Year 9 Mathematics | Victorian Curriculum 2.0
Linear graphs
Topic 04 | Number & Algebra | Answer key

Year 9 core - answers

Fluency

Tier 1: gradient, intercept and sketching

    1. Gradient =5= 5, yy-intercept =−2= -2.
    2. Gradient =−3= -3, yy-intercept =7= 7.
    3. y=−2x+10y = -2x + 10.
    4. y=2x−3y = 2x - 3.
    5. xx-intercept: (4,0)(4, 0); yy-intercept: (0,10)(0, 10).
    6. m=11−24−1=93=3m = \dfrac{11 - 2}{4 - 1} = \dfrac{9}{3} = 3.
    7. m=−7−51−(−3)=−124=−3m = \dfrac{-7 - 5}{1 - (-3)} = \dfrac{-12}{4} = -3.
    8. M=(2+102,  6+142)=(6,10)M = \left(\dfrac{2+10}{2},\;\dfrac{6+14}{2}\right) = (6, 10).
    9. d=25+144=169=13d = \sqrt{25 + 144} = \sqrt{169} = 13.
    10. d=9+16=25=5d = \sqrt{9 + 16} = \sqrt{25} = 5.
Reasoning

Tier 2: mixed practice

    1. y=23x−4y = \dfrac{2}{3}x - 4.
    2. Gradient: m=11−53−1=3m = \dfrac{11-5}{3-1} = 3. Using (1,5)(1, 5): 5=3(1)+c5 = 3(1) + c, c=2c = 2. Equation: y=3x+2y = 3x + 2.
    3. Rearrange 4x−2y=74x - 2y = 7: y=2x−72y = 2x - \dfrac{7}{2}. Gradient =2= 2. Same gradient as y=2x+3y = 2x + 3, so the lines are parallel.
    4. Midpoint: (−2+62,  3+(−1)2)=(2,1)\left(\dfrac{-2+6}{2},\;\dfrac{3+(-1)}{2}\right) = (2, 1). Length: 82+42=80=45\sqrt{8^2 + 4^2} = \sqrt{80} = 4\sqrt{5}.
    5. Rearrange: y=−34x+6y = -\dfrac{3}{4}x + 6. Gradient =−34= -\dfrac{3}{4}. yy-intercept: (0,6)(0, 6). xx-intercept: (8,0)(8, 0).
    6. Gradient of perpendicular: m=12m = \dfrac{1}{2}. Using (4,3)(4, 3): 3=12(4)+c3 = \dfrac{1}{2}(4) + c, c=1c = 1. Equation: y=12x+1y = \dfrac{1}{2}x + 1.
    7. AB=4AB = 4, BC=3BC = 3, AC=16+9=5AC = \sqrt{16 + 9} = 5. Check: 32+42=9+16=25=523^2 + 4^2 = 9 + 16 = 25 = 5^2. Right-angled at BB.
    8. Midpoint xx-coordinate: a+52=4\dfrac{a + 5}{2} = 4, so a+5=8a + 5 = 8, a=3a = 3.
Reasoning

Tier 3: explain and apply

    1. A vertical line has the form x=kx = k. In the gradient formula, x1=x2x_1 = x_2 makes the denominator zero, so mm is undefined. Since y=mx+cy = mx + c requires a defined mm, vertical lines cannot be written in this form.
    2. Hiker 1: d=62+82=100=10d = \sqrt{6^2 + 8^2} = \sqrt{100} = 10 km. Hiker 2: d=82+32=73≈8.54d = \sqrt{8^2 + 3^2} = \sqrt{73} \approx 8.54 km. Hiker 1 walks further by (10−73)≈1.46(10 - \sqrt{73}) \approx 1.46 km.
    3. Gradient of PQPQ: 06=0\dfrac{0}{6} = 0. Gradient of SRSR: 06=0\dfrac{0}{6} = 0. Gradient of PSPS: 42=2\dfrac{4}{2} = 2. Gradient of QRQR: 42=2\dfrac{4}{2} = 2. Opposite sides have equal gradients, so PQRSPQRS is a parallelogram.
    4. Perpendicular gradients: k×4=−1k \times 4 = -1, so k=−14k = -\dfrac{1}{4}.
    5. Midpoint formula: 3=1+xB23 = \dfrac{1 + x_B}{2} gives xB=5x_B = 5; 5=2+yB25 = \dfrac{2 + y_B}{2} gives yB=8y_B = 8. So B=(5,8)B = (5, 8).
Reasoning

Challenge

    1. Midpoint of ACAC: (0+82,  0+62)=(4,3)\left(\dfrac{0+8}{2},\;\dfrac{0+6}{2}\right) = (4, 3). Midpoint of BDBD: (8+02,  0+62)=(4,3)\left(\dfrac{8+0}{2},\;\dfrac{0+6}{2}\right) = (4, 3). Both midpoints are the same, so the diagonals bisect each other.
    2. Gradient of 3x−5y=103x - 5y = 10 is 35\dfrac{3}{5}. Perpendicular gradient: −53-\dfrac{5}{3}. Through (2,−1)(2, -1): y+1=−53(x−2)y + 1 = -\dfrac{5}{3}(x - 2). Multiply by 33: 3y+3=−5x+103y + 3 = -5x + 10. Rearrange: 5x+3y=75x + 3y = 7.
    3. AB=16+4=20=25AB = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5}. BC=4+4=8=22BC = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}. AC=4+16=20=25AC = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5}. Perimeter =45+22= 4\sqrt{5} + 2\sqrt{2}.
    4. Midpoint of QRQR: (5+32,  0+82)=(4,4)\left(\dfrac{5+3}{2},\;\dfrac{0+8}{2}\right) = (4, 4). Distance from P(−1,4)P(-1, 4) to (4,4)(4, 4): 25+0=5\sqrt{25 + 0} = 5.
Year 9 Mathematics study companion | Answer key