#### Topics

##### Number Systems

##### Real Numbers

##### Algebra

##### Pair of Linear Equations in Two Variables

- Linear Equation in Two Variables
- Graphical Method of Solution of a Pair of Linear Equations
- Substitution Method
- Elimination Method
- Cross - Multiplication Method
- Equations Reducible to a Pair of Linear Equations in Two Variables
- Consistency of Pair of Linear Equations
- Inconsistency of Pair of Linear Equations
- Algebraic Conditions for Number of Solutions
- Simple Situational Problems
- Pair of Linear Equations in Two Variables
- Relation Between Co-efficient

##### Arithmetic Progressions

##### Quadratic Equations

- Quadratic Equations
- Solutions of Quadratic Equations by Factorization
- Solutions of Quadratic Equations by Completing the Square
- Nature of Roots of a Quadratic Equation
- Relationship Between Discriminant and Nature of Roots
- Situational Problems Based on Quadratic Equations Related to Day to Day Activities to Be Incorporated
- Application of Quadratic Equation

##### Polynomials

##### Geometry

##### Circles

- Concept of Circle - Centre, Radius, Diameter, Arc, Sector, Chord, Segment, Semicircle, Circumference, Interior and Exterior, Concentric Circles
- Tangent to a Circle
- Number of Tangents from a Point on a Circle
- Concept of Circle - Centre, Radius, Diameter, Arc, Sector, Chord, Segment, Semicircle, Circumference, Interior and Exterior, Concentric Circles

##### Triangles

- Similar Figures
- Similarity of Triangles
- Basic Proportionality Theorem (Thales Theorem)
- Criteria for Similarity of Triangles
- Areas of Similar Triangles
- Right-angled Triangles and Pythagoras Property
- Similarity of Triangles
- Application of Pythagoras Theorem in Acute Angle and Obtuse Angle
- Triangles Examples and Solutions
- Angle Bisector
- Similarity of Triangles
- Ratio of Sides of Triangle

##### Constructions

- Division of a Line Segment
- Construction of Tangents to a Circle
- Constructions Examples and Solutions

##### Trigonometry

##### Heights and Distances

##### Trigonometric Identities

##### Introduction to Trigonometry

- Trigonometry
- Trigonometry
- Trigonometric Ratios
- Trigonometric Ratios and Its Reciprocal
- Trigonometric Ratios of Some Special Angles
- Trigonometric Ratios of Complementary Angles
- Trigonometric Identities
- Proof of Existence
- Relationships Between the Ratios

##### Statistics and Probability

##### Probability

##### Statistics

##### Coordinate Geometry

##### Lines (In Two-dimensions)

##### Mensuration

##### Areas Related to Circles

- Perimeter and Area of a Circle - A Review
- Areas of Sector and Segment of a Circle
- Areas of Combinations of Plane Figures
- Circumference of a Circle
- Area of Circle

##### Surface Areas and Volumes

- Concept of Surface Area, Volume, and Capacity
- Surface Area of a Combination of Solids
- Volume of a Combination of Solids
- Conversion of Solid from One Shape to Another
- Frustum of a Cone
- Concept of Surface Area, Volume, and Capacity
- Surface Area and Volume of Different Combination of Solid Figures
- Surface Area and Volume of Three Dimensional Figures

##### Internal Assessment

## Notes

Here what will be the distance between points PQ? To find this we will use distance formula.

Distance formula says, PQ= `sqrt[ (x_2-x_1)^2 + (y_2-y_1)^2]`

Example1- Find the distance between the points P(6,-6) and O(0,0)

Solution: `x_2=0, x_1=6, y_2=0 and y_1=-6`

PO= `sqrt[ (x_2-x_1)^2 + (y_2-y_1)^2]`

= `sqrt[ (0-6)^2 + (0+6)^2]`

= `sqrt[ (-6)^2 + (6)^2]`

= `sqrt[ 36+36]`

= `sqrt72`

PO = 6 `sqrt2` units

Example2- Find the values of x for which the distance between the points P(4,-5) and Q(12,x) is 10 units

Solution: PQ=10

`x_2= 12, x_1=4, y_2=x and y_1=-5`

PQ= `sqrt[ (x_2-x_1)^2 + (y_2-y_1)^2]`

10 = `sqrt[ (12-4)^2 + (x+5)^2]`

= `sqrt[ (8)^2 + (x+5)^2]`

= `sqrt[ 64 + (x+5)^2]`

Sqauring both the sides

`100= 64+ (x+5)^2`

`36= (x+5)^2`

`+or-6 = x+5`

`x= 6-5 or x= -6-5`

`x=1 or x= -11`

Example3- Find the point P on x-axis which is equidistance from the points A(5,-2) and B(-3,2)

Solution: Let the point be P(x,0)

Now, PA=PB [Given as point P on x-axis is equidistance from the points A(5,-2) and B(-3,2)]

`"PA"^2="PB"^2`

`(x_2-x_1)^2 + (y_2-y_1)^2 = (x_2-x_1)^2 + (y_2-y_1)^2`

`(5-x)^2 + (-2-0)^2 = (-3-x)^2 + (2-0)^2`

`25+x^2-10x+4 = (-3)^2 +x^2 -2(-3)(x) +4`

`25+x^2-10x+4 = 9+x^2+6x+4`

`16x= 16`

`x=1`

Therefore, point P is (1,0)

#### Shaalaa.com | Coordinate Geometry part 2 (Distance Formula)

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