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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
C. 13
D. 3
Key Outcome Grade Facility Disc. Calculator Content SourceC 1.1 C 0.7 0.62 NC G2, G5 HSN 089
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2. What is the distance, in units, between the points (−1, 2) and (4, 5)?
A.√
8
B.√
16
C.√
34
D.√
58
Key Outcome Grade Facility Disc. Calculator Content SourceC 1.1 C 0.64 0.5 NC G1 HSN 054
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3. What is the distance, in units, between the points (a, b) and (−b, a)?
A.√
2√
a2 + b2
B.√
2(a + b)
C.√
2(√
a +√
b)
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
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4. The line through the points (−2, 5) and (7, a) has gradient 3.
What is the value of a?
A. 8
B. 22
C. 28
D. 32
Key Outcome Grade Facility Disc. Calculator Content SourceD 1.1 C 0.58 0.16 NC G2 HSN 05
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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
C. 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
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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.
What is the value of a?
A. −6
B. − 32
C. 32
D. 6
Key Outcome Grade Facility Disc. Calculator Content SourceC 1.1 C 0.53 0.41 NC G2, G5 HSN 151
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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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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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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
Key Outcome Grade Facility Disc. Calculator Content SourceB 1.1 C 0.82 0.43 NC G6 HSN 079
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10. Triangle ABC is shown below.PSfrag replacementsOxy
A
B
CD
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
Key Outcome Grade Facility Disc. Calculator Content SourceD 1.1 C 0.74 0.37 NC G7 HSN 068
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11. Triangle ABC with vertices A(−4, 1) , B(4, 3) and C(1, 5) is shown below.PSfrag replacements
O x
y
A
B
C
M
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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12. Triangle ABC with vertices A(6, 7) , B(7, 0) and C(−1,−2) is shown below.
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Ox
y A
BC
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
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[END OF PAPER 1 SECTION A]
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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
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14.[SQA] The points A and B have coordinates (a, a2) and (2b, 4b2) respectively. Determinethe gradient of AB in its simplest form. 2
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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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16.[SQA]
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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
O x
yA(1, 8)
B(0,−2)
C(−3,−4)
D(
− 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
10 )
•7 dBD = 75√
10•8 dCM = 9
10√
10•9 Area = 63
10 sq. units
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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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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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Ox
y
A(1, 2) B(8, 2)
C(4, 6)
D
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
3 )
•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
3 )
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 = −
√3
a = 1√3
•3 a = −3•4 (0, 1
3 )
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21. The line AC is the diameter of a circle with B lying on the circumference as shownbelow.
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B(3, 7)
C
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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22. The diagram below shows the right-angled triangle OPQ and a circle with centreC(0, 5) and diameter OS.
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O x
y
R
S
C
P
Q
3
6
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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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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O x
y
C
A
B
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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24. Triangle ABC is shown in the diagram below.
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O x
y
A
B(
2√
3, 7)
C
A and C lie on the x -axis, and B is the point(
2√
3, 7)
.
(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√
33
•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)
•3 A(2√
3 − 14, 0), C(− 1√3 , 0)
•4 AC = 14 − 7√
33
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
2
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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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O x
y
C
A
B
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
5
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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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28. Triangle PQR is shown in the diagram.
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O x
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
16
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30. Triangle ABC is shown in the diagram.
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O x
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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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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2.[SQA]
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3.[SQA]
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4.[SQA]
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5.[SQA]
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6.[SQA]
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7.[SQA]
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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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9.[SQA]
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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
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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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12.[SQA]
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13.[SQA]
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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
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Ox
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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15.[SQA]
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16.[SQA]
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17.[SQA]
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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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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 =
√20
•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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O x
y
A
B
C
D
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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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
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O x
y
A
B
Ca
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
3
•4 tan a = − 13
•5 a = 5·96 to 2 d.p.
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22. The line ` passes through points A(−2, 6) and B(5,−1) , as shown below.
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O x
y
A(−2, 6)
B(5,−1)
C
P
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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23. Find the equation of the straight line passing through the point(
2√
3, 7)
, with agradient of 1
2 . 2
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
√3)
24. The line with equation x + 2y − 6 = 0 makes an angle of φ◦ with the x -axis asshown below.
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O x
y
φ◦
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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25. The points R(1, 2) , S(5, 8) and T(11, 4) lie on the circumference of a circle.
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O x
y
R
S
T
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
•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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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
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P(1, 4)
Q(6, 14)
R(7, 6)
T
S
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
•7 ss: juxtaposition of two equations•8 pd: solve for one variable•9 pd: solve for other variable
•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
4
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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
•7 ss: juxtaposition of two equations•8 pd: solve for one variable•9 pd: solve for other variable
•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
4
[END OF PAPER 2]
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