Important Questions for Class 9 Maths Chapter 2 Introduction to Linear Polynomials (2026-27)

Important Questions Class 9 Maths Chapter 2 Introduction to Linear Polynomials

This page gives Important Questions Class 9 Maths Chapter 2 Introduction to Linear Polynomials. A linear polynomial is an expression like ax + b, where the highest power of x is 1. The questions cover drawing and comparing straight-line graphs, slope and y-intercept, growing number patterns, and simple word problems that lead to linear equations.

Some questions use daily-life ideas, like saving money every month and changing temperature from Kelvin to Fahrenheit. Every answer is solved step by step, so students can check each step and learn the method. Teachers can also use these questions for class practice and revision.

Q1. Differentiate between the graphs of the equations y = 3x + 1, and y = −3x + 1.

Answer: For y = 3x + 1:

Taking a few values of x:

When x = −1,

y = 3(−1) + 1 = −2

So, one point is (−1, −2).

When x = 0,

y = 3(0) + 1 = 1

So, another point is (0, 1).

When x = 1,

y = 3(1) + 1 = 4

So, another point is (1, 4).

Thus, the graph of y = 3x + 1 passes through (−1, −2), (0, 1) and (1, 4).

For y = −3x + 1:

When x = −1,

y = −3(−1) + 1 = 4

So, one point is (−1, 4).

When x = 0,

y = −3(0) + 1 = 1

So, another point is (0, 1).

When x = 1,

y = −3(1) + 1 = −2

So, another point is (1, −2).

Thus, the graph of y = −3x + 1 passes through (−1, 4), (0, 1) and (1, −2).

Difference between the graphs:

Both graphs are straight lines.

Both have the same y-intercept, 1, so both pass through (0, 1).

The graph of y = 3x + 1 has slope 3, so it rises from left to right.

The graph of y = −3x + 1 has slope −3, so it falls from left to right.

Both lines have the same steepness because the magnitudes of their slopes are equal, |3| = |−3|.

The two graphs intersect at (0, 1).

Q2. Does this help you to conclude anything about the linear equation y = ax + b when a is fixed but b varies?

Answer: For the linear equation

y = ax + b

if a is fixed and b varies, the slope of the line remains unchanged because the slope is determined by a.

However, b represents the y-intercept. Therefore, changing b changes the point at which the line cuts the y-axis.

For example, if a = 2, then

y = 2x + 1

y = 2x + 3

y = 2x − 2

all have the same slope, 2, but different y-intercepts.

Hence, their graphs are parallel straight lines.

So, we can conclude that when a is fixed and b varies in y = ax + b, the lines remain parallel and shift upward or downward depending on the value of b.

Q3. If the congruent polygons of a prism have 10 sides, how many faces, edges and vertices does the prism have? What if the polygons have n sides?

Answer: Number of sides of a congruent polygon of a prism = 10

Number of faces = n + 2

= 10 + 2

= 12

Number of edges = 3 × n

= 3 × 10

= 30

Number of vertices = 2 × n

= 2 × 10

= 20

A polygon with n sides will have

Number of faces = n + 2

Number of edges = 3n

Number of vertices = 2n

Q4. Find the area of the region remaining at the n^th step in each of the shape sequences that lead to the Sierpinski fractals. Take the area of the starting square/triangle to be 1 sq. unit.

Answer: (a) Let the side of the square of Sierpinski's Carpet be 1 unit, so its area is 1 square unit.

As one-ninth of the square is removed, eight-ninths remains in each step.

Step 0: Area of whole square = (1)² = 1

Step 1: Area of remaining region = (8/9) × (1)² = 8/9

Step 2: Area of remaining region = (8/9) × (8/9) = 64/81

Step 3: Area of remaining region = (8/9) × (64/81) = 8³/9³ = 512/729

Area of remaining region after n^th step = (8/9)ⁿ sq. units

(b) Let the area of the triangle of Sierpinski's Gasket be 1 square unit.

One-fourth of the triangle is removed in each step, and three-fourths remains.

Step 0: Area of the whole triangle = 1 sq. unit.

Step 1: Area of remaining region = 3/4 sq. units.

Step 2: Area of remaining region = (3/4) × (3/4) = 9/16 sq. units

Area of remaining region after n^th step = (3/4)ⁿ sq. units

Q5. Look at the first three stages of a growing pattern of hexagons made using matchsticks. A new hexagon gets added at every stage which shares a side with the last hexagon of the previous stage.

(i) Draw the next two stages of the pattern. How many matchsticks will be required at these stages?

(ii) Complete the following table.

(iii) Find a rule to determine the number of matchsticks required for the n^th stage.

(iv) How many matchsticks will be required for the 15^th stage of the pattern?

(v) Can 200 matchsticks form a stage in this pattern? Justify your answer.

Answer: A single hexagon needs 6 matchsticks.

Since each new hexagon shares one side with the previous hexagon, only 5 new matchsticks are added at every new stage.

So the pattern is:

Stage 1 = 6

Stage 2 = 6 + 5 = 11

Stage 3 = 11 + 5 = 16

(i) Next two stages:

Stage 4: Number of matchsticks = 16 + 5 = 21

Stage 5: Number of matchsticks = 21 + 5 = 26

Therefore:

Stage 4 requires 21 matchsticks

Stage 5 requires 26 matchsticks

(ii) Complete table:

(iii) Rule for the n^th stage:

The number of matchsticks forms an arithmetic pattern: 6, 11, 16, 21, 26, ...

First term = 6

Difference = 5

So, Number of matchsticks in the n^th stage

= 6 + (n − 1) × 5

= 6 + 5n − 5

= 5n + 1

Hence, the rule is M_n = 5n + 1

(iv) Matchsticks required for the 15^th stage:

M₁₅ = 5(15) + 1

= 75 + 1

= 76

Therefore, 76 matchsticks will be required for the 15^th stage.

(v) Can 200 matchsticks form a stage in this pattern?

For some stage n,

5n + 1 = 200

⇒ 5n = 199

⇒ n = 199/5

⇒ n = 39.8

Since n is not a whole number, 200 matchsticks cannot form any stage in this pattern.

Therefore, 200 matchsticks cannot form a stage in this pattern.

Q6. What do all linear functions of the form f(x) = ax + a, a > 0, have in common?

Answer: Given: f(x) = ax + a, where a > 0

We can write it as: f(x) = a(x + 1)

Common properties of all such linear functions are:

1. Slope:

The slope is a, and since a > 0, all the lines have positive slope.

So, all these lines rise from left to right.

2. y-intercept:

Putting x = 0,

f(0) = a

So, the y-intercept is (0, a).

Since a > 0, all the lines cut the y-axis above the origin.

3. x-intercept:

To find where the line cuts the x-axis, put f(x) = 0

ax + a = 0

a(x + 1) = 0

Since a > 0, a is not zero.

So, x + 1 = 0

or x = −1

Thus, every line cuts the x-axis at the same point (−1, 0).

Therefore, all linear functions of the form f(x) = ax + a, a > 0, have the following in common:

all have positive slope

all cut the y-axis above the origin

all pass through the fixed point (−1, 0).

Q7. If the temperature of a liquid can be measured in Kelvin units as x K and in Fahrenheit units as y °F, the relation between the two systems of measurement of temperature is given by the linear equation y = (9/5)(x − 273) + 32.

(i) Find the temperature of the liquid in Fahrenheit if the temperature of the liquid is 313 K.

(ii) If the temperature is 158 °F, then find the temperature in Kelvin.

Answer: (i) When x = 313 K, we have to find y:

Using equation y = (9/5)(x − 273) + 32

y = (9/5)(313 − 273) + 32

⇒ y = (9/5)(40) + 32

⇒ y = 72 + 32

⇒ y = 104

Therefore, the temperature is 104 °F.

(ii) When y = 158 °F, we have to find x:

Using equation y = (9/5)(x − 273) + 32

158 = (9/5)(x − 273) + 32

Subtracting 32 from both sides, we get:

158 − 32 = (9/5)(x − 273)

⇒ 126 = (9/5)(x − 273)

Multiply both sides by 5/9

126 × (5/9) = x − 273

⇒ 70 = x − 273

⇒ x = 343

Therefore, the temperature is 343 K.

Q8. The digits of a two-digit number differ by 3. If the digits are interchanged, and the resulting number is added to the original number, we get 143. Find both the numbers.

Answer: Let the tens digit be x and the units digit be y.

Then the original number = 10x + y

The interchanged number = 10y + x

According to the question: (10x + y) + (10y + x) = 143

⇒ 11x + 11y = 143

⇒ 11(x + y) = 143

⇒ x + y = 13 ...(1)

The digits differ by 3

So, x − y = 3 ...(2)

Now adding (1) and (2), we get:

2x = 16

⇒ x = 8

Substitute x = 8 in (1), we have:

8 + y = 13

⇒ y = 5

Therefore, the original number = 85 and the interchanged number = 58

Hence, the two numbers are 85 and 58.

Q9. If you have ₹800 and you save ₹250 every month, find the amount you have after (i) 6 months (ii) 2 years. Express this as a linear pattern.

Answer: Initial amount = ₹800

Saving every month = ₹250

Linear pattern: Amount after n months = 800 + 250n

(i) Amount after 6 months:

= 800 + 250(6)

= 800 + 1500

= ₹2300

(ii) Amount after 2 years: 2 years = 24 months

Amount after 24 months:

= 800 + 250(24)

= 800 + 6000

= ₹6800

Therefore:

Amount after 6 months = ₹2300

Amount after 2 years = ₹6800

Linear pattern: A = 800 + 250n, where n is the number of months.

Q10. A positive number is 5 times another number. If 21 is added to both the numbers, then one of the new numbers becomes twice the other new number. What are the numbers?

Answer: Let the smaller number be x.

Then the larger number = 5x

After adding 21 to both numbers: New numbers are x + 21 and 5x + 21

According to the question: 5x + 21 = 2(x + 21)

⇒ 5x + 21 = 2x + 42

⇒ 5x − 2x = 42 − 21

⇒ 3x = 21

⇒ x = 7

So, the smaller number = 7 and the larger number = 5 × 7 = 35

Therefore, the two numbers are 7 and 35.

Q11. If we multiply a number by 5/2 and add 2/3 to the product, we get −7/12. Find the number.

Answer: Let the number be x.

According to the question:

(5/2)x + 2/3 = −7/12

Subtract 2/3 from both sides, we get:

(5/2)x = −7/12 − 2/3

⇒ (5/2)x = −7/12 − 8/12

⇒ (5/2)x = −15/12

⇒ (5/2)x = −5/4

Now multiply both sides by 2/5, we get:

x = −(5/4) × (2/5)

⇒ x = −1/2

Therefore, the number is −1/2.

Q12. Identify other points on the line by completing the following table.

Equation of the straight line: y = 2x + 1

Answer: Given the equation of the straight line:

y = 2x + 1

Substitute each value of x to find the corresponding value of y.

For x = 1:

y = 2(1) + 1 = 3

For x = 2:

y = 2(2) + 1 = 5

For x = 5:

y = 2(5) + 1 = 11

For x = 7:

y = 2(7) + 1 = 15

For x = 9:

y = 2(9) + 1 = 19

For x = 12:

y = 2(12) + 1 = 25

For x = 20:

y = 2(20) + 1 = 41

Therefore, the completed table is:

Hence, the points on the line are:

(1, 3), (2, 5), (5, 11), (7, 15), (9, 19), (12, 25), (20, 41)

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