Lines in Context | Math Coach Amy

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Key Ideas Easy Medium Hard

m = riserun

What to Remember

Gradient

m = vertical changehorizontal change

For a cone or a hill, the run to the peak is half the base width when the peak is centred.

Point on a line

Substitute the coordinates. The point is on the line when the result is 0.

Cost and profit

A fixed charge plus a rate is a line, C = (rate) × (amount) + fixed charge.

Perpendicular in context

Find the gradient of the given line, take the negative reciprocal, then use the point.

Easy

A gradient, intercepts, or a cost.

Example

A service road rises 9 m over a horizontal distance of 24 m. Find the gradient.

m = vertical changehorizontal change

m = 924

m = 38

Answer: 38

A footbridge rises 8 m while it runs 20 m horizontally. Find the gradient.

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m = vertical changehorizontal change

m = 820

m = 25

Answer: 25

A path has equation 2x + y − 16 = 0. Find the axis intercepts.

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2x + y − 16 = 0

On the x-axis, y = 0.

2x = 16

x = 162

x = 8

On the y-axis, x = 0.

y = 16

Answer: x-intercept (8, 0), y-intercept (0, 16)

Does the point (4, 3) lie on 3x + 2y − 18 = 0?

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3x + 2y − 18 = 0

Substitute x = 4 and y = 3.

3(4) + 2(3) − 18 = 12 + 6 − 18

12 + 6 − 18 = 0

The point lies on the line.

Answer: yes

A taxi charges $4 plus $2 per kilometre. The cost in dollars is C = 4 + 2k. Find the cost of a 7 km ride.

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C = 4 + 2k

C = 4 + 2(7)

C = 4 + 14

C = 18

Answer: $18

A ramp rises 12 cm for each 30 cm of horizontal distance. Find the gradient.

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m = vertical changehorizontal change

m = 1230

m = 25

Answer: 25

Medium

A hill, a point check, and a boundary line.

Example

A hill is modelled as a cone of height 3600 m and base width 12000 m. The peak is above the centre of the base. Find the gradient of the slope from the edge of the base to the peak.

The horizontal run from the edge to the centre is half the base width.

Run = 120002 = 6000

m = heightrun

m = 36006000

m = 35

Answer: 35

A studio charges a $15 booking fee plus $8 per hour. Write the cost C dollars for h hours, then find the cost of 3 hours.

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C = 15 + 8h

C = 15 + 8(3)

C = 15 + 24

C = 39

Answer: C = 15 + 8h, $39

Does (1, 5) lie on 4x − y − 3 = 0?

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4x − y − 3 = 0

Substitute x = 1 and y = 5.

4(1) − 5 − 3 = 4 − 5 − 3

4 − 5 − 3 = −4

The result is not 0, so the point is not on the line.

Answer: no

A cable runs in a straight line from (2, 11) to (8, −1), in metres. Find its gradient.

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m = y₂ − y₁x₂ − x₁

m = −1 − 118 − 2

m = −126

m = −2

Answer: −2

A boundary has equation 5x + 2y − 20 = 0. Find its gradient and its x-intercept.

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5x + 2y − 20 = 0

2y = −5x + 20

Divide by 2.

y = −52x + 10

The gradient is −52.

The y-intercept is 10.

On the x-axis, y = 0.

5x − 20 = 0

5x = 20

x = 4

The x-intercept is (4, 0).

Answer: gradient −52, x-intercept (4, 0)

A ferry charges $6 plus $3 per kilometre. Write C for a trip of k kilometres, and find C when k = 9.

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C = 6 + 3k

C = 6 + 3(9)

C = 6 + 27

C = 33

Answer: C = 6 + 3k, $33

Hard

Compare models, a perpendicular line, and a length.

Example

Plan A costs $10 plus $1 for every 20 minutes. Plan B has no fixed charge and costs $1 for every 12 minutes. For how many minutes do the plans cost the same? Which plan costs less after that?

Plan A: C = 10 + m20

Plan B: C = m12

10 + m20 = m12

10 = m12 − m20

10 = 5m − 3m60

10 = m30

m = 300

Before 300 minutes, Plan B costs less. After 300 minutes, Plan A costs less.

Answer: 300 minutes; Plan A costs less after that

A shop's old profit model is P = 15n − 1800, and a new model is P = 12n − 900. n is the number of items sold, and P is the profit in dollars. Find n when the two models give the same profit. Then find how many items give a profit of $1500 under each model.

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15n − 1800 = 12n − 900

15n − 12n = 1800 − 900

3n = 900

n = 300

Old model, profit 1500: 15n − 1800 = 1500

15n = 3300

n = 220

New model, profit 1500: 12n − 900 = 1500

12n = 2400

n = 200

The new model reaches a profit of $1500 with fewer items.

Answer: same at 300 items; old 220 items; new 200 items

The line L has equation 5x + 2y − 20 = 0. Show that (2, 5) lies on L, then find the line through (2, 5) that is perpendicular to L.

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Substitute x = 2 and y = 5.

5(2) + 2(5) − 20 = 10 + 10 − 20 = 0

The point lies on L.

Perpendicular gradients multiply to −1.

m₂ = −1−52

m₂ = 25

y − y₁ = m(x − x₁)

y − 5 = 25(x − 2)

y − 5 = 25x − 45

y = 25x + 215

Answer: y = 25x + 215

A quadrilateral has vertices A(−2, 1), B(4, 4), C(5, 0), and D(1, −2). Are AB and DC parallel? Is ABCD a parallelogram?

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m = y₂ − y₁x₂ − x₁

m = 4 − 14 − (−2)

m = 36

m = 12

Gradient of AB is 12.

m = y₂ − y₁x₂ − x₁

m = 0 − (−2)5 − 1

m = 24

m = 12

Gradient of DC is 12.

The gradients match, so AB is parallel to DC.

m = y₂ − y₁x₂ − x₁

m = −2 − 11 − (−2)

m = −33

m = −1

Gradient of AD is −1.

m = y₂ − y₁x₂ − x₁

m = 0 − 45 − 4

m = −41

m = −4

Gradient of BC is −4.

Those gradients are not equal, so AD is not parallel to BC.

ABCD is not a parallelogram.

Answer: AB ∥ DC, but ABCD is not a parallelogram

A plank is at a gradient of 0.6, and the higher end is 2.4 m above the lower end. Gradient means vertical change divided by horizontal change. Find the length of the plank.

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Gradient = 35

Vertical change = 125

125run = 35

Run = 125 ÷ 35 = 4

Length = square root of (125)² + (4)²

Length = square root of 54425

Length = 4square root of 345

As a decimal, the length is about 4.66 m.

Answer: 4square root of 345 m, about 4.66 m

A line has equation 7x + y − 28 = 0. Find the gradient and the x-intercept.

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7x + y − 28 = 0

y = −7x + 28

The gradient is −7.

The y-intercept is 28.

On the x-axis, y = 0.

7x − 28 = 0

x = 4

The x-intercept is (4, 0).

Answer: gradient −7, x-intercept (4, 0)

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