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Relations and Functions - Ordered Pairs-advanced

Grade 9CBSE

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

πŸ”‘Concepts

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An ordered pair consists of two elements grouped in a specific order, denoted as (a,b)(a, b), where aa is the first element and bb is the second. Crucially, (a,b)β‰ (b,a)(a, b) \neq (b, a) unless a=ba = b. This property allows us to represent positions uniquely on a Cartesian plane.

Cartesian plane showing that the ordered pair (2, 4) is distinct from (4, 2).
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Two ordered pairs (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) are equal if and only if their corresponding components are equal, i.e., x1=x2x_1 = x_2 and y1=y2y_1 = y_2. This equality is the basis for solving algebraic equations involving coordinate geometry.

Visual representation of equality of ordered pairs.
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The Cartesian Product AΓ—BA \times B is the set of all possible ordered pairs (a,b)(a, b) where a∈Aa \in A and b∈Bb \in B. If n(A)=pn(A) = p and n(B)=qn(B) = q, then the total number of ordered pairs is n(AΓ—B)=pΓ—qn(A \times B) = p \times q.

Flowchart showing elements of Set A and Set B combining to form ordered pairs.
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Ordered pairs can be visualized as points in the 2D plane. The first component corresponds to the horizontal displacement (abscissa), and the second component corresponds to the vertical displacement (ordinate).

Plotting points across different quadrants using ordered pairs.

πŸ“Formulae

(x1,y1)=(x2,y2)β€…β€ŠβŸΊβ€…β€Šx1=x2Β andΒ y1=y2(x_1, y_1) = (x_2, y_2) \iff x_1 = x_2 \text{ and } y_1 = y_2

AΓ—B={(a,b):a∈A,b∈B}A \times B = \{ (a, b) : a \in A, b \in B \}

n(AΓ—B)=n(A)Γ—n(B)n(A \times B) = n(A) \times n(B)

AΓ—(BβˆͺC)=(AΓ—B)βˆͺ(AΓ—C)A \times (B \cup C) = (A \times B) \cup (A \times C)

πŸ’‘Examples

Problem 1:

Find the values of xx and yy if (x+3,5)=(6,2yβˆ’1)(x + 3, 5) = (6, 2y - 1).

Solution:

Given the equality of ordered pairs: (x+3,5)=(6,2yβˆ’1)(x + 3, 5) = (6, 2y - 1)

By equating the first components: x+3=6x + 3 = 6 x=6βˆ’3x = 6 - 3 x=3x = 3

By equating the second components: 5=2yβˆ’15 = 2y - 1 2y=5+12y = 5 + 1 2y=62y = 6 y=62y = \frac{6}{2} y=3y = 3

Therefore, x=3x = 3 and y=3y = 3.

Explanation:

To find the variables, we use the definition that two ordered pairs are equal if their corresponding coordinates are equal.

Problem 2:

If (a3+1,bβˆ’23)=(53,13)(\frac{a}{3} + 1, b - \frac{2}{3}) = (\frac{5}{3}, \frac{1}{3}), determine the values of aa and bb.

Solution:

Equating the first elements: a3+1=53\frac{a}{3} + 1 = \frac{5}{3} a3=53βˆ’1\frac{a}{3} = \frac{5}{3} - 1 a3=5βˆ’33\frac{a}{3} = \frac{5 - 3}{3} a3=23\frac{a}{3} = \frac{2}{3} a=2a = 2

Equating the second elements: bβˆ’23=13b - \frac{2}{3} = \frac{1}{3} b=13+23b = \frac{1}{3} + \frac{2}{3} b=33b = \frac{3}{3} b=1b = 1

Final values: a=2,b=1a = 2, b = 1.

Explanation:

This involves solving simple linear equations derived from the property of equality of ordered pairs.

Problem 3:

If A={1,2}A = \{1, 2\} and B={3,4}B = \{3, 4\}, find AΓ—BA \times B and show that n(AΓ—B)=n(A)Γ—n(B)n(A \times B) = n(A) \times n(B).

Solution:

Set A={1,2}A = \{1, 2\} has n(A)=2n(A) = 2 elements. Set B={3,4}B = \{3, 4\} has n(B)=2n(B) = 2 elements.

The Cartesian product AΓ—BA \times B is: AΓ—B={(1,3),(1,4),(2,3),(2,4)}A \times B = \{ (1, 3), (1, 4), (2, 3), (2, 4) \}

Counting the elements in AΓ—BA \times B, we get n(AΓ—B)=4n(A \times B) = 4. Calculating n(A)Γ—n(B)n(A) \times n(B): 2Γ—2=42 \times 2 = 4

Hence, n(AΓ—B)=n(A)Γ—n(B)=4n(A \times B) = n(A) \times n(B) = 4.

Explanation:

The Cartesian product forms all possible pairs by taking the first element from set A and the second from set B.

Problem 4:

Determine the values of kk and mm such that the ordered pair (2kβˆ’5,m+3)(2k - 5, m + 3) lies at the origin (0,0)(0, 0). Visualize the position of the point if k=4k=4 and m=1m=1.

Graph showing the origin and the calculated point (3, 4).

Solution:

Given (2kβˆ’5,m+3)=(0,0)(2k - 5, m + 3) = (0, 0). By the equality of ordered pairs:

  1. 2kβˆ’5=0β€…β€ŠβŸΉβ€…β€Š2k=5β€…β€ŠβŸΉβ€…β€Šk=52=2.52k - 5 = 0 \implies 2k = 5 \implies k = \frac{5}{2} = 2.5
  2. m+3=0β€…β€ŠβŸΉβ€…β€Šm=βˆ’3m + 3 = 0 \implies m = -3 For the second part, if k=4k=4 and m=1m=1: x=2(4)βˆ’5=8βˆ’5=3x = 2(4) - 5 = 8 - 5 = 3 y=1+3=4y = 1 + 3 = 4 The point is (3,4)(3, 4).

Explanation:

To find the variables, we equate the corresponding elements of the ordered pairs. A point at the origin must have both its xx and yy coordinates equal to zero.

Problem 5:

Given the set A={x:x2βˆ’5x+6=0}A = \{x : x^2 - 5x + 6 = 0\} and B={1,2}B = \{1, 2\}, find the Cartesian product AΓ—BA \times B and represent it on a coordinate plane.

Discrete points representing the Cartesian product of two sets.

Solution:

First, solve for elements of AA: x2βˆ’5x+6=0β€…β€ŠβŸΉβ€…β€Š(xβˆ’2)(xβˆ’3)=0β€…β€ŠβŸΉβ€…β€Šx=2,3x^2 - 5x + 6 = 0 \implies (x - 2)(x - 3) = 0 \implies x = 2, 3. So, A={2,3}A = \{2, 3\}. B={1,2}B = \{1, 2\}. AΓ—B={(2,1),(2,2),(3,1),(3,2)}A \times B = \{(2, 1), (2, 2), (3, 1), (3, 2)\}.

Explanation:

We first identify the elements of set A by solving the quadratic equation. Then, we pair every element of A with every element of B to form the Cartesian product set.

Ordered Pairs-advanced Class 9 Notes & Examples | CBSE Maths