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Last updated: October 22, 2026 · Written by the Math Made Easy math teachers
Standards: F-IF.C.7a, F-LE.A.1a, F-LE.A.2, A-CED.A.2 · Level: Grade 9 Algebra 1
Problem 2 Negative slope
Slope = (−2 − 4) ÷ (5 − (−1)) = −6 ÷ 6 = −1.
−1
Problem 3 Read slope and intercept
The equation has the form y = mx + b with m = −8 and b = 15.
Slope −8; y-intercept 15
Problem 4 Write from slope and intercept
Substitute m = 6 and b = −11 into y = mx + b.
y = 6x − 11
Problem 5 Equation from a graph
The line crosses the y-axis at −2, so b = −2. Slope = (1 − (−2)) ÷ (3 − 0) = 1.
y = x − 2
Problem 6 Point-slope form
Point-slope: y − (−1) = 3(x − 2), which is y + 1 = 3(x − 2).
Distribute: y + 1 = 3x − 6, so y = 3x − 7.
Check: 3(2) − 7 = −1.
y = 3x − 7
Problem 7 Line through two points
Slope = (−11 − 9) ÷ (6 − (−4)) = −20 ÷ 10 = −2.
Use (−4, 9): y − 9 = −2(x + 4), so y = −2x − 8 + 9 = −2x + 1.
Check (6, −11): −2(6) + 1 = −11.
y = −2x + 1
Problem 8 Standard to slope-intercept
2y = −4x + 10, so y = −2x + 5.
y-intercept: 5. x-intercept: 0 = −2x + 5 gives x = 2.5.
y = −2x + 5; intercepts (2.5, 0) and (0, 5)
Problem 9 Slope-intercept to standard
Multiply by 3: 3y = 2x − 12. Move the x term: −2x + 3y = −12, or 2x − 3y = 12.
2x − 3y = 12
Problem 10 Graph from intercepts
x = 0: −4y = 12 so y = −3. y = 0: 3x = 12 so x = 4.
Intercepts (4, 0) and (0, −3)
Problem 11 Parallel line
Parallel lines have equal slopes, so m = 2.
y − 9 = 2(x − 3), so y = 2x + 3.
Check: 2(3) + 3 = 9.
y = 2x + 3
Problem 12 Perpendicular line
The negative reciprocal of 3 is −1/3.
y − 1 = −(1/3)(x − 6), so y = −(1/3)x + 2 + 1 = −(1/3)x + 3.
Check: −(1/3)(6) + 3 = 1.
y = −13x + 3
Problem 13 Parallel, perpendicular or neither
(a) 6x − 2y = 5 gives y = 3x − 5/2, slope 3: the same slope, so parallel.
(b) 3y = −4x + 7 gives slope −4/3. And (3/4)(−4/3) = −1, so perpendicular.
(a) parallel (b) perpendicular
Problem 14 Is the table linear?
For equal steps of 2 in x, y increases by 6 each time, so the rate of change is 6 ÷ 2 = 3 and the table is linear.
The y-intercept is 5, so y = 3x + 5.
Linear; y = 3x + 5
Problem 15 Fitness membership
C(m) = 22m + 40. C(7) = 22(7) + 40 = 154 + 40 = 194.
The slope 22 is the monthly fee in dollars per month; the intercept 40 is the joining fee at m = 0.
C(m) = 22m + 40; $194 after 7 months
Problem 16 Hiker’s elevation
Slope = (2,950 − 1,900) ÷ (5 − 2) = 1,050 ÷ 3 = 350 feet per hour.
E − 1,900 = 350(t − 2), so E = 350t + 1,200 (start at 1,200 ft).
4,000 = 350t + 1,200 gives 350t = 2,800 and t = 8.
E = 350t + 1,200; 8 hours
Problem 17 Burning candle
H(t) = 12 − 0.75t.
Burns out when 12 − 0.75t = 0, so t = 16 hours.
Domain 0 ≤ t ≤ 16 hours; range 0 ≤ H ≤ 12 inches, matching the graph from (0, 12) to (16, 0).
H(t) = 12 − 0.75t; burns out at 16 hours
Problem 18 School store budget
0.5p + 2n = 20. Multiply by 2: p + 4n = 40.
p-intercept: n = 0 gives p = 40. n-intercept: p = 0 gives n = 10.
For p = 12: 12 + 4n = 40, so 4n = 28 and n = 7.
p + 4n = 40; intercepts (40, 0) and (0, 10); 7 notebooks
Problem 19 Draining a pool
Slope = (200 − 450) ÷ (8 − 3) = −250 ÷ 5 = −50 gallons per minute.
G − 450 = −50(t − 3), so G = −50t + 600.
At t = 0 the pool held 600 gallons. Empty when −50t + 600 = 0, so t = 12 minutes.
G = −50t + 600; 600 gallons; empty at 12 minutes
Problem 20 Right angle test
Slope of AB = (2 − 0) ÷ (6 − 0) = 1/3. Slope of BC = (5 − 2) ÷ (5 − 6) = 3 ÷ (−1) = −3.
The product is (1/3)(−3) = −1, so AB ⊥ BC.
Yes, the angle at B is a right angle.