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Solutions — Full Answer Key

Ecolint Campus des NationsMathematics
Year 10 · Coordinate Geometry

Solutions · Full Answer Key

Pack A answers · Pack B answers · Problem-solving worked solutions

Pack A — Answers

Bronze
1.(5,4)(5, 4)
2.(6,6)(6, 6)
3.m=2m = 2
4.m=−32m = -\dfrac{3}{2}
5.7
6.yy-intercept = 5 (point (0,5)(0, 5))
7.m=4m = 4
8.y=2x+5y = 2x + 5
9.Yes
10.y=11y = 11
Silver
11.5
12.5
13.(1,1)(1, 1)
14.m=−1m = -1
15.y=3x−1y = 3x - 1
16.y=2x+1y = 2x + 1
17.Gradient = 4
18.Gradient = −12-\dfrac{1}{2}
19.y=−2x+6y = -2x + 6
20.x=3x = 3 (point (3,0)(3, 0))
Gold
21.10
22.Undefined (vertical line)
23.y=7y = 7
24.y=−12x+8y = -\dfrac{1}{2}x + 8
25.B=(8,7)B = (8, 7)
26.m=2m = 2
27.Yes (all on y=3x−1y = 3x - 1)
28.y=−23x+173y = -\dfrac{2}{3}x + \dfrac{17}{3}
29.(2,5)(2, 5)
30.mCD=34m_{CD} = \dfrac{3}{4} (parallel to ABAB)
Platinum
31.y=−34x+9y = -\dfrac{3}{4}x + 9
32.AB⊥ACAB \perp AC since ABAB is horizontal and ACAC is vertical
33.Isosceles (AC=BC=5AC = BC = 5); area = 12
34.k=11k = 11
35.D=(3,5)D = (3, 5)
36.x=2x = 2
37.k=8k = 8
38.P=(3,0)P = (3, 0)
39.(a) y=−32x+6y = -\dfrac{3}{2}x + 6 (b) 12
40.B=(7,1)B = (7, 1) or (−3,1)(-3, 1)

Pack B — Answers

Bronze
1.(6,5)(6, 5)
2.(4,7)(4, 7)
3.m=3m = 3
4.m=−2m = -2
5.8
6.yy-intercept = −4-4 (point (0,−4)(0, -4))
7.m=−3m = -3
8.y=3x−1y = 3x - 1
9.Yes
10.y=2y = 2
Silver
11.10
12.29\sqrt{29}
13.(−1,−3)(-1, -3)
14.m=2m = 2
15.y=2x−5y = 2x - 5
16.y=3x−5y = 3x - 5
17.Gradient = −2-2
18.Gradient = −13-\dfrac{1}{3}
19.y=−32x+6y = -\dfrac{3}{2}x + 6
20.x=4x = 4 (point (4,0)(4, 0))
Gold
21.10
22.Undefined (vertical line)
23.y=−3y = -3
24.y=−13x+7y = -\dfrac{1}{3}x + 7
25.B=(7,10)B = (7, 10)
26.m=3m = 3
27.Yes (all on y=x+1y = x + 1)
28.y=−34x+192y = -\dfrac{3}{4}x + \dfrac{19}{2}
29.(2,4)(2, 4)
30.mCD=2m_{CD} = 2 (parallel to ABAB)
Platinum
31.y=−13x+9y = -\dfrac{1}{3}x + 9
32.AB⊥ACAB \perp AC since ABAB is horizontal and ACAC is vertical
33.Isosceles (AC=BC=35AC = BC = 3\sqrt{5}); area = 18
34.k=−9k = -9
35.D=(2,7)D = (2, 7)
36.x=2x = 2
37.k=7k = 7
38.P=(2,0)P = (2, 0)
39.(a) y=−43x+8y = -\dfrac{4}{3}x + 8 (b) 24
40.B=(7,1)B = (7, 1) or (−3,1)(-3, 1)

Problem-solving — Worked Solutions

1Problem 1
Answer
(a) AB=6AB = 6, BC=5BC = 5, AC=5AC = 5 (b) Isosceles (c) 12
Full working
(a) AB=∣7−1∣=6AB = |7 - 1| = 6 (horizontal). BC=(4−7)2+(6−2)2=9+16=5BC = \sqrt{(4-7)^2 + (6-2)^2} = \sqrt{9 + 16} = 5. AC=(4−1)2+(6−2)2=9+16=5AC = \sqrt{(4-1)^2 + (6-2)^2} = \sqrt{9 + 16} = 5.

(b) AC=BC=5AC = BC = 5, AB=6AB = 6. Two sides equal → **isosceles**.

(c) Base AB=6AB = 6 on line y=2y = 2. Height = ∣6−2∣=4|6 - 2| = 4 (vertical distance from CC). Area =12(6)(4)=12= \frac{1}{2}(6)(4) = 12.
2Problem 2
Answer
(a) mAB=1/2m_{AB} = 1/2; mBC=2m_{BC} = 2 (b) D=(3,5)D = (3, 5) (c) 858\sqrt{5}
Full working
(a) mAB=3−15−1=12m_{AB} = \frac{3-1}{5-1} = \frac{1}{2}; mBC=7−37−5=2m_{BC} = \frac{7-3}{7-5} = 2.

(b) In parallelogram ABCDABCD (vertices in order), AD→=BC→\overrightarrow{AD} = \overrightarrow{BC}. So D=A+(C−B)=(1,1)+(2,4)=(3,5)D = A + (C - B) = (1, 1) + (2, 4) = (3, 5).

(c) ∣AB∣=16+4=20=25|AB| = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5}. ∣BC∣=4+16=25|BC| = \sqrt{4 + 16} = 2\sqrt{5}. (Both pairs equal in parallelogram.) Perimeter =2(∣AB∣+∣BC∣)=2(25+25)=85= 2(|AB| + |BC|) = 2(2\sqrt{5} + 2\sqrt{5}) = \mathbf{8\sqrt{5}}.
3Problem 3
Answer
y=−2x+9y = -2x + 9
Full working
Midpoint: M=(−2+62,3+72)=(2,5)M = \left(\frac{-2+6}{2}, \frac{3+7}{2}\right) = (2, 5). Gradient of ABAB: mAB=7−36−(−2)=48=12m_{AB} = \frac{7-3}{6-(-2)} = \frac{4}{8} = \frac{1}{2}. Perpendicular gradient: −2-2. Line through (2,5)(2, 5) with gradient −2-2: y−5=−2(x−2)=−2x+4y - 5 = -2(x - 2) = -2x + 4, so y=−2x+9\mathbf{y = -2x + 9}.
4Problem 4
Answer
Right-angled at QQ
Full working
Compute gradients:
- mPQ=1−(−1)5−(−1)=26=13m_{PQ} = \frac{1 - (-1)}{5 - (-1)} = \frac{2}{6} = \frac{1}{3}.
- mQR=4−14−5=−3m_{QR} = \frac{4 - 1}{4 - 5} = -3.
- mPR=4−(−1)4−(−1)=1m_{PR} = \frac{4 - (-1)}{4 - (-1)} = 1.

Check products:
- mPQ×mQR=13×(−3)=−1m_{PQ} \times m_{QR} = \frac{1}{3} \times (-3) = -1 ✓ → perpendicular at QQ.
- mPQ×mPR=13≠−1m_{PQ} \times m_{PR} = \frac{1}{3} \neq -1.
- mQR×mPR=−3≠−1m_{QR} \times m_{PR} = -3 \neq -1.

So the right angle is at **QQ**.
5Problem 5
Answer
(a) y=3x−7y = 3x - 7 (b) xx-int (7/3,0)(7/3, 0); yy-int (0,−7)(0, -7) (c) 49/6≈8.1749/6 \approx 8.17
Full working
(a) Point-slope: y−(−1)=3(x−2)y - (-1) = 3(x - 2), so y+1=3x−6y + 1 = 3x - 6, giving y=3x−7y = 3x - 7.

(b) yy-intercept: x=0⇒y=−7x = 0 \Rightarrow y = -7 → (0,−7)(0, -7). xx-intercept: y=0⇒3x=7⇒x=73y = 0 \Rightarrow 3x = 7 \Rightarrow x = \frac{7}{3} → (73,0)(\frac{7}{3}, 0).

(c) Triangle with legs 73\frac{7}{3} (on xx-axis) and 7 (on yy-axis, taking magnitudes): Area =12⋅73⋅7=496≈8.17= \frac{1}{2} \cdot \frac{7}{3} \cdot 7 = \mathbf{\frac{49}{6} \approx 8.17}.
6Problem 6
Answer
(a) (3,4)(3, 4) (b) y=−34x+254y = -\frac{3}{4}x + \frac{25}{4} (c) B=(7,1)B = (7, 1) and D=(−1,7)D = (-1, 7)
Full working
(a) Midpoint of ACAC: (0+62,0+82)=(3,4)(\frac{0+6}{2}, \frac{0+8}{2}) = (3, 4).

(b) Gradient of ACAC: 86=43\frac{8}{6} = \frac{4}{3}. Perpendicular gradient: −34-\frac{3}{4}. Through (3,4)(3, 4): y−4=−34(x−3)y - 4 = -\frac{3}{4}(x - 3), so y=−34x+94+4=−34x+254y = -\frac{3}{4}x + \frac{9}{4} + 4 = -\frac{3}{4}x + \frac{25}{4}.

(c) Diagonal ACAC length: 36+64=10\sqrt{36 + 64} = 10. So half-diagonal = 5 from centre. BB and DD lie on line BDBD at distance 5 from (3,4)(3, 4). Direction along BDBD: (cos⁡θ,sin⁡θ)(\cos\theta, \sin\theta) where gradient =−34= -\frac{3}{4}, so direction (45,−35)(\frac{4}{5}, -\frac{3}{5}) (unit vector). Points: B=(3+4,4−3)=(7,1)B = (3 + 4, 4 - 3) = (7, 1); D=(3−4,4+3)=(−1,7)D = (3 - 4, 4 + 3) = (-1, 7). Verify ∣AB∣=49+1=50=52|AB| = \sqrt{49 + 1} = \sqrt{50} = 5\sqrt{2} — wait, side of square should be 102=52\frac{10}{\sqrt{2}} = 5\sqrt{2} ✓.
7Problem 7
Answer
(a) y=−12x+5y = -\frac{1}{2}x + 5 (b) (1.6,4.2)(1.6, 4.2) (c) 3.2≈1.79\sqrt{3.2} \approx 1.79
Full working
(a) Perpendicular to gradient 2 → gradient −12-\frac{1}{2}. Through (0,5)(0, 5): y=−12x+5y = -\frac{1}{2}x + 5.

(b) Set 2x+1=−12x+52x + 1 = -\frac{1}{2}x + 5, so 52x=4\frac{5}{2}x = 4, x=85=1.6x = \frac{8}{5} = 1.6. Then y=2(1.6)+1=4.2y = 2(1.6) + 1 = 4.2. Intersection: (1.6,4.2)(1.6, 4.2).

(c) Distance from (0,5)(0, 5) to (1.6,4.2)(1.6, 4.2): 1.62+0.82=2.56+0.64=3.2≈1.79\sqrt{1.6^2 + 0.8^2} = \sqrt{2.56 + 0.64} = \sqrt{3.2} \approx \mathbf{1.79}.
8Problem 8
Answer
P′=(5,3)P' = (5, 3)
Full working
Reflection in y=xy = x swaps xx and yy coordinates. So P(3,5)→P′(5,3)P(3, 5) \to P'(5, 3).

**Check:** midpoint of PP′PP' is (4,4)(4, 4), which lies on y=xy = x ✓. Also, gradient of PP′PP' is 3−55−3=−1\frac{3-5}{5-3} = -1, perpendicular to y=xy = x (gradient 1) since 1×−1=−11 \times -1 = -1 ✓.
9Problem 9
Answer
(a) Both 5200≈72.1\sqrt{5200} \approx 72.1 m (b) (30,20)(30, 20) (c) 200 m
Full working
(a) Diagonal AC=602+402=3600+1600=5200≈72.1AC = \sqrt{60^2 + 40^2} = \sqrt{3600 + 1600} = \sqrt{5200} \approx 72.1 m. By symmetry, BDBD has the same length.

(b) In a rectangle, diagonals bisect each other, meeting at the centre. Centre = (0+602,0+402)=(30,20)(\frac{0+60}{2}, \frac{0+40}{2}) = (30, 20).

(c) Perimeter = 2(60+40)=2002(60 + 40) = 200 m.
10Problem 10
Answer
(a) y=3x−1y = 3x - 1 and y=−13x+173y = -\frac{1}{3}x + \frac{17}{3} (b) 90° (perpendicular) (c) ∣PQ∣=503≈16.67|PQ| = \frac{50}{3} \approx 16.67
Full working
(a) Line 1: y−5=3(x−2)⇒y=3x−1y - 5 = 3(x - 2) \Rightarrow y = 3x - 1. Line 2: y−5=−13(x−2)⇒y=−13x+23+5=−13x+173y - 5 = -\frac{1}{3}(x - 2) \Rightarrow y = -\frac{1}{3}x + \frac{2}{3} + 5 = -\frac{1}{3}x + \frac{17}{3}.

(b) Product of gradients: 3×(−13)=−13 \times (-\frac{1}{3}) = -1 → **perpendicular** (90°).

(c) Line 1 xx-intercept: 0=3x−1⇒x=130 = 3x - 1 \Rightarrow x = \frac{1}{3}. So P=(13,0)P = (\frac{1}{3}, 0).
Line 2 xx-intercept: 0=−13x+173⇒x=170 = -\frac{1}{3}x + \frac{17}{3} \Rightarrow x = 17. So Q=(17,0)Q = (17, 0).
∣PQ∣=17−13=51−13=503≈16.67|PQ| = 17 - \frac{1}{3} = \frac{51-1}{3} = \frac{50}{3} \approx 16.67.
11Problem 11
Answer
Parallelogram (opposite sides parallel & equal); NOT a rhombus (sides not all equal)
Full working
**Parallelogram check.**
- AB→=(4,0)\overrightarrow{AB} = (4, 0); DC→=(5−1,3−3)=(4,0)\overrightarrow{DC} = (5-1, 3-3) = (4, 0). So ABAB and DCDC are parallel and equal length.
- AD→=(1,3)\overrightarrow{AD} = (1, 3); BC→=(1,3)\overrightarrow{BC} = (1, 3). Parallel and equal.

Both pairs of opposite sides parallel and equal → **parallelogram** ✓.

**Rhombus check.** All four sides must be equal. ∣AB∣=4|AB| = 4; ∣AD∣=1+9=10≈3.16|AD| = \sqrt{1 + 9} = \sqrt{10} \approx 3.16. Not equal → **not a rhombus**.
12Problem 12
Answer
All three are at distance 5 from PP; they lie on a circle centred at PP with radius 5.
Full working
Distances from P(0,0)P(0, 0):
- ∣PA∣=0+25=5|PA| = \sqrt{0 + 25} = 5.
- ∣PB∣=9+16=25=5|PB| = \sqrt{9 + 16} = \sqrt{25} = 5.
- ∣PC∣=16+9=25=5|PC| = \sqrt{16 + 9} = \sqrt{25} = 5.

All three points are exactly 5 units from P(0,0)P(0, 0). **Significance:** the three points lie on a circle centred at the origin with radius 5, i.e. the circle x2+y2=25x^2 + y^2 = 25.