Mathematics Class 9 — Ganita…
🏠 Home Log in
TOPIC 6 OF 25

Graphs of Linear Polynomials, Exercises & Summary

🎓 Class 9 Mathematics CBSE Theory Ch 2 — Introduction to Linear Polynomials ⏱ ~35 min
🌐 Language: [gtranslate]

This MCQ module is based on: Graphs of Linear Polynomials, Exercises & Summary

This mathematics assessment will be based on: Graphs of Linear Polynomials, Exercises & Summary
Targeting Class 9 level in Algebra, with Intermediate difficulty.

Upload images, PDFs, or Word documents to include their content in assessment generation.

2.5 Graphs of Linear Polynomials

Every linear polynomial \(p(x) = ax + b\) corresponds to a straight line? on the Cartesian plane: plot points \((x, y)\) where \(y = ax + b\).

Plotting \(y = x + 3\), \(y = 3x\) and \(y = 2x\)

xy y = x + 3 y = 3x y = 2x -4-3-2-11234
Fig 2.8: Straight-line graphs of three linear polynomials
Key Geometric Facts
For \(y = ax + b\):
• Slope = \(a\) (rise per unit run). Positive \(a\) → line rises (linear growth). Negative \(a\) → line falls (linear decay).
• y-intercept = \(b\) (where the line crosses the y-axis at \((0, b)\)).
• When \(b = 0\), the line passes through the origin.
• Lines with the same \(a\) but different \(b\) are parallel.
Example 13: From Fig 2.14, observe where the lines \(y = x + 3\), \(y = 2x + 5\), \(y = 3x - 2\) cut the y-axis.
y-intercepts are at \((0, 3)\), \((0, 5)\), \((0, -2)\) respectively. In each case the y-intercept is exactly the constant term \(b\) — the value of the line when \(x = 0\).

End-of-Chapter Exercises (Exercise Set 2.4)

Q1. Suppose a plant has height 1.75 feet and grows by 0.5 feet each month. (i) Find the height after 7 months. (ii) Make a table of values for \(n\) varying from 0 to 10 months and show how the height \(h\) increases each month. (iii) Find an expression that relates \(h\) and \(n\), and explain why this represents a linear pattern.
(i) \(h(7) = 1.75 + 0.5(7) = 5.25\) ft. (ii) Heights for n=0..10: 1.75, 2.25, 2.75, 3.25, 3.75, 4.25, 4.75, 5.25, 5.75, 6.25, 6.75. (iii) \(h = 0.5n + 1.75\); each month the height increases by exactly 0.5 ft (a constant rate) → linear pattern.
Q2. After 5 years, a plant is 3 ft tall. After 9 years it is 5 ft tall. (i) Find the value of the plant's height at \(n = 0\). (ii) Make a table for \(n = 0\) to \(8\) and show how it increases. (iii) Find an expression relating height and \(n\), and explain why it represents linear growth.
Slope = \(\dfrac{5-3}{9-5} = 0.5\). At \(n=5\) height is 3, so \(3 = 0.5(5) + b \Rightarrow b = 0.5\). (i) \(h(0) = 0.5\) ft. (ii) n=0..8: 0.5, 1, 1.5, 2, 2.5, 3, 3.5, 4, 4.5. (iii) \(h = 0.5n + 0.5\) — a constant +0.5 ft per year, hence linear growth.
Q3. The initial population of a village is 750. Every year, 50 people move from a nearby city. (i) Find the population after 6 years.
\(P(t) = 750 + 50t\). \(P(6) = 750 + 300 = 1050\) people.
Q4. The graphs of two-digit numbers differ by 3. If the digits are interchanged, and the resulting number is added to the original, we get \(11x + 11y\). Show that any such sum is divisible by 11.
Original number = \(10x + y\); after swap = \(10y + x\). Sum = \(11x + 11y = 11(x + y)\) — manifestly divisible by 11. ✓
Q5. If you have ₹500 and save ₹250 every two months, find an expression for total savings after \(n\) months. Express it as a linear pattern.
Per-month savings rate = ₹125. So \(S(n) = 500 + 125n\) rupees — linear growth with slope 125 and y-intercept 500.
Q6. Find numbers whose digits add to 11 if they differ by 3 and the digits are interchanged etc. (Q6, p. 37 NCERT). Express as linear equations.
Let digits be \(x\) (tens) and \(y\) (units). \(x + y = 11\), \(x - y = 3\). Solving: \(x = 7, y = 4\). Number: 74.
Q7. Draw the graphs of \(y = -3x + 4\) and \(y = 4x + 7\). Identify the coordinates of points where these graphs cross the x-axis. Are they parallel?
For \(y = -3x + 4\): set \(y = 0 \Rightarrow x = 4/3\). x-intercept = \((4/3, 0)\). For \(y = 4x + 7\): \(x = -7/4\). x-intercept = \((-7/4, 0)\). Slopes are −3 and 4 — different, so not parallel.
Q8. If the temperature of a liquid in Fahrenheit (°F) is related to that in Kelvin (K) by the system \(F = \tfrac{9}{5}(K - 273) + 32\): (i) Find the temperature in Fahrenheit if K = 313. (ii) If the temperature is 158 °F, find K.
(i) \(F = 1.8(40) + 32 = 72 + 32 = 104\)°F. (ii) Solve \(158 = 1.8(K-273) + 32 \Rightarrow 126 = 1.8(K-273) \Rightarrow K-273 = 70 \Rightarrow K = 343\) K.
Q9. The work done by a body on the application of constant force is directly proportional to the distance travelled. Express this as a linear polynomial. If \(W = 25\) units when \(d = 5\) units, find the constant of proportionality and the work for \(d = 12\).
\(W = kd\) (linear, no y-intercept). \(25 = 5k \Rightarrow k = 5\). For \(d = 12\): \(W = 60\) units.
Q10. The graph of a linear polynomial \(p(x)\) passes through the points \((1, 5)\) and \((3, 15)\). (i) Find \(p(x)\). (ii) Find the coordinates where the graph cuts the axes. (iii) Draw the graph of \(p(x)\) and verify your answers.
Slope \(a = (15-5)/(3-1) = 5\). Using \((1,5)\): \(5 = 5(1) + b \Rightarrow b = 0\). So \(p(x) = 5x\). (ii) Crosses both axes at the origin \((0, 0)\). (iii) Straight line through O with slope 5 — passes through \((1, 5)\) and \((3, 15)\) as required. ✓
Q11. Let \(p(x) = ax + b\) and \(q(x) = cx + d\) be two linear polynomials such that: (i) the graph of \(p(x)\) passes through (2, 3) and (6, 1); (ii) the sum \(p(x) + q(x)\) is equal to 4 for all real \(x\). Find \(p(x)\) and \(q(x)\).
Slope of p: \((1-3)/(6-2) = -0.5\). Using (2,3): \(3 = -0.5(2) + b \Rightarrow b = 4\). So \(p(x) = -0.5x + 4\). For \(p+q = 4\) constant: \(q(x) = 4 - p(x) = 0.5x\). Check: degree 1, so linear. ✓
Q12. Look at the first three stages of a growing pattern of hexagons made using matchsticks. A new hexagon adds 5 matches on a side with the last hexagon of the previous group: (i) Draw the next stage. (ii) Complete the table. (iii) Derive a formula for the matchsticks at the \(n^{\text{th}}\) stage.
Stage 1: 6 sticks. Stage 2: 11 sticks (+5). Stage 3: 16 (+5). General: \(M(n) = 5n + 1\). Stage 15: \(M(15) = 76\). For 56 sticks: \(5n + 1 = 56 \Rightarrow n = 11\) → 11th stage.
Q13. If the polynomial \(p(x) = ax + b\) and \(q(x) = cx + d\) are two linear polynomials such that: (i) graphs pass through (2, 3) and (6, 11); (ii) the difference \(p(x) - q(x)\) equals \(-4\) for all real \(x\). Find \(p(x)\) and \(q(x)\).
Slope: \((11-3)/(6-2) = 2\). \(3 = 2(2)+b \Rightarrow b = -1\). So \(p(x) = 2x - 1\). \(p - q = -4 \Rightarrow q(x) = p(x) + 4 = 2x + 3\). Verify: q(2) = 7, q(6) = 15 (q passes through different points, but the constraint asks p-q = const ✓).
Q14. What do all linear functions of the form \(f(x) = ax\), with \(a \neq 0\), have in common?
All such functions: pass through the origin, have y-intercept 0, are directly proportional (\(f(x)/x = a\) is constant), and their zero is \(x = 0\) for every choice of \(a\).
Activity: Walk the Line
L3 Apply
Materials: Open ground, chalk, measuring tape, stopwatch
Predict: If you walk at a constant pace, plot your distance vs. time. What kind of line do you expect?
  1. Mark a 20 m straight line on the ground at 1 m intervals.
  2. Walk from start to end at a steady pace; have a partner record your position every 2 seconds.
  3. Plot \(t\) (time, x-axis) versus \(d\) (distance, y-axis).
  4. Use any two points to find the slope (your speed).
  5. Write the linear polynomial \(d(t) = vt + d_0\) for your motion.

The slope you find equals your average walking speed. Your y-intercept \(d_0\) tells where you were at \(t = 0\). If your line is curved, your pace was not steady — physics calls this non-uniform motion.

Competency-Based Questions

Scenario: A solar water-heater starts at 18 °C in the morning and the temperature rises linearly during sunshine. Two readings: at 9:00 AM (t = 0) the water is at 18 °C; at 12:00 noon (t = 3 h) it is at 60 °C. Let \(T(t) = at + b\).
Q1. Find the linear polynomial \(T(t)\).
L3 Apply
Slope \(a = (60-18)/3 = 14\) °C/h. \(b = T(0) = 18\). So \(T(t) = 14t + 18\).
Q2. Predict when the water will reach 95 °C and analyse whether linear extrapolation is realistic.
L4 Analyse
\(14t + 18 = 95 \Rightarrow t = 5.5\) h after 9 AM = 2:30 PM. However, real heating saturates as it approaches the surrounding-air ceiling — linear extrapolation is unrealistic at higher temperatures. The model breaks down.
Q3. Another solar heater has \(T_2(t) = 12t + 25\). Evaluate which heater becomes hotter sooner.
L5 Evaluate
Equal temperatures: \(14t + 18 = 12t + 25 \Rightarrow 2t = 7 \Rightarrow t = 3.5\) h. Before 3.5 h, \(T_2\) is hotter (higher start). After 3.5 h, \(T_1\) overtakes (steeper slope). So heater 2 leads early, heater 1 wins later in the day.
Q4. Design a third heater with linear polynomial \(T_3(t) = at + b\) such that it equals heater 1 at exactly \(t = 2\) hours and heater 2 at \(t = 4\) hours. Find \(a\) and \(b\).
L6 Create
At t=2: \(T_1 = 46\). At t=4: \(T_2 = 73\). So \(T_3\) passes through (2, 46) and (4, 73). Slope = \((73-46)/2 = 13.5\). \(46 = 13.5(2) + b \Rightarrow b = 19\). So \(T_3(t) = 13.5t + 19\).

Assertion–Reason Questions

A: The graph of \(y = 3x + 2\) and \(y = 3x - 5\) are parallel lines.
R: Lines with the same slope are parallel.
(a) Both true, R explains A.
(b) Both true, R doesn't explain.
(c) A true, R false.
(d) A false, R true.
Answer: (a) — Both have slope 3, hence parallel. R explains A.
A: The y-intercept of \(y = -2x + 7\) is at the point \((0, 7)\).
R: The y-intercept is the value of \(y\) when \(x = 0\).
(a) Both true, R explains A.
(b) Both true, R doesn't explain.
(c) A true, R false.
(d) A false, R true.
Answer: (a)
A: Every linear polynomial \(p(x) = ax\) (with \(b = 0\)) has its graph passing through the origin.
R: When \(b = 0\), \(p(0) = a(0) = 0\), so \((0, 0)\) is on the graph.
(a) Both true, R explains A.
(b) Both true, R doesn't explain.
(c) A true, R false.
(d) A false, R true.
Answer: (a) — Crisp algebraic reasoning. R explains A.

Chapter Summary

Key Take-aways
  • An algebraic expression combines variables, constants and operations. Each piece separated by + or − is a term; the numerical multiplier of a variable is its coefficient.
  • A polynomial in one variable is an algebraic expression of the form \(a_n x^n + \dots + a_1 x + a_0\). The degree is the highest power of \(x\).
  • A linear polynomial is \(p(x) = ax + b\) with \(a \neq 0\). It always has degree 1 and exactly one zero: \(x = -b/a\).
  • A linear pattern is a sequence with constant differences between successive terms.
  • The graph of \(y = ax + b\) is a straight line. Slope = \(a\); y-intercept = \(b\). When \(b = 0\), the line passes through the origin.
  • Linear growth (\(a > 0\)) is a straight line with positive slope; linear decay (\(a < 0\)) has negative slope.
  • Lines with the same slope but different y-intercepts are parallel.

Frequently Asked Questions

How do you draw the graph of a linear polynomial in Class 9 Maths?

Write the polynomial as y = ax + b, pick two convenient x-values, compute their matching y-values to get two points, plot the points on the Cartesian plane and join them with a straight line.

What is the slope of the graph of y = ax + b?

The slope is the coefficient a. It measures how steeply the line rises (a > 0) or falls (a < 0). A larger absolute value of a means a steeper line; a = 0 would give a horizontal line.

What is the y-intercept of a linear polynomial graph?

The y-intercept is the value of y when x = 0, which is just b. So the line y = ax + b crosses the y-axis at the point (0, b).

Why is the graph of a linear polynomial always a straight line?

Because for any two points on y = ax + b, the rise over run between them is always the constant a. A constant rate of change is exactly the geometric definition of a straight line.

How do parallel lines relate to linear polynomials?

Two graphs y = a1*x + b1 and y = a2*x + b2 are parallel if and only if their slopes are equal: a1 = a2 and b1 is not equal to b2. Equal slope and equal intercept means the same line, not just parallel.

How do you find where a linear polynomial graph cuts the x-axis?

Set y = 0 and solve ax + b = 0, giving x = -b/a. The graph crosses the x-axis at the single point (-b/a, 0), which is also the unique zero of the polynomial.

Keyword

AI Tutor
Mathematics Class 9 — Ganita Manjari
Ready
Hi! 👋 I'm Gaura, your AI Tutor for Graphs of Linear Polynomials, Exercises & Summary. Take your time studying the lesson — whenever you have a doubt, just ask me! I'm here to help.