NCERT Ganita Manjari Part II • Linear Equations & Graphical Coordinate Geometry
1. Standard Form of a Linear Equation in Two Variables
Core Definition
An equation that can be put in the form:
$$ax + by + c = 0$$
where $a, b, c$ are real numbers and $a$ and $b$ are not both zero ($a^2 + b^2 \neq 0$), is called a linear equation in two variables ($x$ and $y$).
2. Solutions of a Linear Equation
A solution is an ordered pair of numbers $(\alpha, \beta)$ such that substituting $x = \alpha$ and $y = \beta$ satisfies the equation ($a\alpha + b\beta + c = 0$).
Fundamental Theorem: A linear equation in two variables has infinitely many solutions!
Geometrically, every single solution $(\alpha, \beta)$ corresponds to a unique point on the Cartesian plane that lies on the straight line representing the equation.
HOW TO FIND MULTIPLE SOLUTIONS
Express one variable explicitly in terms of the other:
$$y = \frac{-c - ax}{b}$$
Substitute arbitrary values for $x$ (such as $x = 0, 1, 2, -1$) to compute the corresponding unique values for $y$.
3. Graph of a Linear Equation: The Straight Line
Every first-degree polynomial equation in two variables produces a continuous, unbroken straight line on the Cartesian plane.
Key Features of the Graph:
X-intercept: The point where the line crosses the X-axis (set $y = 0 \implies x = -\frac{c}{a}$). Point: $\left(-\frac{c}{a}, 0\right)$.
Y-intercept: The point where the line crosses the Y-axis (set $x = 0 \implies y = -\frac{c}{b}$). Point: $\left(0, -\frac{c}{b}\right)$.
Number of points required: Only two distinct points are needed to determine a line (Euclid's Postulate 1). A third point is always calculated as a check for collinearity.
4. The Concept of Slope (Gradient)
The slope ($m$) measures the rate of vertical rise per unit of horizontal run:
$$\text{Slope } m = \frac{\text{Vertical Change } (\Delta y)}{\text{Horizontal Change } (\Delta x)} = \frac{y_2 - y_1}{x_2 - x_1}$$
$$\text{Slope-Intercept Form: } y = mx + c$$
where $m$ is the slope (gradient) and $c$ is the Y-intercept.
Figure 13.1: Geometric representation of slope $m = \frac{\Delta y}{\Delta x}$ on a Cartesian straight line.
Slope Value ($m$)
Visual Direction of Line
Example Equation
Positive ($m > 0$)
Rises upwards from left to right ($\nearrow$)
$y = 2x + 1$
Negative ($m < 0$)
Falls downwards from left to right ($\searrow$)
$y = -3x + 4$
Zero ($m = 0$)
Horizontal line parallel to X-axis ($\rightarrow$)
$y = 5$
Undefined ($\Delta x = 0$)
Vertical line parallel to Y-axis ($\uparrow$)
$x = 3$
5. Special Equations of Lines Parallel to Axes
Equation of X-axis: $y = 0$ (all points on the X-axis have a Y-coordinate of zero).
Equation of Y-axis: $x = 0$ (all points on the Y-axis have an X-coordinate of zero).
Line parallel to X-axis: $y = k$ (where $k$ is a constant distance from the X-axis).
Line parallel to Y-axis: $x = k$ (where $k$ is a constant distance from the Y-axis).
6. Pairs of Linear Equations (Systems of Equations)
When considering two lines simultaneously:
$$a_1 x + b_1 y + c_1 = 0 \quad \text{and} \quad a_2 x + b_2 y + c_2 = 0$$
Four solutions: $(0, 4), (6, 0), (3, 2), (-3, 6)$.
Example 2: Real-World Word Problem & Graph HOTS
Problem: In a city, the taxi fare is calculated as follows: For the first kilometre, the fare is ₹15, and for the subsequent distance, it is ₹8 per km. Taking the total distance covered as $x\text{ km}$ and total fare as ₹$y$, write a linear equation and determine the fare for a trip of $12\text{ km}$.
Solution:
Total distance $= x\text{ km}$.
Distance for the first kilometre $= 1\text{ km}$ at ₹15.
Remaining distance $= (x - 1)\text{ km}$ at ₹8 per km.
Total fare $y$:
$$y = 15 + 8(x - 1) = 15 + 8x - 8 = 8x + 7$$
Standard form:
$$8x - y + 7 = 0$$
For a distance of $x = 12\text{ km}$:
$$y = 8(12) + 7 = 96 + 7 = 103\text{ rupees}$$
Answer: The total fare is ₹103.
Example 3: Solving a Pair by Elimination NCERT
Problem: Solve the following system of linear equations algebraically:
$$3x + 2y = 11$$
$$2x + 3y = 4$$
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