Ch-2 Introduction to Linear Polynomials Class 9th Maths NCERT Solution (Ganita Manjari )

May 21, 2026 • Admin ETeam • Class 9 NCERT Solution

Chapter 2 – Introduction to Linear Polynomials | Solutions

Ganita Manjari · Grade 9 · Part I

Chapter 2 · Introduction to
Linear Polynomials

Complete step-by-step solutions to all Exercise Sets 2.1 – 2.6 and End-of-Chapter Exercises

Exercise Set 2.1 Exercise Set 2.2 Exercise Set 2.3 Exercise Set 2.4 Exercise Set 2.5 Exercise Set 2.6 End-of-Chapter ★

Exercise Set 2.1 — Degrees of Polynomials

Q1 Find the degrees of the following polynomials: (i) 2x² – 5x + 3 (ii) y³ + 2y – 1 (iii) –9 (iv) 4z – 3

Solution

The degree of a polynomial is the highest power of the variable present.

(i) 2x² – 5x + 3

Identify powers of x: x² (power 2), x¹ (power 1), and constant (power 0).

Highest power = 2. This is a Quadratic Polynomial.

Degree = 2

(ii) y³ + 2y – 1

Powers present: y³ (power 3), y (power 1), –1 (power 0).

Highest power = 3. This is a Cubic Polynomial.

Degree = 3

(iii) –9

–9 is a constant. We can write it as –9·x⁰, so the highest power = 0.

Degree = 0 (Constant Polynomial)

(iv) 4z – 3

Powers present: z¹ (power 1), –3 (power 0). Highest power = 1.

Degree = 1 (Linear Polynomial)

Q2 Write polynomials of degrees 1, 2 and 3.

Solution

Many answers are possible. Here are standard examples:

Degree 1 (Linear): 3x + 5 Degree 2 (Quadratic): x² – 4x + 7 Degree 3 (Cubic): 2x³ + x² – x + 1

📌 Any polynomial where the highest power of the variable matches the required degree is correct.

Q3 What are the coefficients of x² and x³ in the polynomial x⁴ – 3x³ + 6x² – 2x + 7?

Solution

Write the polynomial term-by-term:

x⁴

-3x³

+6x²

-2x

+7

The coefficient of x³ is the number multiplied with x³, which is –3.

The coefficient of x² is the number multiplied with x², which is 6.

Coefficient of x² = 6 | Coefficient of x³ = –3

Q4 What is the coefficient of z in the polynomial 4z³ + 5z² – 11?

Solution

Look for the term containing

z¹

in 4z³ + 5z² – 11.

There is no z term — it is absent (coefficient = 0).

Coefficient of z = 0

Q5 What is the constant term of the polynomial 9x³ + 5x² – 8x – 10?

Solution

The constant term is the term that has no variable (x⁰ term).

In 9x³ + 5x² – 8x – 10, the constant term is –10.

Constant term = –10

Exercise Set 2.2 — Evaluating Polynomials & Linear Equations

Q1 Find the value of the linear polynomial 5x – 3 if: (i) x = 0 (ii) x = –1 (iii) x = 2

Solution

Substitute each value in $p(x) = 5x - 3$ :

(i) x = 0

p(0) = 5(0) – 3 = 0 – 3 = –3

Value = –3

(ii) x = –1

p(–1) = 5(–1) – 3 = –5 – 3 = –8

Value = –8

(iii) x = 2

p(2) = 5(2) – 3 = 10 – 3 = 7

Value = 7

Q2 Find the value of the quadratic polynomial 7s² – 4s + 6 if: (i) s = 0 (ii) s = –3 (iii) s = 4

Solution

Substitute each value in $q(s) = 7s² - 4s + 6$ :

(i) s = 0

q(0) = 7(0)² – 4(0) + 6 = 0 – 0 + 6 = 6

Value = 6

(ii) s = –3

q(–3) = 7(–3)² – 4(–3) + 6 = 7(9) + 12 + 6 = 63 + 12 + 6 = 81

Value = 81

(iii) s = 4

q(4) = 7(4)² – 4(4) + 6 = 7(16) – 16 + 6 = 112 – 16 + 6 = 102

Value = 102

Q3 The present age of Salil’s mother is three times Salil’s present age. After 5 years, their ages will add up to 70 years. Find their present ages.

Solution

Let Salil’s present age = x years. Then mother’s present age = 3x years.

After 5 years: Salil’s age = x + 5, Mother’s age = 3x + 5.

Form the equation:

(x + 5) + (3x + 5) = 70 4x + 10 = 70 4x = 60 x = 15

Salil’s age = 15 years, Mother’s age = 3 × 15 = 45 years.

Salil = 15 years | Mother = 45 years

Q4 The difference between two positive integers is 63. The ratio of the two integers is 2 : 5. Find the two integers.

Solution

Since ratio is 2 : 5, let the integers be 2k and 5k.

The difference:

5k – 2k = 63 3k = 63 k = 21

Smaller integer = 2 × 21 = 42, Larger = 5 × 21 = 105.

Verify: 105 – 42 = 63 ✓ 42 : 105 = 2 : 5 ✓

The two integers are 42 and 105.

Q5 Ruby has 3 times as many two-rupee coins as five-rupee coins. If she has a total ₹88, how many coins of each type does she have?

Solution

Let number of five-rupee coins = x.
Then number of two-rupee coins = 3x.

Total amount:

5(x) + 2(3x) = 88 5x + 6x = 88 11x = 88 x = 8

Five-rupee coins = 8, Two-rupee coins = 3 × 8 = 24.

Verify: 8 × 5 + 24 × 2 = 40 + 48 = ₹88 ✓

Five-rupee coins: 8 | Two-rupee coins: 24

Q6 A farmer cuts a 300 feet fence into two pieces. The longer piece is four times as long as the shorter piece. How long are the two pieces?

Solution

Let shorter piece = x feet, then longer piece = 4x feet.

x + 4x = 300 5x = 300 x = 60

Shorter piece = 60 ft, Longer piece = 4 × 60 = 240 ft.

Shorter piece = 60 feet | Longer piece = 240 feet

Q7 If the length of a rectangle is three more than twice its width and its perimeter is 24 cm, what are the dimensions of the rectangle?

Solution

Let width = w cm. Then length = 2w + 3 cm.

Perimeter = 2(length + width):

2(2w + 3 + w) = 24 2(3w + 3) = 24 6w + 6 = 24 6w = 18 w = 3

Width = 3 cm, Length = 2(3) + 3 = 9 cm.

Rectangle with width = 3 cm and length = 9 cm

Width = 3 cm | Length = 9 cm

Exercise Set 2.3 — Exploring Linear Patterns

Q1 A student has ₹500 in her savings account. She gets ₹150 every month as pocket money. Find a linear expression for the amount she has in the n^th month.

Solution

Initial amount = ₹500. She adds ₹150 each month.

After n months, total amount = 500 + 150n.

Amount in n-th month = 500 + 150n (rupees)

Month (n)	1	2	3	4	5
Amount (₹)	650	800	950	1100	1250

Linear expression: A(n) = 500 + 150n

Q2 A rally starts with 120 members. Each hour, 9 members drop out. Find a linear expression for the number of members at the end of the n^th hour.

Solution

Start = 120, decrease each hour = 9.

Members after n hours = 120 – 9n

Hour (n)	0	1	2	3	4	5
Members	120	111	102	93	84	75

M(n) = 120 – 9n (This is a linear decay pattern)

Q3 Suppose the length of a rectangle is 13 cm. Find the area if breadth is (i) 12 cm, (ii) 10 cm, (iii) 8 cm. Find the linear pattern for area.

Solution

Area = length × breadth = 13 × b.

(i) b = 12 cm

Area = 13 × 12 = 156 cm²

(ii) b = 10 cm

Area = 13 × 10 = 130 cm²

(iii) b = 8 cm

Area = 13 × 8 = 104 cm²

Breadth b (cm)	8	10	12
Area (cm²)	104	130	156

Difference between consecutive areas = 26 (constant when b increases by 2). The linear pattern is A = 13b.

Areas: 156, 130, 104 cm² | Linear pattern: A = 13b

Q4 A rectangular box has length 7 cm, breadth 11 cm. Find the volume if height is (i) 5 cm, (ii) 9 cm, (iii) 13 cm. Find the linear pattern.

Solution

Volume = l × b × h = 7 × 11 × h = 77h.

(i) h = 5 cm

V = 77 × 5 = 385 cm³

(ii) h = 9 cm

V = 77 × 9 = 693 cm³

(iii) h = 13 cm

V = 77 × 13 = 1001 cm³

Height h (cm)	5	9	13
Volume (cm³)	385	693	1001

As h increases by 4, V increases by 308 (= 77 × 4). Linear pattern: V = 77h.

Volumes: 385, 693, 1001 cm³ | Linear pattern: V = 77h

Q5 Sarita is reading a book of 500 pages. She reads 20 pages every day. How many pages will be left after 15 days? Express this as a linear pattern.

Solution

Pages remaining after n days = 500 – 20n.

After 15 days:

Pages left = 500 – 20(15) = 500 – 300 = 200 pages

Day (n)	0	5	10	15	20	25
Pages left	500	400	300	200	100	0

After 15 days: 200 pages left | Linear pattern: P(n) = 500 – 20n

Exercise Set 2.4 — Linear Growth & Linear Decay

Q1 A plant has height 1.75 feet and grows 0.5 feet each month. (i) Find height after 7 months. (ii) Table for t = 0 to 10. (iii) Expression relating h and t.

Solution

(i) Height after 7 months

h = 1.75 + 0.5(7) = 1.75 + 3.5 = 5.25 feet

Height after 7 months = 5.25 feet

(ii) Table of values (t = 0 to 10)

t (months)	0	1	2	3	4	5	6	7	8	9	10
h (feet)	1.75	2.25	2.75	3.25	3.75	4.25	4.75	5.25	5.75	6.25	6.75

(iii) Expression & why Linear Growth

h(t) = 1.75 + 0.5t

This represents linear growth because as t increases by 1 (one month), h increases by a fixed constant (0.5 feet). The slope is positive (+0.5).

Q2 A mobile phone is bought for ₹10,000 and its value decreases by ₹800 every year. (i) Value after 3 years. (ii) Table for t = 0 to 8. (iii) Expression for v and t.

Solution

(i) Value after 3 years

v = 10000 – 800(3) = 10000 – 2400 = ₹7,600

Value after 3 years = ₹7,600

(ii) Table of values (t = 0 to 8)

t (years)	0	1	2	3	4	5	6	7	8
v (₹)	10000	9200	8400	7600	6800	6000	5200	4400	3600

(iii) Expression & why Linear Decay

v(t) = 10000 – 800t

This represents linear decay because as t increases by 1 year, v decreases by a fixed amount (₹800). The slope is negative (–800).

Q3 Initial population of a village is 750. Every year 50 people move from a city to the village. (i) Population after 6 years. (ii) Table for t = 0 to 10. (iii) Expression for P and t.

Solution

(i) Population after 6 years

P = 750 + 50(6) = 750 + 300 = 1050

Population after 6 years = 1,050

(ii) Table (t = 0 to 10)

t (years)	0	1	2	3	4	5	6	7	8	9	10
P	750	800	850	900	950	1000	1050	1100	1150	1200	1250

(iii) Expression & why Linear Growth

P(t) = 750 + 50t

This is linear growth: each year P increases by a fixed 50. Slope = +50 (positive).

Q4 A telecom company charges ₹600 recharge. Balance reduces ₹15 each day. (i) Equation for b(x). (ii) After how many days will balance run out? (iii) Table for x = 1 to 10.

Solution

(i) Equation for remaining balance b(x)

b(x) = 600 – 15x

This is linear decay: balance decreases by a fixed ₹15 per day (constant rate of decrease, negative slope –15).

(ii) When will balance run out?

600 – 15x = 0 15x = 600 x = 40 days

Balance runs out after 40 days.

(iii) Table (x = 1 to 10)

Day x	1	2	3	4	5	6	7	8	9	10
b(x) (₹)	585	570	555	540	525	510	495	480	465	450

Exercise Set 2.5 — Linear Relationships

Q1 A learning platform charges a fixed fee + cost per module. 10 modules → ₹400; 14 modules → ₹500. Find a and b in y = ax + b.

Solution

Using y = ax + b:

When x = 10, y = 400 → 400 = 10a + b … (1) When x = 14, y = 500 → 500 = 14a + b … (2)

Subtract equation (1) from (2):

500 – 400 = (14a + b) – (10a + b) 100 = 4a a = 25

Substitute a = 25 in (1):

400 = 10(25) + b 400 = 250 + b b = 150

a = 25 → cost per module = ₹25; b = 150 → fixed monthly fee = ₹150.

a = 25, b = 150 | y = 25x + 150

Q2 A gym charges fixed fee + cost per hour of badminton court. 10 hrs → ₹800; 15 hrs → ₹1100. Find a and b in y = ax + b.

Solution

When x = 10, y = 800 → 800 = 10a + b … (1) When x = 15, y = 1100 → 1100 = 15a + b … (2)

Subtract (1) from (2):

300 = 5a → a = 60

From (1): b = 800 – 10(60) = 800 – 600 = 200.

a = 60 → ₹60 per hour; b = 200 → fixed monthly fee = ₹200.

a = 60, b = 200 | y = 60x + 200

Q3 °C = a°F + b. Ice melts: 0°C = 32°F. Water boils: 100°C = 212°F. Find a and b.

Solution

°C = 0, °F = 32 → 0 = 32a + b … (1) °C = 100, °F = 212 → 100 = 212a + b … (2)

Subtract (1) from (2):

100 = 180a a = 100/180 = 5/9

From (1):

b = –32a = –32 × (5/9) = –160/9

Therefore:

°C = (5/9)°F – 160/9 = (5/9)(°F – 32)

This is the well-known Celsius–Fahrenheit conversion formula!

a = 5/9, b = –160/9 | °C = (5/9)(°F – 32)

Exercise Set 2.6 — Visualising Linear Relationships

Q1 Draw the graphs of the following sets of lines and reflect on the role of ‘a’ and ‘b’ in each case.

Solution

(i) y = 4x, y = 2x, y = x

Fig: y = x, y = 2x, y = 4x all pass through origin. Steeper as slope increases.

All lines y = ax pass through the origin (0,0). As a increases, the line becomes steeper. Since b = 0, the y-intercept is 0.

(ii) y = –6x, y = –3x, y = –x

Fig: y = –x, y = –3x, y = –6x — all pass through origin, sloping downward (negative slope).

All lines with negative slope fall from left to right. Steepness increases as |a| increases.

(iii) y = 5x, y = –5x

Fig: y = 5x rises steeply; y = –5x is its mirror image, falling steeply. Both pass through origin.

y = 5x and y = –5x are reflections of each other in the x-axis (or y-axis). Equal magnitude, opposite direction of slope.

(iv) y = 3x – 1, y = 3x, y = 3x + 1

Fig: Parallel lines y = 3x–1, y = 3x, y = 3x+1 — same slope (3), different y-intercepts.

Key insight: When the slope ‘a’ is fixed but ‘b’ varies, the lines are parallel. They shift up/down by the value of b without changing direction.

(v) y = –2x – 3, y = –2x, y = 2x + 3

Fig: y=–2x–3 and y=–2x are parallel (same slope –2). y=2x+3 has opposite slope.

y=–2x and y=–2x–3 are parallel (both have slope –2). y=2x+3 has slope +2, opposite direction, so it’s not parallel to them.

End-of-Chapter Exercises

Q1 Write a polynomial of degree 3 in variable x, where the coefficient of x² is –7.

Solution

A degree-3 polynomial has highest power x³. The coefficient of x² must be –7.

p(x) = x³ – 7x² + 2x + 1

📌 Many answers are possible; any polynomial with degree 3 and coefficient of x² equal to –7 is correct.

Example: x³ – 7x² + 2x + 1

Q2 Find the values of: (i) 5x² – 3x + 7 if x = 1 (ii) 4t³ – t² + 6 if t = a

Solution

(i) 5x² – 3x + 7 at x = 1

= 5(1)² – 3(1) + 7 = 5 – 3 + 7 = 9

Value = 9

(ii) 4t³ – t² + 6 at t = a

= 4(a)³ – (a)² + 6 = 4a³ – a² + 6

Value = 4a³ – a² + 6

Q3 If we multiply a number by 5/2 and add 2/3 to the product, we get –7/12. Find the number.

Solution

Let the number = n. Form the equation:

(5/2)n + 2/3 = –7/12

Isolate the term with n:

(5/2)n = –7/12 – 2/3 = –7/12 – 8/12 [converting 2/3 = 8/12] = –15/12 = –5/4

n = (–5/4) ÷ (5/2) = (–5/4) × (2/5) = –10/20 = –1/2

Verify: (5/2)(–1/2) + 2/3 = –5/4 + 2/3 = –15/12 + 8/12 = –7/12 ✓

The number = –1/2

Q4 A positive number is 5 times another number. If 21 is added to both, one of the new numbers becomes twice the other. Find the numbers.

Solution

Let smaller number = n. Then larger = 5n.

After adding 21: n + 21 and 5n + 21.

The larger becomes twice the smaller:

5n + 21 = 2(n + 21) 5n + 21 = 2n + 42 3n = 21 n = 7

Numbers: 7 and 5 × 7 = 35.

Verify: 35 + 21 = 56 = 2 × (7 + 21) = 2 × 28 = 56 ✓

The two numbers are 7 and 35.

Q5 If you have ₹800 and save ₹250 every month, find the amount after (i) 6 months and (ii) 2 years. Express as a linear pattern.

Solution

Linear pattern: A(n) = 800 + 250n, where n = number of months.

(i) After 6 months

A(6) = 800 + 250(6) = 800 + 1500 = ₹2,300

₹2,300

(ii) After 2 years = 24 months

A(24) = 800 + 250(24) = 800 + 6000 = ₹6,800

₹6,800

Linear pattern: A(n) = 800 + 250n (rupees after n months)

★Q6 The digits of a two-digit number differ by 3. If the digits are interchanged and the resulting number is added to the original, we get 143. Find both numbers.

Solution

Let tens digit = a, units digit = b, where a – b = 3 (so a = b + 3).

Original number = 10a + b. Reversed = 10b + a.

(10a + b) + (10b + a) = 143 11a + 11b = 143 a + b = 13

Solve the system: a + b = 13 and a – b = 3:

2a = 16 → a = 8 b = 13 – 8 = 5

Original number = 85, reversed = 58.

Verify: 85 + 58 = 143 ✓ | Digits differ by 8 – 5 = 3 ✓

The two numbers are 85 and 58.

★Q7 Draw graphs of (i) y = –3x+4 (ii) 2y = 4x+7 (iii) 5y = 6x–10 (iv) 3y = 6x–11. Identify slopes and y-intercepts. Are any lines parallel?

Solution

Rewrite all equations in slope-intercept form y = ax + b:

Equation	Slope-Intercept Form	Slope (a)	y-intercept (b)	y-axis point
(i) y = –3x + 4	y = –3x + 4	–3	4	(0, 4)
(ii) 2y = 4x + 7	y = 2x + 3.5	2	3.5	(0, 3.5)
(iii) 5y = 6x – 10	y = (6/5)x – 2	6/5	–2	(0, –2)
(iv) 3y = 6x – 11	y = 2x – 11/3	2	–11/3	(0, –11/3)

Lines (ii) and (iv) both have slope = 2 but different y-intercepts (3.5 and –11/3). Therefore they are parallel to each other.

Fig: Lines (ii) and (iv) shown as parallel (dashed). Both have slope = 2.

Lines (ii) y = 2x + 3.5 and (iv) y = 2x – 11/3 are parallel (same slope = 2).

★Q8 y = (9/5)(x – 273) + 32 relates Kelvin (x) to Fahrenheit (y). (i) Find °F when x = 313 K. (ii) Find K when y = 158°F.

Solution

(i) x = 313 K → y = ?

y = (9/5)(313 – 273) + 32 = (9/5)(40) + 32 = 72 + 32 = 104 °F

Temperature = 104°F

(ii) y = 158°F → x = ?

158 = (9/5)(x – 273) + 32 158 – 32 = (9/5)(x – 273) 126 = (9/5)(x – 273) x – 273 = 126 × (5/9) x – 273 = 70 x = 343 K

Temperature = 343 K

★Q9 Work = force × distance. Express in linear form with constant force = 3 units. What is work done when distance = 2 units?

Solution

Work w = F × d. With F = 3:

w = 3d

When d = 2:

w = 3 × 2 = 6 units

Fig: w = 3d passes through origin. Point (2, 6) is verified on the graph.

w = 3d | Work done when d = 2 is 6 units.

★Q10 Graph of linear polynomial p(x) passes through (1, 5) and (3, 11). (i) Find p(x). (ii) Find where it cuts the axes. (iii) Draw graph and verify.

Solution

Let p(x) = ax + b. Use the two given points:

At (1, 5): a(1) + b = 5 → a + b = 5 …(1) At (3, 11): a(3) + b = 11 → 3a + b = 11 …(2)

Subtract (1) from (2):

2a = 6 → a = 3 From (1): b = 5 – 3 = 2

p(x) = 3x + 2

Where it cuts axes:

x-axis (y=0): 3x + 2 = 0 → x = –2/3 → Point (–2/3, 0) y-axis (x=0): p(0) = 2 → Point (0, 2)

Fig: p(x) = 3x + 2. Cuts x-axis at (–2/3, 0) and y-axis at (0, 2).

p(x) = 3x + 2 | Cuts x-axis at (–2/3, 0) | Cuts y-axis at (0, 2)

★Q11 p(x) = ax+b and q(x) = cx+d such that: (i) p(0)=5 (ii) p(x)–q(x) cuts x-axis at (3,0) (iii) p(x)+q(x) = 6x+4 for all x. Find p(x) and q(x).

Solution

From condition (i): p(0) = b = 5.

From condition (iii): p(x) + q(x) = (a+c)x + (b+d) = 6x + 4.

a + c = 6 …(A) b + d = 4 → 5 + d = 4 → d = –1

From condition (ii): p(x)–q(x) = (a–c)x + (b–d) passes through (3, 0):

(a–c)(3) + (b–d) = 0 3(a–c) + (5–(–1)) = 0 3(a–c) + 6 = 0 a – c = –2 …(B)

Solve (A) and (B):

From (A): a + c = 6 From (B): a – c = –2 Add: 2a = 4 → a = 2 Then c = 6 – 2 = 4

So p(x) = 2x + 5 and q(x) = 4x – 1.

Verify: p(0)=5 ✓ | p(x)+q(x)=6x+4 ✓ | p(x)–q(x)=–2x+6; at x=3: –6+6=0 ✓

p(x) = 2x + 5 | q(x) = 4x – 1

★Q12 Growing pattern of hexagons made with matchsticks. A new hexagon sharing a side is added each stage. Find the rule for the n^th stage.

Solution

Fig: Hexagon matchstick pattern — each new hexagon shares one side (saves 1 matchstick).

Stage 1 (1 hexagon): 6 matchsticks.
Each new hexagon added shares 1 side, so requires 5 new matchsticks (not 6).

Stage (n)	1	2	3	4	5	n
Matchsticks	6	11	16	21	26	5n + 1

Rule: Matchsticks at stage n = 5n + 1.

M(n) = 5n + 1

(iv) Stage 15:

M(15) = 5(15) + 1 = 75 + 1 = 76 matchsticks

(v) Can 200 matchsticks form a stage?

5n + 1 = 200 5n = 199 n = 39.8 (not a whole number)

Since n must be a positive integer, 200 matchsticks cannot form any stage of this pattern.

Rule: M(n) = 5n + 1 | Stage 15 → 76 matchsticks | 200 matchsticks → Not possible.

★Q13 p(x) passes through (2,3) and (6,11). q(x) passes through (4,–1) and is parallel to p(x). Find p(x), q(x), and where they cut the x-axis.

Solution

Find p(x) = ax + b using (2,3) and (6,11):

3 = 2a + b …(1) 11 = 6a + b …(2) Subtract: 8 = 4a → a = 2 Then b = 3 – 2(2) = –1

p(x) = 2x – 1

q(x) is parallel to p(x) → same slope: q(x) = 2x + d.
Using point (4, –1):

–1 = 2(4) + d –1 = 8 + d d = –9

q(x) = 2x – 9

x-axis intercepts:

p(x) = 0: 2x – 1 = 0 → x = 1/2 → Point (1/2, 0) q(x) = 0: 2x – 9 = 0 → x = 9/2 → Point (9/2, 0)

p(x) = 2x–1 (cuts x-axis at (1/2, 0)) | q(x) = 2x–9 (cuts x-axis at (9/2, 0))

★Q14 What do all linear functions of the form f(x) = ax + a, a > 0, have in common?

Solution

Rewrite: f(x) = ax + a = a(x + 1).

Find the x-intercept (where f(x) = 0):

a(x + 1) = 0 Since a ≠ 0 (a > 0), we get x + 1 = 0 → x = –1

All such lines pass through the point (–1, 0), regardless of the value of a.

Fig: All lines f(x) = ax + a (for different values of a > 0) pass through the common point (–1, 0).

All functions of the form f(x) = ax + a pass through the common point (–1, 0).

Chapter Summary

POLYNOMIAL TYPES

Degree 0 → Constant | Degree 1 → Linear

Degree 2 → Quadratic | Degree 3 → Cubic

KEY FORMULAS

y = ax + b (slope a, y-intercept b)

Parallel lines: same slope a, different b

LINEAR GROWTH / DECAY

Growth: +constant per step → positive slope

Decay: –constant per step → negative slope

GRAPH RULES

y = ax passes through origin (0,0)

y = ax + b cuts y-axis at (0, b)

Ch-2 Introduction to Linear Polynomials Class 9th Maths NCERT Solution (Ganita Manjari )

Chapter 2 · Introduction toLinear Polynomials

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Chapter 2 · Introduction to
Linear Polynomials