s5 unit 1- the straight line

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Higher Outcome 1

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Higher Unit 1Higher Unit 1Distance Formula

The Midpoint Formula

Gradients Collinearity

Gradients of Perpendicular Lines The Equation of a Straight LineMedian, Altitude & Perpendicular BisectorConcurrencyExam Type Questions

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Higher Outcome 1

Distance FormulaDistance FormulaLength of a straight lineLength of a straight line

A(x1,y1)

B(x2,y2)

x2 – x1

y2 – y1

C

x

y

O

This is just Pythagoras’ Theorem

2 2 2(AB) =(AC) +(BC)

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Higher Outcome 1Distance FormulaDistance Formula

The length (distance ) of ANY line can be given by the formula :

2 2tan 2 1 2 1(x ) (y )dis ceAB x y

Just Pythagoras Theorem in

disguise

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Higher Outcome 1

2 2distance 2 1 2 1(x ) (y )AB x y

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Higher Outcome 1CollinearityCollinearity

A

C

x

y

O x1 x2

B

Points are said to be collinear if they lie on the same straight.

The coordinates A,B C are collinear since they lie on

the same straight line.D,E,F are not collinear they do not lie on the

same straight line.D

EF

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Higher Outcome 1Straight Line TheoryStraight Line Theory

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Higher Outcome 1Finding Mid-Point of a lineFinding Mid-Point of a line

A(x1,y1)

B(x2,y2)

x

y

O

1 21 2 , ,2 2y yx xM

x1 x2

M

y1

y2

The mid-point (Median) between 2 points is given by

Simply add both x coordinates together

and divide by 2.Then do the same

with the y coordinates.

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Higher Outcome 1

1 21 2 , ,2 2y yx xM

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Higher Outcome 1Straight line FactsStraight line Facts

Y – axis Intercept

2 1

2 1

y - yGradient = x - x

y = mx + c

Another version of the straight line general formula is:ax + by + c = 0

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May 3, 2023www.mathsrevision.com 10

m < 0

m > 0

m = 0

x = a

y = c

Sloping left to right up has +ve gradient

Sloping left to right down has -ve gradient

Horizontal line has zero gradient.

Vertical line has undefined gradient.

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May 3, 2023www.mathsrevision.com 11

m = tan θ

m > 0Lines with the same gradient

means lines are Parallel

The gradient of a line is ALWAYSequal to the tangent of the angle

made with the line and the positive x-axisθ

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60o

60o

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Higher Outcome 1

Gradient of perpendicular linesGradient of perpendicular lines

If 2 lines with gradients m1 and m2 are perpendicular then m1 × m2 = -1 When rotated through 90º about the origin A (a, b) → B (-b, a)

-aB(-b,a)

-b

A(a,b)

aO

y

x

- 0- 0

OAb bma a

- 0 -- - 0

OBa amb b

- -1-

OA OBb a abm ma b ab

Conversely:If m1 × m2 = -1 then the two lines with gradients m1 and m2 are

perpendicular.

-b

Investigation

Demo

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Higher Outcome 1

=

The Equation of the Straight LineThe Equation of the Straight Liney – b = m (x - a)

The equation of any line can be found if we know the gradient and one point on the line.

O

y

xx - a

P (x, y)

m

A (a, b)y - by - b

x – a

m = y - b(x – a)m

Gradient, m Point (a,

b)y – b = m ( x – a ) Point on the line ( a, b )

a x

y

b

Demo

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Higher Outcome 1

A

B C

D

May 3, 2023 19www.mathsrevision.com

AMedian means a line from a vertex tothe midpoint of the base.

Altitude means a perpendicular linefrom a vertex to the base.

B D C

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May 3, 2023 20www.mathsrevision.com

A

B D C

Perpendicular bisector - a line that cuts another lineinto two equal parts at an angle of 90o

Any number of lines are said to be concurrent if there is a point through which they all

pass.For three lines to be concurrent,

they must all pass through a single point.

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Higher Outcome 1

Find the equation of the line which passes through the point (-1, 3) and is perpendicular to the line with equation 4 1 0x y

Find gradient of given line: 4 1 0 4 1 4x y y x m

Find gradient of perpendicular: 1 (using formula 1)1 24

m mm

Find equation: 1 31 4( 3) 1 4 12

4 ( 1)y

x y x yx

4 13 0y x

Typical Exam Typical Exam QuestionsQuestions

Finding the Equation of an Altitude

A

BTo find the equation of an altitude:

• Find the gradient of the side it is

perpendicular to ( ).mAB C

• To find the gradient of the altitude, flip the gradient

of AB and change from positive to negative:maltitude = mAB

–1

• Substitute the gradient

and the point C into

y – b = m ( x – a )

ImportantWrite final equation in the formA x + B y + C = 0with A x positive.

Common Straight Strategies for Exam Questions

Finding the Equation of a Median

P

Q

To find the equation of a median:

• Find the midpoint of the side itbisects, i.e. O

• Calculate the gradient of the median OM.• Substitute the gradient and either

point on the line (O or M) intoy – b = m ( x – a )

ImportantWrite answer in the formA x + B y + C = 0with A x positive.

=

=M

( )

M = 2y2 y1

2x2 x1 ,+ +

Common Straight Strategies for Exam Questions

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Higher Outcome 1

A triangle ABC has vertices A(4, 3), B(6, 1)

and C(–2, –3) as shown in the diagram.

Find the equation of AM, the median from B to C

Find mid-point of BC: 1 2 1 2(2, 1) Using M ,2 2

x x y y

Find equation of median AM

Find gradient of median AM

2 1

2 1

-2 Using -

y ym mx x

2 5 Using - ( - ) y x y b m x a


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Higher Outcome 1

P(–4, 5), Q(–2, –2) and R(4, 1) are the vertices

of triangle PQR as shown in the diagram.

Find the equation of PS, the altitude from P.

Find gradient of QR: 2 1

2 1

-1 Using 2 -

y ym mx x

Find equation of altitude PS

Find gradient of PS (perpendicular to QR)

1 22 ( 1) m m m

2 3 0 Using ( ) y x y b m x a


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Higher Outcome 1Triangle ABC has vertices A(–1, 6), B(–3, –2) and C(5, 2)Find: a) the equation of the line p, the median from C of triangle ABC. b) the equation of the line q, the perpendicular bisector of BC. c) the co-ordinates of the point of intersection of the lines p and q.

Find mid-point of AB Find equation of

p2y

Find gradient of p(-2, 2)

Find mid-point of BC

(1, 0) Find gradient of BC12

m

0m

Find gradient of q 2m Find equation of q

2 2 y x

Solve p and q simultaneously for intersection

(0, 2)

Exam Type Exam Type QuestionsQuestions p

q

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Higher Outcome 1Find the equation of the straight line which is parallel to the line with equation and which passes through the point

(2, –1) . Find gradient of given line:

Knowledge: Gradient of parallel lines are the same. Therefore for line we want to find has gradient

23

m

Find equation: Using y – b =m(x - a)

3 2 1 0 y x

2 3 5x y

2 23 3

3 2 5 5y x y x m


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Higher Outcome 1

Find gradient of the line:

1tan3

Use table of exact values 1 1tan 303

2 ( 1) 3 13 3 0 3 3 3

m

tanm

Find the size of the angle a° that the line

joining the points A(0, -1) and B(33, 2)

makes with the positive direction of the x-axis.

Exam Type QuestionsExam Type Questions

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Higher Outcome 1A and B are the points (–3, –1) and (5, 5).Find the equation of

a) the line AB.

b) the perpendicular bisector of AB Find gradient of the AB: 4 3 5y x

Find mid-point of AB 1, 2

34

m Find equation of AB

Gradient of AB (perp):43

m

Use y – b = m(x – a) and point ( 1, 2) to obtain line of perpendicular bisector of AB we get

3 4 10 0 y x

Exam Type QuestionsExam Type Questions

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Higher Outcome 1

The line AB makes an angle of 60o with

the y-axis, as shown in the diagram. Find the exact value of the gradient of AB.

Find angle between AB and x-axis: 090 60 30o o

Use table of exact values

tanm tan 30om

13

m

(x and y axes are perpendicular.)


60o

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Higher Outcome 1The lines and make angles of a and b with the positive direction of the x-axis, as shown in the diagram.a) Find the values of a and bb) Hence find the acute angle between the two given lines.

2m

Find supplement of b 180 135 45

2 4y x 13x y

2 4y x

13x y 1m

Find a° tan 2 63a a

Find b° tan 1 135b b

angle between two lines

Use angle sum triangle = 180°

72°


45o

72o

63o 135o

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Higher Outcome 1

Triangle ABC has vertices A(2, 2), B(12, 2) and C(8, 6). a) Write down the equation of l1, the perpendicular bisector of AB b) Find the equation of l2, the perpendicular bisector of AC.

c) Find the point of intersection of lines l1 and l2. 7, 2Mid-point AB

Find mid-point AC

(5, 4) Find gradient of AC 23

m

Equation of perp. bisector AC

Gradient AC perp. 32

m 2 3 23y x

Point of intersection (7, 1)

7x Perpendicular bisector AB

Exam Type Exam Type QuestionsQuestions l1

l2

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Higher Outcome 1

A triangle ABC has vertices A(–4, 1), B(12,3) and C(7, –7). a) Find the equation of the median CM. b) Find the equation of the altitude AD. c) Find the co-ordinates of the point of intersection of CM and AD

4, 2Mid-point AB

Equation of median CM using y – b = m(x – a)

Gradient of perpendicular AD

Gradient BC 2m 12

m

Equation of AD using y – b = m(x – a)

3m Gradient CM (median) 3 14 0 y x

Solve simultaneously for point of intersection(6, -4)2 2 0y x

Exam TypeExam TypeQuestionsQuestions

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Higher Outcome 1A triangle ABC has vertices A(–3, –3), B(–1, 1) and C(7,–3). a) Show that the triangle ABC is right angled at B. b) The medians AD and BE intersect at M. i) Find the equations of AD and BE. ii) Find find the co-ordinates of M.

2m Gradient AB

Product of gradients

Gradient of median AD

Mid-point BC 3, 1 13

m Equation AD

12

m Gradient BC12 1

2

Solve simultaneously for M, point of intersection

3 6 0y x

Hence AB is perpendicular to BC, so B = 90°

Gradient of median BE

Mid-point AC 2, 3 43

m Equation AD

3 4 1 0y x

51,3

Exam Type Exam Type QuestionsQuestions M

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Higher Outcome 1Are you on Target !

• Update you log book

• Make sure you complete and correct ALL of the Straight Line questions in the past paper booklet.

s5 unit 1- the straight line

Education