradient of a line y

A formula to memorise

B (x2 , y2)

A (x1 , y1)

mAB =y2 - y1

x2 - x1

A formula to memorise

B (x2 , y2)

A (x1 , y1)

mAB =y2 - y1

x2 - x1

1 Calculate the gradient of line AB

B (6 , 5)

A (2 , 3)mAB =y2 - y1

x2 - x1

=5 - 36 - 2

Did you getthis answer?

2 Calculate the gradient of line CD. D (6 , 2)

C (2 , -1)mCD =

y2 - y1

x2 - x1

=2 - (-1) 6 - 2

3 Calculate the gradient of line EF.

F (5, -1)

E (-3 , 3)

mEF = y2 - y1

x2 - x1

=-1 - 35 - (-3)

= -4 8

= - 12 End of Section 2

Section 3

3 The equation of a linegiven its gradient and theintercept on the y - axis

3 The equation of a line given gradient and intercept

(0, 3) m = ½

(x, y)

LFind the equation of line KL which has a gradient of ½ and passes through the point (0, 3).

mKL =y - 3x - 0

y - 3 = ½ x

y = ½ x + 3The equation of KL is y = ½ x + 3

(0, 3) m = ½

(x, y)

LFind the equation of line KL which has a gradient of ½ and passes through the point (0, 3).

The equation of KL is y = ½ x + 3

Formula: y = m x + c

(0, c) m

(x, y)

LThe equation of line with gradient m and intercept c is:

y = m x + c

Memorise this

1 Find the equation of line PQ which has a gradient of -2 and passes through the point (0, 5).

(0, 5)

m = -2

The equation of PQ is y = -2 x + 5

(x, y)Use the formula

(0, -3)

m = ¾

F2 Find the equation of line EF which has a gradient of ¾ and passes through the point (0, -3).

The equation of EF is y = ¾ x - 3

(x, y)

Use the formula

You should now do Section A1 questions 1 - 10 on page 3 of

the Basic Skills booklet.

End of Section 3

Section 4

4 The equation of a linegiven one point on the line and the gradient

4 The equation of a line given one point and the gradient

(4, 3)

L (x, y)

Find the equation of the linethrough the point (4, 3) with gradient 3.

mKL =y - 3x - 4

y - 3 =

y - 3 = 3x - 12

y = 3x The equation of KL is y = 3x - 9

3(x - 4)

(4, 3)

L (x, y)

Find the equation of the linethrough the point (4, 3) with gradient 3.

The equation of KL is y = 3x - 9

Formula: y - b = m (x - a)

(a, b)

L (x, y)

The equation of the linethrough the point (a, b) with gradient m is :

y - b = m (x - a)

Memorise this

4 The equation of a line given one point and the gradienty

(-1, 2)

(x, y)

1 Find the equation of the linethrough the point (-1, 2) with gradient 2.

The equation of PQ is y = 2 x + 4

Use the formula

(-1, 2)

(x, y)

1 Find the equation of the linethrough the point (-1, 2) with gradient 2.

The equation of PQ is y = 2 x + 4

m = 2 y - b = m (x - a)

y - 2 =

y - 2 = 2 (x + 1)

y - 2 = 2 x + 2

y = 2 x

(a, b)

(x - (-1))2

2 Find the equation of the linethrough the point (6, -2) with gradient ½.

x (6, -2)

(x, y)

Om = ½

Use the formula

The equation of MN is 2y = x - 10

x (6, -2)

(x, y)O

2 Find the equation of the linethrough the point (6, -2) with gradient ½.

The equation of MN is 2y = x - 10

m = ½

y - b = m (x - a)

y - (-2) =

y + 2 = ½ (x - 6)

2y + 4 =

2y = x

(a, b)

or x - 2y - 10 = 0

Multiply both sides by 2to clear the fraction.

½ (x - 6)

(-1, 4)

(x, y)

3 Find the equation of the linethrough the point (-1, 4) with gradient 2/3 .

The equation of RS is 3y = -2x + 10

m = -2/3

Use the formula

(-1, 4)

(x, y)

3 Find the equation of the linethrough the point (-1, 4) with gradient 2/3 .

The equation of RS is 3y = -2 x + 10

m = -2/3

y - b = m (x - a)

y - 4 =

3y - 12 =

(a, b)

or 2 x + 2y - 10 = 0

Multiply both sides by 3to clear the fraction.

-2/3(x - (-1))

y- 4 = -2/3 (x + 1)

-2(x + 1)

-2 x + 10

You should now do Section A1 questions 11 - 20 on page 3 of

End of Section 4

Section 5

5 The equation of a linegiven two points on the line

Find the equation of the linejoining the points A (3, 1) and B (6, 4) .

Step 1 Calculate the gradient

mAB = y2 - y1

x2 - x1

=4 - 16 - 3

Step 2 Calculate the equation

y - b = m (x - a)

y - 1 =

y - 1 = x - 3

y = x - 2

Choose A (3, 1) as thepoint on the line.i.e. a = 3, b = 1

(You get exactly thesame answer if youchoose B.)

(6, 4)

(3, 1)

(a, b)

1 (x - 3)

The equation of CD is y = 2x

Use the formula

1 Find the equation of the linejoining the points C (1, 2) and D (5, 10) .

(5, 10)

(1, 2)

Answer coming up!

mAB = y2 - y1

x2 - x1

=10 - 2 5 - 1

y - b = m (x - a)

y - 2 =

y - 2 = 2 x - 2

y = 2 x

Choose C (1, 2) as thepoint on the line.i.e. a = 1, b = 2

(You get exactly thesame answer if youchoose B.)

(a, b)1 Find the equation of the linejoining the points C (1, 2) and D (5, 10) .

(5, 10)

(1, 2)

2 (x - 1)

2 Find the equation of the linejoining the points G (-3, 1) and H (5, -3) .

(5, -3)

(-3, 1)

The equation of GH is 2y = - x - 1

Use the formula

Answer coming up!

mGH = y2 - y1

x2 - x1

= -3 - 15 - (-3)

= -4 8

y - b = m (x - a)

y - 1 =

2y - 2 =

2y = - x

Choose G (-3, 1) as the point on the line.i.e. a = -3, b = 1

(You get exactly thesame answer if youchoose H.)

(a, b)

or x + 2y +1 = 0

H (5, -3)

(-3, 1)

-½(x - (-3))

- x - 3

y - b = m (x - a)

y - 1 = -½(x - (-3))

2y - 2 = - x - 3 2y = - x - 1

Multiply both sides by2 to clear the fraction.

A fuller explanation

y - 1 = -½(x + 3)

(a, b)

(5, -3)

mAB =y2 - y1

x2 - x1

(x2 , y2)

A (x1 , y1)

y = m x + c

(0, c)

y - b = m (x - a)

(a , b)

(x , y)

(x1 , y1)

(x2 , y2)

1 Calculate m

m =y2 - y1

x2 - x1

2 y - b = m (x - a) (a, b)

Summary

You should now do Sections A2 and A3 on page 3 of

End of Section 5

PC(b) Find the gradient of a straight line using m = tan

Gradientsand

angles

Mathematics 1(Higher) 1.1 Outcome 1 Use the properties of the straight line

mAB =pq

= 0.70 (to 2 dp)

tan 35

35= -0.70 (to 2 dp)

tan 145

Line EF is downhill,so its gradient is nottan 35.

Always take the angle

between the line and the positive directionof the x-axis.

= 0.53 (to 2 dp)

tan 28

1 What is the gradient of the line GH (to 2 dp)?

mKL =48

= -1.11 (to 2 dp)

tan 132132

2 What is the gradient of the line KL (to 2 dp)?

You should now do the questions on page 7 of

End of PC(b)

PC(c) Find the equation of a line parallel to and perpendicular to a line

Index Click on the section you want

1 Parallel lines

2 Perpendicular lines

3 Equations

Section 1

1 Parallel lines

These lines are all parallel to each other

If one of the lines has agradient m, they all havea gradient m.

Parallel lines have

equal gradients

The line y = 2x + 10 has a gradient of 2.

So any line parallel to this one has a gradient of 2.

y = 2x + 10

y = 2x + 5

y = 2x

y = 2x - 5

y = 2x - 10

The line 2x - y + 5 = 0 also belongs to this set of parallel lines.Can you see why?

2x - y + 5 = 0 2x + 5 = y y = 2x + 5

1 Which of the following lines is/ are parallel to the line y = 3x - 5?

y = 3x - 1 y = -3x + 3 y = 3x

3x + y = 3 3x - y = 3

Click on the letterof a correct answer

NB There could be more than one rightanswer .

y = 3x - 1

Correct!This line has a gradient of 3.

Have another go!

Wrong!This line has a gradient of -3.

y = -3x + 3B

y = 3x

Have another go!

3x + y = 3

Wrong!This line has a gradient of -3.

Have another go!

y = -3x +3

Click here to seeall the answers

Have another go!

y = 3x +3

3x - y = 3E

Parallel toy = 3 x - 5

Not parallel toy = 3 x - 5

y = -3x +3

y = 3x - 1 y = -3x + 3 y = 3x

3x + y = 3 3x - y = 3

y = 3x +3

2 Which of the following lines is/ are parallel to the line x + y = 8?

y = x + 5y = - x +

1 y = x

x + y = 10 x - y = 7

Wrong!This line has a gradient of +1.

Have another go

y = x + 5A

y = - x + 1

Correct!This line has a gradient of -1.

Have another go

y = x C

Correct!This line has a gradient of -1.

Have another go

y = -x +10

x + y = 10D

Have another go

y = x - 7

x - y = 7E

Parallel tox + y = 8

Not parallel tox + y = 8

y = -x +10

y = x + 5y = - x +

1 y = x

x + y = 10 x - y = 7

y = x - 7

3 Which of the following lines is/ are parallel to the line y = ½ x - 3?

y = 2x - 1

y = ½ x + 1 2y = x

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

Wrong!This line has a gradient of 2.

Have another go

y = 2x - 1A

Correct!This line has a gradient of ½.

Have another go

y = ½ x + 1B

Have another go

y = ½x

2y = x C

Have another go

y = ½ x - 2x - 2y = 4D

Have another go

y = ½ x + 3 ½

x - 2y + 7= 0E

Parallel toy = ½ x - 3

Not parallel to y = ½ x - 3

y =½x

y = ½x - 2

y = 2x - 1 y = ½ x + 1 2y = x

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

y = ½ x + 3 ½

Continue with Section 2Perpendicular lines

End of Section 1

Section 2

2 Perpendicular lines

mAB =32

CD is perpendicularto AB.

mCD =23

mAB mCD = 32

mEF =34

GH is perpendicularto EF.

mGH =43

mEF mGH = 34

mPQ =31

RS is perpendicularto PQ.

mRS =13

mPQ mRS = 31

If two lines with gradientsm1 and m2 are perpendicularthen m1 × m2 = -1

Memorise this

If two lines with gradientsm1 and m2 are perpendicularthen m1 × m2 = -1.

Parallel lines haveequal gradients.

Summarym

y 1 For each line write down the gradient of any linea parallel to the lineb perpendicular to the line

1 Answers1 ½ , -2

2 -3, 1/3

3 3/4, -4/3

4 -1/3, 3

Here are the answers

Answers1 4 , -¼ 2 ¾, -4/3

3 -5, 1/5 4 -1, 1

5 ½, -2 6 -3/5, 5/3

1 y = 4x - 1

2 y = ¾ x + 5

6 3x + 5y = 15

3 y = -5x

4 x + y = 15

5 x - 2y + 3 = 0

Here are the answers

2 For each line write down the gradient of any linea parallel to the lineb perpendicular to the line

You should now do Section C1 on page 11 of the Basic Skills booklet.

End of Section 2

Section 3

3 Equations

AB has equation y = 3x + 5.Find the equation of the line parallel to AB through (1, -2) perpendicular to AB through (1, -2)

Parallel linemAB = 3So mparallel = 3Point on line is (1, -2) y - b = m (x - a) y - (-2) = 3(x - 1) y + 2 = 3x - 3 y = 3x - 5

Perpendicular linemAB = 3So mperp = -1/3Point on line is (1, -2) y - b = m (x - a) y - (-2) = -1/3 (x - 1) 3y + 6 = - x + 3 x + 3y + 3 = 0

Click here for revisionof finding equationsof straight lines

Find the equation of the line:

1 Through (0, 3), parallel to y = 2x +1

2 Through (1, 5), perp to y = ¼ x - 3

3 Through (-2, 2), parallel to x + y = 10

4 Through (5, -3), perp to y = -½ x +75 Through (3, -1), parallel to 2x + 3y + 5 =0

Answers

1 y = 2x +3

2 y = -4x + 9

3 y = -x

4 y = 2x -13

5 3x + 2y -11 = 0

You should now do Sections C2 and C3 on page 11 of the Basic Skills

booklet.

End of PC(c)

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