Straight Line - mrsomersmaths · the gradient of AB in its simplest form. 2 PSfrag replacements O x...

Higher Mathematics

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Straight Line

Paper 1 Section A

Each correct answer in this section is worth two marks.1. The line with equation y = ax + 4 is perpendicular to the line with equation

3x + y + 1 = 0.

What is the value of a?

A. −3

B. − 13

Key Outcome Grade Facility Disc. Calculator Content SourceC 1.1 C 0.7 0.62 NC G2, G5 HSN 089

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Questions marked ‘[SQA]’ c© SQAAll others c© Higher Still Notes

Higher Mathematics

2. What is the distance, in units, between the points (−1, 2) and (4, 5)?

Key Outcome Grade Facility Disc. Calculator Content SourceC 1.1 C 0.64 0.5 NC G1 HSN 054

3. What is the distance, in units, between the points (a, b) and (−b, a)?

a2 + b2

2(a + b)

a +√

D. 2√

a2 + b2

Key Outcome Grade Facility Disc. Calculator Content SourceA 1.1 C 0.3 0.23 CN G1 HSN 011

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Higher Mathematics

4. The line through the points (−2, 5) and (7, a) has gradient 3.

Key Outcome Grade Facility Disc. Calculator Content SourceD 1.1 C 0.58 0.16 NC G2 HSN 05

5. The equation of a line is 3y = ax + 1 where a 6= 0 is a constant.

Given that the line has a gradient of 75 , what is the value of a?

A. − 215

B. − 75

D. 215

Key Outcome Grade Facility Disc. Calculator Content SourceD 1.1 C 0.5 0.64 NC G2, G4 HSN 162

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Higher Mathematics

6. The line with equation y = − 3a x + 4, where a 6= 0 is a constant, is perpendicular

to the line with equation y =12 x + 1.

A. −6

B. − 32

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Higher Mathematics

7. The line l passes through (3,−2) and is parallel to the line with equationy = 1

2 x + 5.

What is the equation of l ?

A. x − 2y + 1 = 0

B. x − 2y − 7 = 0

C. x − 2y + 7 = 0

D. x − 2y − 5 = 0

Key Outcome Grade Facility Disc. Calculator Content SourceB 1.1 C 0.57 0.39 CN G3 HSN 06

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Higher Mathematics

8. Find the equation of the line passing through (6,−4) and parallel to the line withequation 2x − 3y − 1 = 0.

A. 2x − 3y − 24 = 0

B. 3x + 2y − 10 = 0

C. 2x − y − 16 = 0

D. 2x − 3y − 18 = 0

Key Outcome Grade Facility Disc. Calculator Content SourceA 1.1 C 0.63 0.33 NC G3, G2 HSN 158

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Higher Mathematics

9. Given that (1, 0) is the midpoint of A(−3, a) and B(b, 2) , what are the values of aand b?

a bA. −2 4

B. −2 5

C. 2 −5

D. 4 −2

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10. Triangle ABC is shown below.PSfrag replacementsOxy

Here are two statements about the line BD:

I. BD is an altitude of triangle ABC

II. BD is the perpendicular bisector of AC

Which of the following is true?

A. neither statement is correct

B. only statement I is correct

C. only statement II is correct

D. both statements are correct

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Higher Mathematics

11. Triangle ABC with vertices A(−4, 1) , B(4, 3) and C(1, 5) is shown below.PSfrag replacements

Point M(0, 2) is the midpoint of AB. What is the equation of the median throughC?

A. 3x − y + 2 = 0

B. x − 4y + 8 = 0

C. 4x + y − 2 = 0

D. 3x − y − 1 = 0

Key Outcome Grade Facility Disc. Calculator Content SourceA 1.1 C 0.78 0.38 CN G7 HSN 138

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Higher Mathematics

12. Triangle ABC with vertices A(6, 7) , B(7, 0) and C(−1,−2) is shown below.

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The line through C and B has gradient 14 . Find the equation of the altitude

through A.

A. 4x + y − 11 = 0

B. x − 4y + 22 = 0

C. 4x + y − 31 = 0

D. 8x − 3y − 27 = 0

Key Outcome Grade Facility Disc. Calculator Content SourceC 1.1 C 0.55 0.67 CN G7, G5 HSN 128

[END OF PAPER 1 SECTION A]

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Higher Mathematics

Paper 1 Section B13.[SQA] Three lines have equations 2x + 3y − 4 = 0, 3x − y − 17 = 0 and x − 3y − 10 = 0.

Determine whether or not these lines are concurrent. 4

14.[SQA] The points A and B have coordinates (a, a2) and (2b, 4b2) respectively. Determinethe gradient of AB in its simplest form. 2

15.[SQA] Find the equation of the straight line which is parallel to the line with equation2x + 3y = 5 and which passes through the point (2,−1) . 3

Part Marks Level Calc. Content Answer U1 OC13 C CN G3, G2 2x + 3y = 1 2001 P1 Q1

•1 ss: express in standard form•2 ic: interpret gradient•3 ic: state equation of straight line

•1 y = − 23 x + 5

3 stated or implied by •2

•2 mline = − 23 stated or implied by •3

•3 y − (−1) = − 23 (x − 2)

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Higher Mathematics

16.[SQA]

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Higher Mathematics

17. The kite ABCD has vertices A(1, 8) , B(0,−2) ,C(−3,−4) and D

− 215 , k

as shown in thediagram.(a) Determine the value of k . 5

(b) Find the area of triangle BCD. 4PSfrag replacements

yA(1, 8)

B(0,−2)

C(−3,−4)

− 215 , k

Part Marks Level Calc. Content Answer U1 OC1(a) 5 C CN CGD, G2, G5 k = −3/5 AT100(b) 4 C CN CGD, G6, G1 63

10 sq. units

•1 pd: compute mAC•2 pd: compute mBD•3 ss: know to use m × m⊥ = −1•4 pd: form equation•5 pd: process for k

•6 ss: find midpoint of BD•7 pd: find length of base•8 pd: find length of altitude•9 ic: complete

•1 mAC = 3•2 mBD = − 5

21 (k + 2)•3 mAC × mBD = −1•4 −3 × 5

21 (k + 2) = −1•5 k = −3/5

•6 M = midptBD = (− 2110 ,− 13

•7 dBD = 75√

10•8 dCM = 9

10•9 Area = 63

10 sq. units

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Higher Mathematics

18. (a) The line l1 passes through the point (1, 10) and is perpendicular to the linewith equation x + 2y = 1.Find the equation of l1 . 3

(b) The line l2 passes through the point (6, 5) and makes an angle a with thepositive direction of the x -axis, where tan a = 1

3 .Find the equation of l2 . 2

(c) Determine the coordinates of the point of intersection of l1 and l2 . 3

Part Marks Level Calc. Content Answer U1 OC1(a) 3 C CN G5, G4, G3 2x − y + 8 = 0 AT001(b) 2 C CN G2, G3 x − 3y + 9 = 0(c) 3 C CN G8 (−3, 2)

•1 ic: extract gradient•2 pd: use m × m⊥ = −1•3 ic: state equation of line

•4 ss: use m = tan θ

•5 ic: state equation of line

•6 ss: know to solve simultaneously•7 pd: solve for one unknown•8 pd: solve for second unknown

•1 y = − 12 x + 1

2•2 ml1 = 2•3 y − 10 = 2(x − 1)

•4 ml2 = 13

•5 y − 5 = 13 (x − 6)

•6 2x − y + 8 = 0; x − 3y + 9 = 0•7 x = −3•8 y = 2

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Higher Mathematics

19. Triangle ABC has vertices A(1, 2) , B(8, 2) andC(4, 6) .(a) Find the equation of the line through the

collinear points A, C and D. 2

(b) Triangle ABD is right-angled at B. Find theequation of BD. 1

(c) Find the coordinates of D. 2

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A(1, 2) B(8, 2)

C(4, 6)

Part Marks Level Calc. Content Answer U1 OC1(a) 2 C CN G2, G3 4x − 3y + 2 = 0 OB 01-011(b) 1 C CN G3 x = 8(c) 2 C NC G8 D(8, 11 1

•1 ss: find mAC•2 pd: find equation of side AC/AD

•3 pd: equation of BD

•4 ss: know to substitute•5 ic: state coordinates

•1 mAC = 43

•2 y − 2 = 43 (x − 1)

•3 x = 8

•4 4(8) − 3y + 2 = 0•5 D(8, 11 1

20. The equation of a straight line is√

3x + ay + 1 = 0, where a 6= 0 is a constant.

Given that this line has a gradient of 1√3 , find the value of a , and hence state the

coordinates of the point where the line cuts the y-axis. 4

Part Marks Level Calc. Content Answer U1 OC14 C NC G2, G4 a = −3, (0, 1

3 ) OB 01-007

•1 ss: express in standard form•2 ic: interpret gradient•3 pd: complete for a•4 ic: interpret intercept

•1 y = −√

3a x − 1

a•2 m = −

a = 1√3

•3 a = −3•4 (0, 1

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Higher Mathematics

21. The line AC is the diameter of a circle with B lying on the circumference as shownbelow.

PSfrag replacementsOxy A(−1, 5)

B(3, 7)

A is the point (−1, 5) and B(3, 7) .

Find the equation of the chord BC. 3

Part Marks Level Calc. Content Answer U1 OC13 C CN G2, G5, G3 2x + y − 13 = 0 OB 01-003

•1 pd: compute mAB•2 ss: use mAB × mBC = −1•3 ic: state equation

•1 mAB = 12

•2 mBC = −2•3 y − 7 = −2(x − 3)

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Higher Mathematics

22. The diagram below shows the right-angled triangle OPQ and a circle with centreC(0, 5) and diameter OS.

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Find the equation of the chord RS. 5

Part Marks Level Calc. Content Answer U1 OC15 B CN G2, G5, G3 2x + y − 10 = 0 Ex 1-1-7

•1 pd: find gradient of OQ•2 ic: use the fact that OQ ⊥ RS•3 pd: find gradient of RS•4 ic: coordinates of S•5 ic: state equation of RS

•1 mOQ = 12

•2 OQ ⊥ RS•3 so mRS = −2•4 S(0, 10)•5 y − 10 = −2(x − 0)

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Higher Mathematics

23. Triangle ABC has vertices A(−3, 5) , B(9, 9) andC(9,−3) .(a) Write down the equation of BC. 1

(b) Find the equation of the altitude from A. 2

(c) Find the equation of the perpendicularbisector of AB. 4

(d) Find where the perpendicular bisector of ABand the altitude from A intersect. 2

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Part Marks Level Calc. Content Answer U1 OC1(a) 1 C CN G3 x = 9 AT077(b) 2 C CN G7 y = 5(c) 4 C CN G7 3x + y − 16 = 0(d) 2 C CN G8

( 113 , 5

•1 ic: state equation of vertical line

•2 ss: use m × m⊥ = −1•3 ic: state equation of line

•4 pd: find gradient of AB•5 ss: use m × m⊥ = −1•6 ss: find midpoint•7 ic: state equation of line

•8 ss: start to solve equation•9 pd: complete

•1 x = 9

•2 malt. = 0•3 y = 5

•4 mAB = 1/3•5 m⊥ = −3•6 midptAB = (3, 7)•7 y − 7 = −3(x − 3)

•8 −3x + 16 = 5•9 ( 11

3 , 5)

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Higher Mathematics

24. Triangle ABC is shown in the diagram below.

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A and C lie on the x -axis, and B is the point(

(a) The line AB has a gradient of 12 . Find the equation of AB. 2

(b) BC is part of the line with equation√

3x − y + 1 = 0.Find the length of AC. 2

Part Marks Level Calc. Content Answer U1 OC1(a) 2 C CN G3 x − 2y + 14 − 2

√3 = 0 OB 01-006

(b) 2 C CN G4, G1 14 − 7√

•1 ic: identify point and gradient•2 ic: state equation

•3 ss: find A and C•4 pd: compute distance

•1 m = 12 , pt (2

√3, 7) stated or implied

by •2

•2 y − 7 = 12 (x − 2

•3 A(2√

3 − 14, 0), C(− 1√3 , 0)

•4 AC = 14 − 7√

25. The line with equation 3x + ay + 1 = 0, where a is a constant, is perpendicular tothe line with equation 2y − x = 2.

Find the value of a . 3

Part Marks Level Calc. Content Answer U1 OC13 C CN G4, G5 a = 3

2 OB 01-009

•1 ic: express in standard form•2 ss: use m × m⊥ = −1•3 pd: complete for a

•1 y = 12 x + 1 and y = − 3

a x − 13

•2 12 × 3

a = −1•3 a = 3

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Higher Mathematics

26. Triangle ABC has vertices A(4, 7) , B(−2, 1)and C(6,−3) .(a) Find the equation of line p , the median

from A. 3

(b) Find the equation of line q , the altitudefrom C. 3

(c) Find the point of intersection of the linesp and q . 3

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Part Marks Level Calc. Content Answer U1 OC1(a) 3 C CN G7 4x − y − 9 = 0 AT075(b) 3 C CN G7 x + y − 3 = 0(c) 3 C CN G8 ( 12

5 , 35 )

•1 ss: find midpoint of BC•2 pd: calculate gradient•3 ic: state equation of line

•4 pd: find gradient of AB•5 ss: use m × m⊥ = −1•6 ic: state equation of line

•7 ss: start to solve simultaneousequations

•8 pd: solve for x•9 pd: solve for y

•1 midptBC = (2,−1)•2 mp = 4•3 y − 7 = 4(x − 4)

•4 mAB = 1•5 mq = −1•6 y + 3 = −1(x − 6)

•7 4x − 9 = −x + 3•8 x = 12

5•9 y = 3

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Higher Mathematics

27. The line with equation 4y + 3x − 24 = 0 intersects the x -axis at P and the y-axisat R.

(a) Write down the coordinates of P and R. 1

(b) The perpendicular bisector of PR meets the line y = −1 at Q.Find the coordinates of Q. 5

(c) Show that P, Q and R could be three vertices of a square. 3

Part Marks Level Calc. Content Answer U1 OC1(a) 1 C CN A6 P(8, 0), R(0, 6) AT046(b) 5 C CN G7, G8 Q(1,−1)

(c) 3 B CN G1, G2, G5 proof

•1 ic: find coordinates

•2 ss: find midpoint•3 pd: find gradient•4 ss: use m × m⊥ = −1•5 ic: state equation of line•6 pd: find intersection

•7 ss: show sides have same length•8 pd: find gradients of sides•9 ss: use m × m⊥ = −1

•1 P(8, 0), R(0, 6)

•2 midptPR = (4, 3)•3 mPR = − 3

4•4 m⊥ = 4

3•5 y − 3 = 4

3 (x − 4)•6 Q(1,−1)

•7 PQ = QR =√

50•8 mPQ = 1

7 , mQR = −7•9 mPQ × mQR = −1 so ⊥

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Higher Mathematics

28. Triangle PQR is shown in the diagram.

PSfrag replacements

yP(1, 3)

Q(7,−2)R

The line PR has equation 6x + y − 9 = 0, and QR has equation x − 5y − 17 = 0.

(a) Find the coordinates of point R. 2

(b) Hence find the equation of the median through R. 3

Part Marks Level Calc. Content Answer U1 OC1(a) 2 C NC G8 R(2,−3) OB 01-014(b) 3 C NC G6, G7 7x − 4y − 26 = 0

•1 ss: solve simultaneously•2 pd: obtain R

•3 ss: method for medians/findmidpointPQ

•4 pd: hence find mmedian•5 ic: obtain equation of median

•1 start to eliminate a variable•2 R(2,−3)

•3 midpointPQ = (4, 12 )

•4 mmedian = 74

•5 y + 3 = 74 (x − 2)

29. The points A(3, 2) , B(2a, 12) and C(a,−1) are collinear.

Find the value of the constant a . 4

Part Marks Level Calc. Content Answer U1 OC14 C CN G8, G2 a = 39

16 OB 01-001

•1 ss: use collinearity fact•2 pd: expressions for gradients•3 pd: equate and process•4 pd: complete

•1 mAB = mAC•2 mAB = 10

2a−3 , mAC = −3a−3

•3 10(a − 3) = −3(2a − 3)•4 a = 39

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Higher Mathematics

30. Triangle ABC is shown in the diagram.

PSfrag replacements

yA(−4, 6)

B(4,−3)C

The line AC has equation 8x + 2y = −20, and BC has equation x + 6y = −14.

Find the equation of the altitude through C. 5

Part Marks Level Calc. Content Answer U1 OC15 C CN G8, G7 8x − 9y − 2 = 0 OB 01-013

•1 ss: solve simultaneously•2 pd: obtain C•3 ss: method for altitudes/find mAB•4 pd: hence find malt.•5 ic: obtain equation of altitude

•1 e.g. BC − 3AC gives −3x = 46, etc•2 C(−2,−2)•3 mAB = − 9

8•4 malt. = 8

9•5 y + 2 = 8

9 (x + 2)

[END OF PAPER 1 SECTION B]

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Higher Mathematics

Paper 21.[SQA] Find the equation of the perpendicular bisector of the line joining A(2,−1) and

B(8, 3) . 4

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Higher Mathematics

2.[SQA]

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Higher Mathematics

3.[SQA]

4.[SQA]

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Higher Mathematics

5.[SQA]

6.[SQA]

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Higher Mathematics

7.[SQA]

8.[SQA] Find the equation of the line through the point (3,−5) which is parallel to the linewith equation 3x + 2y − 5 = 0. 2

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Higher Mathematics

9.[SQA]

10.[SQA] The vertices of a triangle are P(−1, 1) , Q(2, 1) and R(−6, 2) . Find the equation ofthe altitude of triangle PQR, drawn from P. 3

11.[SQA] Find the equation of the median AD of triangle ABC where the coordinates of A,B and C are (−2, 3) , (−3,−4) and (5, 2) respectively. 3

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Higher Mathematics

12.[SQA]

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Higher Mathematics

13.[SQA]

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Higher Mathematics

14.[SQA] Triangle 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. 3

(b) the equation of the line q , theperpendicular bisector of BC. 4

(c) the coordinates of the point ofintersection of the lines p and q . 1

PSfrag replacements

yA(−1, 6)

B(−3,−2)

C(5, 2)

Part Marks Level Calc. Content Answer U1 OC1(a) 3 C CN G7 y = 2 2002 P2 Q1(b) 4 C CN G7 y = −2x + 2(c) 1 C CN G8 (0, 2)

•1 ss: determine midpoint coordinates•2 pd: determine gradient thro’ 2 pts•3 ic: state equation of straight line

•4 ss: determine midpoint coordinates•5 pd: determine gradient thro’ 2 pts•6 ss: determine gradient perp. to •5

•7 ic: state equation of straight line

•8 pd: process intersection

•1 F = midAB = (−2, 2)•2 mFC = 0 stated or implied by •3

•3 equ. FC is y = 2

•4 M = midBC = (1, 0)•5 mBC = 1

2•6 m⊥ = −2•7 y − 0 = −2(x − 1)

•8 (0, 2)

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Higher Mathematics

15.[SQA]

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Higher Mathematics

16.[SQA]

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Higher Mathematics

17.[SQA]

18.[SQA] P, Q and R have coordinates (1,−2) , (6, 3) and (9, 14) respectively and are threevertices of a kite PQRS.

(a) Find the equations of the diagonals of this kite and the coordinates of the pointwhere they intersect. 7

(b) Find the coordinates of the fourth vertex S. 2

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Higher Mathematics

19. Points A(2, 3) and C(8, 11) are the end points of a diagonal of rectangle ABCD.

The diagonal BD is parallel to the x -axis.

Find the area of the rectangle. 5

Part Marks Level Calc. Content Answer U1 OC15 B CN G1, G6, G8 40 sq. units AT045

•1 pd: find length of diagonals•2 ic: find midpoint•3 ic: interpret midpoint•4 pd: compute side lengths•5 ic: complete

•1 AC = BD = 10•2 midptBD = midptAC = (5, 7)•3 B(0, 7) or D(10, 7)•4 BC =

√80, AB =

•5 Area = 40 sq. units

20. The diagram shows a rhombus ABCD.The points B and D have coordinates(3, 5) and (9, 3) respectively.The equation of AD is x − y = 6.Find the size of angle CAD. 6

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Part Marks Level Calc. Content Answer U1 OC12 A CR G2 AT0244 C CR G5, G2 26·6◦

•1 pd: find mBD•2 ss: use mAC × mBD = −1•3 ic: interpret gradient of AD•4 ss: use m = tan θ

•5 pd: find both angles•6 ic: complete

•1 mBD = − 13

•2 mAC = 3•3 mAD = 1•4 CAx = tan−1 3 = 71·6◦•5 DAx = 45◦•6 26·6◦

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Higher Mathematics

21. The diagram shows triangle ABC.A circle passes through all the vertices of thetriangle.AC is a diameter of the circle.The equation of AB is y − 3x = 2.(a) Find the gradient of BC. 3

(b) BC makes an angle of a radians with thepositive direction of the x -axis.Find the value of a . 2

PSfrag replacements

Part Marks Level Calc. Content Answer U1 OC1(a) 3 C CN G2, G5 mBC = − 1

3 AT047(b) 2 C CR G2 a = 161·6 to 1 d.p.

•1 ic: extract gradient•2 ic: interpret geometry•3 ss: use m × m⊥ = −1

•4 ss: use m = tan θ

•5 pd: complete

•1 mAB = 3•2 ABC = π/2•3 mBC = − 1

•4 tan a = − 13

•5 a = 5·96 to 2 d.p.

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Higher Mathematics

22. The line ` passes through points A(−2, 6) and B(5,−1) , as shown below.

PSfrag replacements

A(−2, 6)

B(5,−1)

The line through O and P is perpendicular to ` , and C lies on the x -axis.

(a) Find the equation of the line through O and P. 3

(b) Hence find the coordinates of P. 3

(c) Calculate the area of the triangle OCP. 3

Part Marks Level Calc. Content Answer U1 OC1(a) 3 C CN G2, G5, G3 y = x OB 01-004(b) 3 C CN G3, G8 P(2, 2)

(c) 3 C CN G1 Area = 4

•1 pd: compute mAB•2 ss: use mAB × mOP = −1•3 ic: complete

•4 ss: find equation of `

•5 pd: find point of intersection•6 pd: complete

•7 ss: find C•8 ic: height comes from P•9 pd: compute area

•1 mAB = −1•2 mOP = 1•3 y = x

•4` has eqn x + y − 4 = 0

•5 Solve x + y − 4 = 0 and y = x•6 P(2, 2)

•7 C(4, 0) so OC = 4 units•8 Height of 4OPC is 2•9 Area = 1

2 × 4 × 2

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Higher Mathematics

23. Find the equation of the straight line passing through the point(

, with agradient of 1

Part Marks Level Calc. Content Answer U1 OC12 C CN G3 x − 2y + 14 − 2

√3 = 0 OB 01-008

•1 ic: identify point and gradient•2 ic: state equation

•1 m = 12 , pt (2

√3, 7) stated or implied

by •2

•2 y − 7 = 12 (x − 2

24. The line with equation x + 2y − 6 = 0 makes an angle of φ◦ with the x -axis asshown below.

PSfrag replacements

x + 2y − 6 = 0

Calculate the value of φ correct to two decimal places. 4

Part Marks Level Calc. Content Answer U1 OC14 B CR G4, G2 φ = 153·43 OB 01-002

•1 ic: extract gradient of line•2 ss: use m = tan θ

•3 pd: compute angle•4 ic: complete

•1 y = − 12 x + 3 ⇒ m = − 1

2•2 tan θ◦ = − 1

2•3 θ = 153·43•4 φ = θ = 153·43

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Higher Mathematics

25. The points R(1, 2) , S(5, 8) and T(11, 4) lie on the circumference of a circle.

PSfrag replacements

3x − 2y − 12 = 0

The line with equation 3x − 2y − 12 = 0 is the perpendicular bisector of ST.

(a) Find the equation of the perpendicular bisector of RS. 4

(b) The centre of the circle is the point where the perpendicular bisectors of RSand ST intersect.Calculate the coordinates of the centre of the circle. 3

Part Marks Level Calc. Content Answer U1 OC1(a) 4 C CN G7, G5, G6 2x + 3y − 21 = 0 Ex 1-1-4(b) 3 C CN G8 (6, 3)

•1 pd: find gradient of line•2 ss: find gradient of perp. line•3 pd: find midpoint•4 ic: state equation of perp. bisector

•5 ss: juxtaposition of two equations•6 pd: solve for one variable•7 pd: solve for other variable

•1 mRS =8 − 25 − 1 = 3

•2 m⊥ = − 23

•3 midpointRS = (3, 5)•4 y − 5 = − 2

3 (x − 3)

•5 2x + 3y − 21 = 0 and3x − 2y − 12 = 0

•6 x = 6•7 y = 3

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Higher Mathematics

26. Triangle PQR has vertices P(1, 4) , Q(6, 14) andR(7, 6) .(a) Find the equation of the median QS. 3

(b) Find the equation of the altitude RT. 3

(c) The median QS and the altitude RT intersect atA.Find the coordinates of A. 3

P(1, 4)

Q(6, 14)

R(7, 6)

Part Marks Level Calc. Content Answer U1 OC1(a) 3 C CN G7, G6, G3 9x − 2y − 26 = 0 WCHS U1 Q1(b) 3 C CN G7, G5, G3 x + 2y − 19 = 0(c) 3 C CN G8 A( 9

2 , 294 )

•1 ic: interpret “median”•2 ss: find gradient of median•3 ic: state equation of median

•4 ss: know to find gradient of “base”•5 ss: find perpendicular gradient•6 ic: state equation of altitude

•1 S(4, 5) (midpoint of PR)•2 mmedian = 14−5

6−4 = 92

•3 y − 14 = 92 (x − 6)

•4 mPQ = 4−141−6 = 2

•5 malt. = − 12

•6 y − 6 = − 12 (x − 7)

•7 9x − 2y − 26 = 0 and x + 2y − 19 = 0•8 xA = 9

2•9 yA = 29

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Higher Mathematics

27. The vertices of the triangle PQR are P(2, 6) , Q(−4,−4) and R(−3, 7) .

(a) Find the equation of the median through R. 3

(b) Find the equation of the altitude through Q. 3

(c) The median through R and the altitude through Q intersect at point T.Calculate the coordinates of T. 3

Part Marks Level Calc. Content Answer U1 OC1(a) 3 C CN G7, G6, G3 y = −3x − 2 Ex 1-1-3(b) 3 C CN G7, G5, G3 y = 5x + 16(c) 3 C CN G8 T(− 9

4 , 194 )

•1 ic: interpret “median”•2 ss: find gradient of median•3 ic: state equation of median

•4 ss: know to find gradient of “base”•5 ss: find perpendicular gradient•6 ic: state equation of altitude

•1 M(−1, 1) (midpoint of PQ)•2 mmedian = 7−1

−3−(−1) = −3•3 y − 7 = −3(x + 3)

•4 mPR = 6−72−(−3) = − 1

5•5 malt. = 5•6 y + 4 = 5(x + 4)

•7 y = −3x − 2 and y = 5x + 16•8 xT = − 9

4•9 yT = 19

[END OF PAPER 2]

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Straight Line - mrsomersmaths · the gradient of AB in its simplest form. 2 PSfrag replacements O x...

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