Definitions [7]
The terms that do not have the same literal coefficients are called unlike terms.
For example:
6a, 6ab and 6ac are unlike terms.
Terms having the same literal part (same variables with the same powers) are called like terms.
For example:
xy, 5xy, -4xy, etc. are like terms
Identity: An identity is an equality, which is true for all values of the variables in equality.
A balanced equation is one in which the value of the left-hand side (LHS) is equal to the value of the right-hand side (RHS).
A Pictograph is a chart that uses pictures or symbols to represent data. Each picture stands for a specific number of items, making the data easy to understand at a glance.
The two mutually perpendicular number lines intersecting each other at their zeroes are called rectangular axes or coordinate axes, or axes of reference.
The position of a point in a plane is expressed by a pair of numbers, one concerning the x-axis and the other concerning the y-axis. called co-ordinates.
-
x → distance from y-axis (abscissa)
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y → distance from x-axis (ordinate)
Formulae [7]
(a + b)2 = a2 + 2ab + b2
(a - b)2 = a2 - 2ab + b2.
(a + b)(a - b) = a2 - b2
- (x + a)(x + b) = x2 + (a + b)x + ab
- (a + b)3 = a3 + 3a2b + 3ab2 + b3.
- (a - b)3 = a3 - 3a2b + 3ab2 - b3.
- (a + b)2 = a2 + 2ab + b2
- (a - b)2 = a2 - 2ab + b2
- (a + b)(a - b) = a2 - b2
Theorems and Laws [4]
Prove that the diagonals of a rectangle ABCD with vertices A(2, –1), B(5, –1), C(5, 6) and D(2, 6) are equal and bisect each other.
The vertices of the rectangle ABCD are A(2, -1), B(5, -1), C(5, 6) and D(2, 6) Now,
`"Coordinates of midpoint of" AC = ((2+5)/2 , (-1+6)/2) = (7/5 ,5/2)`
`"Coordinates of midpoint of " BD = ((5+2)/2 , (-1+6)/2)= (7/2,5/2)`
Since, the midpoints of AC and BD coincide, therefore the diagonals of rectangle ABCD bisect each other.
If the points P(x, y) is equidistant from the points A(5, 1)and B(–1, 5), prove that 3x = 2y.
As per the question, we have
AP = BP
`⇒ sqrt((x -5)^2 +(y-1)^2) = sqrt((x+1)^2 +(y-5)^2)`
`⇒(x-5)^2 +(y-1)^2 = (x+1)^2 +(y-5)^2` (Squaring both sides)
`⇒x^2 - 10x +25 + y^2 -2y +1 = x^2 +2x +1+y^2 -10y+25`
⇒ –10x – 2y = 2x – 10y
⇒ 8y = 12x
⇒ 3x = 2y
If the point (x, y) is equidistant form the points (a + b, b – a) and (a – b, a + b), prove that bx = ay.
As per the question, we have
`sqrt((x-a-b)^2 +(y-b+a)^2 ) = sqrt((x-a+b)^2 +(y-a-b)^2)`
`⇒(x-a-b)^2 +(y-b+a)^2 = (x-a+b)^2 +(y-a-b)^2` (Squaring both sides)
`⇒x^2 + (a+b)^2 -2x (a+b) +y^2 +(a-b)^2 -2y(a-b)=x^2 +(a-b)^2 -2x(a-b)+y^2 +(a+b)^2 -2y (a+b)`
`⇒-x(a+b) - y (a-b) = -x(a-b) -y(a+b)`
`⇒-xa -xb -ay +by = -xa + bx -ya-by`
⇒ by=bx
Hence, . bx = ay
Prove that the points A(–4, –1), B(–2, –4), C(4, 0) and D(2, 3) are the vertices of a rectangle.
Let A (-4,-1); B (-2,-4); C (4, 0) and D (2, 3) be the vertices of a quadrilateral. We have to prove that the quadrilateral ABCD is a rectangle.
So we should find the lengths of opposite sides of quadrilateral ABCD.
`AB = sqrt((-2+4)^2) + (-4 + 1)^2)`
`= sqrt(4 + 9)`
`= sqrt13`
`CD = sqrt((4 - 2)^2 + (0 - 3)^2)`
`= sqrt(4 +9)`
`= sqrt13`
Opposite sides are equal. So now we will check the lengths of the diagonals.
`AC = sqrt((4 + 4)^2 + (0 + 1)^2)`
`= sqrt(64 + 1)`
`= sqrt(65)`
`BD = sqrt((2 + 2)^2 + (3 + 4)^2)`
`= sqrt(16 + 49)`
`= sqrt65`
Opposite sides are equal as well as the diagonals are equal. Hence ABCD is a rectangle.
The given points are A (-4,-1); B (-2,-4); C (4, 0) and D (2, 3) .
`AB = sqrt({-2-(-4)}^2 + { -4-(-1)}^2) = sqrt ((2)^2+(-3)^2) = sqrt(4+9) = sqrt(13) ` units
` BC = sqrt({ 4-(-2)}^2+{0-(-4)}^2) = sqrt((6)^2 +(4)^2) = sqrt(36+16) = sqrt(52) = 2 sqrt(13) units`
`CD = sqrt((2-4)^2 +(3-0)^2) = sqrt((-2)^2 +(3)^2) = sqrt(4+9) = sqrt(13) units`
`AD = sqrt({2-(-4)}^2 + {3-(-1)}^2) = sqrt((6)^2 +(4)^2) = sqrt(36+16) = sqrt(52) = 2 sqrt(13) units`
`Thus , AB = CD = sqrt(13) units and BC = AD = 2 sqrt(13) units`
Also , `AC = sqrt({4-(-4)}^2+{0-(-1)}^2) = sqrt ((8)^2+(1)^2 ) = sqrt(64+1) = sqrt(65) units`
`BD = sqrt({2-(-2)}^2 +{3-(-4)}^2) = sqrt((4)^2 +(7)^2) = sqrt(16+49) = sqrt(65) units`
Also, diagonal AC = diagonal BD
Hence, the given points form a rectanglr
Key Points
Sign Convention
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Right of y-axis → +x
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Left of y-axis → −x
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Above x-axis → +y
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Below x-axis → −y
Standard Line Results
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x = 0 → y-axis
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y = 0 → x-axis
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x = a → line parallel to the y-axis
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y = b → line parallel to the x-axis
Quadrant Reminder
| Quadrant | Sign of (x, y) |
|---|---|
| I | (+, +) |
| II | (−, +) |
| III | (−, −) |
| IV | (+, −) |
| Condition | Nature of Lines | Number of Solutions | Type of Pair |
|---|---|---|---|
| \[\frac{a_1}{a_2}\neq\frac{b_1}{b_2}\] | Intersecting | One (unique) solution | Consistent |
| \[\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2}\] | Parallel | No solution | Inconsistent |
| \[\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\] | Coincident | Infinitely many solutions | Dependent (consistent) |
Concepts [43]
- Algebraic Expressions
- Terms, Factors and Coefficients of Expression
- Classification of Terms in Algebra
- Addition of Algebraic Expressions
- Subtraction of Algebraic Expressions
- Multiplication of Algebraic Expressions
- Multiplying Monomial by Monomials
- Multiplying a Monomial by a Binomial
- Multiplying a Monomial by a Trinomial
- Multiplying a Binomial by a Binomial
- Multiplying a Binomial by a Trinomial
- Division of Algebraic Expressions
- Dividing a Monomial by a Monomial
- Dividing a Polynomial by a Monomial
- Concept of Identity
- Expansion of (a + b)2 = a2 + 2ab + b2
- Expansion of (a - b)2 = a2 - 2ab + b2
- Expansion of (a + b)(a - b) = a2-b2
- Expansion of (x + a)(x + b)
- Expansion of (a + b)3
- Expansion of (a - b)3
- Expansion of (x + a)(x + b)(x + c)
- Factorising Algebraic Expressions
- Factorisation by Taking Out Common Factors
- Factorisation by Taking Out the Common Binomial Factor from Each Term
- Factorisation by Regrouping Terms
- Factorisation Using Identities
- Factors of the Form (x + a)(x + b)
- Factorise Using the Identity (a + b)3
- Factorise Using the Identity (a – b)3
- Concept of Find the Error
- Expressions with Variables
- Equation in Mathematics
- Word Problems on Linear Equations
- Concept of Graph
- Cartesian Coordinate System
- Co-ordinate Geometry
- Quadrants and Sign Convention
- Plotting a Point in the Plane If Its Coordinates Are Given.
- Geometrical Representation of a Linear Equation by Plotting a Straight Line
- Geometrical Representation of a Linear Equation by Line Parallel to the Coordinate Axes
- Linear Pattern
- Graphical Method with Different Cases of Solution
