Concept Scroll LogoConceptScroll
Join or Sign In
MathematicsClass 9Introduction to

Introduction to | Class 9 Mathematics Notes

By ConceptScroll Team · Published on 17 July 2026 · 6 min read

Introduction to | Class 9 Mathematics Notes

Introduction to – this guide gives you a concise, exam-ready overview of Introduction to from Class 9 Mathematics, written by ConceptScroll editors and reviewed against the latest NCERT textbook.

2.6 VISUALISING LINEAR RELATIONSHIPS

This section teaches how to graph linear equations of the form y = ax + b on the coordinate plane. It explains the process of finding points by substituting values of x and calculating corresponding y values.

For example, the line y = 2x + 1 passes through points (0, 1) and (3, 7). Plotting these points and joining them forms the straight line representing the equation.

The section includes exercises to find more points on the line and verify that their coordinates satisfy the equation.

Examples 12 and 13 show plotting points for lines y = 3x and y = -2x respectively, confirming that points lie on straight lines.

Further, graphs of lines y = (1/2)x, y = x, and y = 2x are drawn to compare slopes. It is observed that lines with greater slope values are steeper.

The concept of slope is introduced as the measure of steepness, and its effect on the graph is explained. Positive slopes produce lines rising from left to right, while negative slopes produce lines falling from left to right.

The section also discusses the effect of changing the y-intercept b while keeping slope a fixed, showing that lines shift parallelly.

Examples 16 and accompanying figures illustrate these concepts with graphs of y = 2x - 1, y = 2x + 1, and y = 2x + 5.

The section concludes by summarizing that in y = ax + b, a is the slope and b is the y-intercept, and that lines with the same slope but different y-intercepts are parallel.

📊 Diagram: Fig. 2.5 shows the straight line y = 2x + 1; Fig. 2.6 and Fig. 2.7 show points plotted for lines y = 3x and y = -2x respectively; Fig. 2.8 and Fig. 2.9 illustrate graphs of y = (1/2)x, y = x, y = 2x; Fig. 2.10 and Fig. 2.11 show graphs of lines with negative slopes; Fig. 2.12A, 2.12B, 2.12C and Fig. 2.13 show graphs of y = 2x -1, y = 2x +1, y = 2x +5; Fig. 2.14 shows graphs of y = x + 3, y = 2x + 5, y = 3x - 2.

🧪 Activity: Think and Reflect questions involve identifying points on lines, understanding slopes and y-intercepts, and comparing graphs.

🔗 Connection: This section prepares students for exercises on graphing linear equations and understanding their properties.

Frequently asked questions

1. Suppose a plant has height 1.75 feet and it grows by 0.5 feet each month. (i) Find the height after 7 months. (ii) Make a table of values for $t$ varying from 0 to 10 months and show how the height, $h$, increases every month. (iii) Find an expression that relates $h$ and $t$, and explain why it represents linear growth.

(i) Initial height = 1.75 feet, growth per month = 0.5 feet Height after 7 months = Initial height + 7 × growth per month = 1.75 + 7 × 0.5 = 1.75 + 3.5 = 5.25 feet

(ii) Table of values:

t (months)012345678910
h (feet)1.752.252.753.253.754.254.755.255.756.256.75

(iii) Expression relating height h and time t:

h = 1.75 + 0.5t

This is a linear expression because h changes by a consta

2. A mobile phone is bought for ₹10,000. Its value decreases by ₹800 every year. (i) Find the value of the phone after 3 years. (ii) Make a table of values for $t$ varying from 0 to 8 years and show how the value of the phone, $\nu$, depreciates with time. (iii) Find an expression that relates $\nu$ and $t$, and explain why it represents linear decay.

(i) Initial value = ₹10,000 Depreciation per year = ₹800 Value after 3 years = Initial value - 3 × depreciation = 10000 - 3 × 800 = 10000 - 2400 = ₹7600

(ii) Table of values:

t (years)012345678
ν (₹)1000092008400760068006000520044003600

(iii) Expression relating value ν and time t:

ν = 10000 - 800t

This represents linear decay because the value decreases by a fixed amount (₹800) every y

3. The initial population of a village is 750. Every year, 50 people move from a nearby city to the village. (i) Find the population of the village after 6 years. (ii) Make a table of values for $t$ varying from 0 to 10 years and show how the population, $P$, increases every year. (iii) Find an expression that relates $P$ and $t$, and explain why it represents linear growth.

(i) Initial population = 750 Increase per year = 50 Population after 6 years = 750 + 6 × 50 = 750 + 300 = 1050

(ii) Table of values:

t (years)012345678910
P750800850900950100010501100115012001250

(iii) Expression relating population P and time t:

P = 750 + 50t

This represents linear growth because the population increases by a fixed number (50) every year. The

4. A telecom company charges ₹600 for a certain recharge scheme. This prepaid balance is reduced by ₹15 each day after the recharge. (i) Write an equation that models the remaining balance $b(x)$ after using the scheme for $x$ days. Explain why it represents linear decay. (ii) After how many days will the balance run out? (iii) Make a table of values for $x$ varying from 1 to 10 days and show how the balance $b(x)$, reduces with time.

(i) Initial balance = ₹600 Daily reduction = ₹15 Equation for balance after x days:

b(x) = 600 - 15x

This represents linear decay because the balance decreases by a fixed amount (₹15) each day.

(ii) To find when balance runs out, set b(x) = 0:

0 = 600 - 15x 15x = 600 x = 600 / 15 = 40 days

(iii) Table of values:

x (days)12345678910
b(x) (₹)585570555540525510495480465450

Ready to ace this chapter?

Get the full Introduction to chapter — interactive notes, diagrams, worked solutions, polls and a free practice quiz — in the ConceptScroll app.

Open in ConceptScroll →

Study smarter with ConceptScroll

Daily NCERT-aligned reels, AI doubt solving and chapter quizzes — all free.

Start learning free
#cbse notes#class 9#mathematics#ncert

Continue reading

← All articles