Topics
Compound Interest
- Compound Interest as a Repeated Simple Interest Computation with a Growing Principal
- Use of Compound Interest in Computing Amount Over a Period of 2 Or 3-years
- Use of Formula
- Finding CI from the Relation CI = A – P
Commercial Mathematics
Goods and Services Tax (G.S.T.)
Banking
Algebra
Geometry
Shares and Dividends
Symmetry
Mensuration
Linear Inequations
Quadratic Equations
- Quadratic Equations
- Method of Solving a Quadratic Equation
- Factorisation Method
- Quadratic Formula (Shreedharacharya's Rule)
- Nature of Roots of a Quadratic Equation
- Equations Reducible to Quadratic Equations
Trigonometry
Statistics
Problems on Quadratic Equations
- Method for Solving a Quadratic Word Problem
- Problems Based on Numbers
- Problems on Ages
- Problems Based on Time and Work
- Problems Based on Distance, Speed and Time
- Problems Based on Geometrical Figures
- Problems on Mensuration
- Problems on C.P. and S.P.
- Miscellaneous Problems
Ratio and Proportion
Probability
Remainder Theorem and Factor Theorem
- Function and Polynomial
- Division Algorithm for Polynomials
- Remainder Theorem
- Factor Theorem
- Applications of Factor Theorem
Matrices
Arithmetic Progression
Geometric Progression
Reflection
- Co-ordinate Geometry
- Advanced Concept of Reflection in Mathematics
- Invariant Points
- Combination of Reflections
- Using Graph Paper for Reflection
Section and Mid-Point Formulae
Equation of a Line
Similarity
Loci
- Locus
- Points Equidistant from Two Given Points
- Points Equidistant from Two Intersecting Lines
- Summary of Important Results on Locus
- Important Points on Concurrency in a Triangle
Angle and Cyclic Properties of a Circle
Tangent Properties of Circles
Constructions
Volume and Surface Area of Solids (Cylinder, Cone and Sphere)
- Mensuration of Cylinder
- Hollow Cylinder
- Mensuration of Cones
- Mensuration of a Sphere
- Hemisphere
- Conversion of Solids
- Solid Figures
- Problems on Mensuration
Trigonometrical Identities
Heights and Distances
- Angles of Elevation and Depression
- Problems based on Elevation and Depression
Graphical Representation of Statistical Data
Measures of Central Tendency (Mean, Median, Quartiles and Mode)
Probability
- Theorem
Theorem: Points Equidistant from Two Given Points
Statement:
The locus of a point which is equidistant from two fixed points is the perpendicular bisector of the line segment joining the two fixed points.
Given:
- A and B are two fixed points
- P is a point such that PA = PB
To Prove:
-
P lies on the perpendicular bisector of AB
Construction:
- Join AB
- Let M be the midpoint of AB
- Join PM
Proof:
-
In ΔPMA and ΔPMB,
PA = PB (given), MA = MB (M is the midpoint of AB), PM = PM (common). -
∴ ΔPMA ≅ ΔPMB (SSS).
-
∴ ∠PMA = ∠PMB (c.p.c.t.).
-
∠PMA + ∠PMB = 180° (straight line AB).
-
∴ ∠PMA = ∠PMB = 90°.
-
Hence, PM ⟂ AB passes through midpoint M.
Thus, P lies on the perpendicular bisector of AB.
Converse: Point on Perpendicular Bisector is Equidistant
Statement:
Every point on the perpendicular bisector of a line segment is equidistant from the two fixed points.
Given:
- A and B are two fixed points
- MQ is the perpendicular bisector of AB
- P is any point on MQ
To Prove:
PA = PB
Construction:
-
Join PA and PB
Proof:
-
In ΔPMA and ΔPMB,
MA = MB (M is the midpoint of AB), PM = PM (common),
∠PMA = ∠PMB = 90° (MQ ⟂ AB). -
∴ ΔPMA ≅ ΔPMB (SAS).
-
∴ PA = PB (c.p.c.t.).
Conclusion:
Hence, every point on the perpendicular bisector of AB is equidistant from A and B.
