Functions & Linear Graphs

Shinglee Think Mathematics 1A — Chapter 6 (Sec 1)
Gradients, y-intercepts & equations of lines · No graphing required
Prepared by Miss Clarissa Ng
www.clartutors.com
Section A — Gradients & y-intercepts

Q1. Find the gradient and the y-intercept of each line:

(a)   y = 3x + 5

(b)   y = −12x − 4

(c)   3y + 6x = 9

(d)   5x − 2y = 10

Q2. Write each equation in the form y = mx + c, then state its gradient and y-intercept:

(a)   4x − 3y = 12

(b)   6x + 2y = 8

Q3. Find the gradient of the line passing through each pair of points:

(a)   (1, 2) and (3, 8)

(b)   (−2, 5) and (4, −7)

Q4. State whether each pair of lines is parallel, perpendicular or neither:

(a)   y = 3x + 1 and y = 3x − 7

(b)   y = 5x + 2 and y = −15x + 4

(c)   2y = x + 6 and y = 2x − 3

Q5. Which of the following lines passes through the point (2, −1)? Show your working:

(a)   y = 3x − 7

(b)   y = x − 3

(c)   y = −2x + 5

Section B — Equations of Lines

Q1. Find the equation of the line passing through each pair of points. Give your answer in the form y = mx + c:

(a)   (2, 5) and (4, 11)

(b)   (−3, −1) and (1, 7)

Q2. The line L1 has equation y = 4x + 1. Find the equation of the line which is parallel to L1 and passes through (3, 2). Give your answer in the form y = mx + c.

Q3. The line L1 has equation y = 4x + 1. Find the equation of the line which is perpendicular to L1 and passes through (0, 5). Give your answer in the form y = mx + c.

Q4. Find the equation of each line. Give your answer in the form y = mx + c:

(a)   passes through (1, 4) and is parallel to the line y = −2x + 5

(b)   passes through (−2, 4) and is perpendicular to the line 3y = x + 9

Q5. The line L2 passes through (0, −3) and is parallel to the line 4x + 2y = 10. Find the equation of L2.

Q6. The line L3 passes through (2, −1) and is perpendicular to the line y = −14x + 6. Find the equation of L3.

Section C — Applications

Q1. The cost, C dollars, of a taxi ride for d kilometres is given by the equation C = 2.5d + 4.

(a) Find the cost of a 6 km ride.

(b) What does the value 4 represent in this context?

Q2. The table shows the cost, C dollars, of a taxi ride for d kilometres.

d (km)C ($)
29
518

(a) Find the equation of C in terms of d.

(b) How much would a 10 km ride cost?

Q3. In the figure, ABCD is a rectangle with A(1, 2), B(3, 4) and C(6, 1). Find the equation of the line passing through D and C. Give your answer in the form y = mx + c.

Q4. (OPEN) The line y = kx + 2 is perpendicular to the line y = −3x + 7. Find the value of k.

Answer Key

Section A — Gradients & y-intercepts

Q1(a)m = 3, c = 5
Q1(b)m = −½, c = −4
Q1(c)y = −2x + 3 → m = −2, c = 3
Q1(d)y = (5/2)x − 5 → m = 5/2, c = −5
Q2(a)y = (4/3)x − 4 → m = 4/3, c = −4
Q2(b)y = −3x + 4 → m = −3, c = 4
Q3(a)m = (8−2)/(3−1) = 3
Q3(b)m = (−7−5)/(4+2) = −2
Q4(a)Parallel (both m = 3)
Q4(b)Perpendicular (m₁ × m₂ = 5 × (−1/5) = −1)
Q4(c)Neither (m = ½ and m = 2; not equal, product ≠ −1)
Q5(a) y = 3x − 7 — substituting x = 2 gives y = −1 ✓

Section B — Equations of Lines

Q1(a)m = (11−5)/(4−2) = 3; y − 5 = 3(x − 2) → y = 3x − 1
Q1(b)m = (7+1)/(1+3) = 2; y − 7 = 2(x − 1) → y = 2x + 5
Q2y = 4x − 10 (parallel: m = 4; through (3, 2): c = 2 − 12)
Q3y = −(1/4)x + 5 (perpendicular: m = −1/4; passes through (0, 5) so c = 5)
Q4(a)y = −2x + 6 (parallel keeps m = −2; c = 4 + 2(1))
Q4(b)y = −3x − 2 (perpendicular to m = 1/3 → m = −3; c = 4 − 6)
Q5y = −2x − 3 (parallel to m = −2; through (0, −3) so c = −3)
Q6y = 4x − 9 (perpendicular to m = −1/4 → m = 4; c = −1 − 8)

Section C — Applications

Q1(a)C = 2.5(6) + 4 = $19
Q1(b)The fixed starting fare (cost when d = 0)
Q2(a)m = (18−9)/(5−2) = 3; C − 9 = 3(d − 2) → C = 3d + 3
Q2(b)C = 3(10) + 3 = $33
Q3D(4, 3); m = (1−3)/(6−4) = −1 → y = −x + 7
Q4k × (−3) = −1 → k = ⅓