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Introduction to Linear Polynomials - Linear Relationships

Grade 9CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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A polynomial of degree 11 is called a linear polynomial. Its standard form is p(x)=ax+bp(x) = ax + b, where aa and bb are real numbers and a≠0a \neq 0.

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A linear polynomial has exactly one zero, which is the value of xx that makes p(x)=0p(x) = 0.

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Geometrically, the graph of a linear polynomial y=ax+by = ax + b is a straight line. The zero of the polynomial is the xx-coordinate of the point where the line intersects the xx-axis.

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Linear relationships in two variables are expressed in the form ax+by+c=0ax + by + c = 0. Every point (x,y)(x, y) that satisfies this equation lies on the line representing the relationship.

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A linear equation in two variables has infinitely many solutions, each representing a point on the line.

📐Formulae

p(x)=ax+b(a≠0)p(x) = ax + b \quad (a \neq 0) (Standard form of a Linear Polynomial)

x=−bax = -\frac{b}{a} (Zero of a Linear Polynomial)

ax+by+c=0ax + by + c = 0 (General form of a Linear Equation in two variables)

y=mx+cy = mx + c (Slope-intercept form where mm is the slope and cc is the yy-intercept)

💡Examples

Problem 1:

Find the zero of the linear polynomial p(x)=5x−10p(x) = 5x - 10.

Solution:

To find the zero, we set p(x)=0p(x) = 0: 5x−10=05x - 10 = 0 5x=105x = 10 x=105x = \frac{10}{5} x=2x = 2

Explanation:

The zero of the polynomial is the value of xx for which the expression equals zero. Here, x=2x = 2 is the zero.

Problem 2:

Represent the following statement as a linear equation in two variables: 'The cost of a notebook is Rs 5 more than twice the cost of a pen.'

Solution:

Let the cost of a notebook be xx and the cost of a pen be yy. According to the problem: x=2y+5x = 2y + 5 Rewriting in standard form: x−2y−5=0x - 2y - 5 = 0

Explanation:

We define two variables for the two unknown quantities and create an equation based on the given relationship.

Problem 3:

Check if (3,2)(3, 2) is a solution for the linear relationship 2x−3y=02x - 3y = 0.

Solution:

Substitute x=3x = 3 and y=2y = 2 into the LHS: LHS=2(3)−3(2)LHS = 2(3) - 3(2) LHS=6−6LHS = 6 - 6 LHS=0LHS = 0 Since LHS=RHSLHS = RHS, the point (3,2)(3, 2) is a solution.

Explanation:

A point is a solution to a linear equation if substituting its coordinates into the equation makes the left-hand side equal to the right-hand side.