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We now know how to represent various Geometrical figures in a Cartesian plane. But what forms the most basic of these figures is the line, which forms an important component on any discussion in Co-ordinate geometry. Our attempt to create a general group of lines that satisfy a certain set of properties leads us to the study & formulation of the FAMILY OF LINES.

OBJECTIVE

After studying the chapter on STRAIGHT LINES you will realize that the equations of two or more lines can be expressed together by an equation of degree higher than one. We will learn how and also learn how to find the lines represented by such an equation.

PRE-REQUISITE

The only pre-requisite needed to understand the family of Lines is the knowledge of

Þ         Cartesian Geometry

Þ         Straight lines

CORE CONCEPTS

Family of Straight lines

If L₁ = 0 and L₂ = 0 are two lines then equation of family of lines passing through their intersection is given by

L₁ + λ L₂ = 0                                                          …(A)

where ‘λ’ is any parameter. ( Equation A is satisfied by the point of intersection of L₁ = 0 and L₂ = 0)

Note

Þ      To determine a particular line one more condition is required so as to determine or eliminate λ.

Illustrations

1. If x (2q + p) + y (3q + p) = p + q then what does this equation represent for variables p, q?

Sol. Rearranging the given equation

p (x + y – 1) + q (2x + 3y – 1) = 0

(x + y – 1) +(q/p)(2x + 3y – 1) = 0, p ≠ 0

This equation represents the family of lines passing through the intersection of lines x + y – 1 = 0 and
2x + 3y – 1 = 0 which is a fixed point i.e. (2, – 1).

If p = 0 then equation becomes

q ( 2x + 3y – 1) = 0

this also represents a line which passes through fixed point (2, –1).

Hence the given equation represents family of lines passing through a fixed point (2, –1) for variables p, q.

2. Find the equation of a line, through the intersection of 2x + 3y – 7 = 0 and x + 3y – 5 = 0 and having distance from origin as large as possible.

Sol. Point of intersection of two lines is A(2, 1)

Now, with OA as radius and O itself as centre draw a circle.

There will be infinitely many lines through A and each except one of them produces a chord of circle and hence their distance from origin i.e. centre of circle is less than OA i.e. radius of circle.

But the exceptional one which infact is a tangent to circle at A will be at a distance OA from O.

Thus, tangent to circle at A will be the line through A and is farthest from origin.

Now, OA is perpendicular to tangent at A.

(slope of OA) × (slope of tangent at A)

= – 1

or  {(1-0)/(2-0)} * (slope of tangent at A) = – 2

equation of required line is

(y – 1) = – 2 (x – 2)

or             2x + y – 5 = 0

The equation ax² + 2hxy + by² + 2gx + 2fy + c = 0 represents a second degree equation where a, h, b etc are constants simultaneously.

Let a≠ 0.

Now, the above equation becomes

a² x² + 2ax (hy + g) = – aby² – 2afy – ac

On completing the square on the left side, we get,

a²x² + 2ax (hy + g) + (hy + g)² = y² (h² – ab) + 2y (gh – af) + g² – ac.

i.e.           (ax + hy + g)

We cannot obtain x in terms of y, involving only terms of the first degree, unless the quantity under the radical sign be a perfect square. The condition for this is

(gh – af)² = (h² – ab) (g² – ac)

i.e.     g² h² – 2afgh + a² f² = g² h² – abg² – ach² + a² bc

Canceling and dividing by a, we have the required condition

abc + 2fgh – af² – bg² – ch² = 0

Illustration

1. What is the point of intersection of two straight lines given by general equation ax² + 2hxy + by² + 2gx + 2fy + c = 0?

Sol. The general equation is

ax² + 2hxy + by² + 2gx + 2fy + c = 0          …(1)

Let (a, b) be the point of intersection we consider line paralleled transformation

x = x’ + α,               y = y’ + β

From (1) we have

a(x’ +α)² + 2h (x’ + α) (y’ + β) + b(y’ + β)² +               2g(x’ + α) + 2f(y’ + β) + c = 0

Þ      ax’² + 2hx’y’ + by’² + a a² + 2hab + bb² + 2ga +2 fb + 2 x’ (aα + hβ+ g) +    2y’ + 2y’ (hα + bβ + f) = 0

Þ      ax’²+ 2hx’y’ + by’²+ 2x’ (aα + hβ + g) + 2g’(hα + bβ+ f)=0

which must be in the form

ax’² + 2hx’y’ + by’² = 0

This can not be possible unless

aα + hβ + g = 0

hα + bβ + f = 0

Solving

You must have seen the night-sky composed of stars, planets, moon and you could have also been lucky enough to see a shooting star or meteor as it is termed. If you make a careful study, you could also observe, that the pattern of night-sky keeps on changing with time as well as with seasons in a seemingly complex manner. Our ancient Indian scientists had made careful observation about this changing pattern and had deduced that planets revolve around the sun, a fact which was later rediscovered by Copernicus.

It took the genius of Newton to reduce this complicated motion into a very simple universal law – the law of Gravitation, a law, which not only applies to celestial bodies but also applies to that famous apple which, Newton saw falling.

OBJECTIVE

After learning this chapter we will be able to understand the law of gravitational force, gravitational potential energy. We will also be able to learn the concept of planetary motion of different planets and other heavenly bodies. In this chapter we will revolve around the Universal Law of Gravitation and its applications and try to gain an insight into one of the basic forces in the Universe.

PRE-REQUISITE

Þ            Concept of acceleration

Þ            Earth’s gravity

Þ            Concept of circular motion

Þ      Concept of centripetal and centrifugal forces

Þ      Newton’s third law of motion (Action and Reaction)

CORE CONCEPTS

When you throw a ball up, the ball goes up, its velocity is retarded and it finally comes to rest. Henceforth, it gains acceleration and returns back to the ground. This retardation and subsequent acceleration is a consequence of the universal gravitational force. There is a specific law, which guides the above phenomena, known as Newton’s Law of Gravitation. It states:

Every particle in this Universe attracts every other particle with a force, which is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

Let m1 and m2 be the masses of the two particles and r be the separation between them (see figure given below).

“The space around a body within which its force of gravitational attraction is perceptible (by any other body in this space) is called its gravitational field.”

The intensity E, of the gravitational field of a mass ‘m‘ at a point at distance ‘r‘ from it is the force experienced by a unit mass placed at this point in the field. (Assuming that the presence of unit mass does not affect the gravitational field of the mass m)

The committee will submit its report in January 2010. Some of the points under consideration are — the structure of the present exam and the question of whether weightage could be given to the Class XII results. Most directors were of the view that the proliferation of coaching institutes was skewing the pattern of entrants into the NITs. Another factor to be looked at was the number of vacant seats. The committee would also consider whether the courses were market-friendly or not and could suggest alternate programmes.

A decision to constitute another committee to look into the curriculum was also mooted. This would comprise the directors of the NITs of Trichy, Surathkal and Warangal who would probably assess the present curriculum vis-à-vis the current needs of the industry and redesign it with an eye on the future needs of the country. On the cards is also an interaction with industry organisations to know their requirements so that students have the best possible job prospects.

HRD Minister Kapil Sibal also asked all the 20 NITs to give a vision statement which could also include whether they would like to set up campuses abroad.

In another development, the HRD Ministry is thinking about introducing the Engineers Bill, 2009, that requires obligatory registration /certification of professional engineers to streamline the quality of engineers in the country.

This would then help set some quality standards as “there are many engineering colleges which fail to deliver quality engineers from their institutions,” said an official.

Source: http://www.deccanchronicle.com/tabloids/aieee-get-revamped-019

Out of the many chemical reactions certain reactions are reversible reaction i.e. they proceed in forward as well as backward direction i.e. reactants combine to give products and products gives back the reactants. In such reactions after some time of commencement of reaction system a stage is reached, when the properties of the system remain constant with time, then the reaction system is said to be in equilibrium. There are many kinds of equilibrium but we will focus on Dynamic equilibrium only, then we will study the important laws governing equilibrium, Le Chatelier’s Principle & Law of Mass Actions. Then we will see some important constants associated with the equilibrium & relationship between them. Then we will end with some special features of Equilibrium.

OBJECTIVE

The stage of equilibrium can be reached starting from either reactants or products only or with any amounts of reactants and products taken together. At equilibrium, the processes of conversion of reactant(s) into product(s) and vice versa do not cease but both of them occur simultaneously at equal rates. Thus an equilibrium is actually a dynamic equilibrium. Our aim in this chapter is to learn about Chemical Equilibrium & its properties so that we can manipulate the reactants or product concentrations along with the reaction conditions in order to get a desired effect.

PRE-REQUISITE

The knowledge of the following will help you understand the chapter easily & effectively.

Þ            Gaseous State and Gas laws.

Þ            Stoichiometry

Þ            Mole concept.

Core Concept

Certain reactions are reversible reaction i.e. they proceed in forward as well as backward direction i.e. reactants combine to give products and products give back reactants. In such reactions after some time of commencement of reaction a stage is reached, when the properties of the system remain constant with time, then the reaction system is said to be in equilibrium. The stage of equilibrium can be reached starting from either reactants or products only or with any amounts of reactants and products taken together. At equilibrium, the processes of conversion of reactant(s) into product(s) and vice versa do not cease but both of them occur simultaneously at equal rates. Thus an equilibrium is actually a dynamic equilibrium. Our aim in this chapter is to learn about Chemical Equilibrium.

The concept of Equilibrium:

It is a common observation that clothes dry quicker when there is a breeze or when we keep shaking it. Have you ever thought why does this happen? Well, we can get an answer to this query if we know what an equilibrium is.

In the study of any chemical reaction, two questions are of vital importance viz. How far a reaction would go and how fast will it reach the goal.

Well, equilibrium is that state of a system in which it has minimum energy. If we disturb the equilibrium state (by changing T, p, volume, concentration of reactants, products or adding an inert gas), the system always tries to come back to an equilibrium state. This tendency can be exploited for getting maximum production in industry.

Equilibrium could be of two types

Physical Equilibrium 

It is found between two different states of a substance, e.g., water at its boiling point has got two states, water (liquid) and water vapour (gas). Both of these are in equilibrium with each other.

Chemical Equilibrium

When a reaction starts, reactants are present at some definite concentrations. As reaction proceeds, concentration of reagents decreases and that of products increases. Sooner or later, however, concentration levels off and becomes constant. A state in which concentration no longer changes with time (this state which persists as long as the system is free of external perturbations) is known as the state of chemical equilibrium

We use the symbol to denote equilibrium in contrast to the symbol → where the reaction goes to completion.

The reaction in which products recombine to give back reactants or the reaction which proceeds in forward as well as backward direction is termed as reversible reaction or in other words it is a dynamic state.

ü        Energy is defined to be the capacity to do work. Energy is a scalar quantity and measured in Joules (J).

Mechanical energy is of two types:

Þ            Kinetic Energy

Þ            Potential Energy

Kinetic energy is the energy which is possessed by a body by virtue of its motion. Kinetic energy of a body is the energy due to its motion. If a body of mass ‘m’ moves with a velocity ‘v’, its kinetic energy is given by

For an unaccelerated body, the resultant force acting on the body is zero. But for an accelerated body, the resultant force is not zero and will do some work. We can relate the work done by this force with the consequent increase in kinetic energy of the body. Consider a body being acted on by various forces

ü        The above equation states that the work done by the resultant external force on a body is equal to the change in the kinetic energy of the body. This relation is called Work-Energy Theorem.

Illustrations

1.      A 50 g bullet strikes a wooden plank and comes to rest after penetrating 10 cm into wood. If wood offers a resistance of 10000 N, what is the velocity with which the bullet strikes the plank?

Sol. Work done by the resistant force, acting on the bullet opposite in the direction of displacement of bullet, is given by

W = 10000 N * (10 * 10²) m

= 1000 J

According to Work-Energy Theorem this should be equal to the change in kinetic energy.

W = K₂K₁

i.e.,                          1000 = 0  K1

(K₂ = 0, because the final velocity of bullet is zero)

K1 = 1/2(mv²) = 1000

where v is the initial velocity of bullet

v² = (2 * 1000 J)/ (5 * 10^-3 kg)

= 4000 (m/s)²

v = 200 m/s

ü        We all know that it is hard work lifting a heavy box from one platform to another in a railway station. Similarly we know that children need lots of energy as they grow up.

ü        We feel tired if we run a hundred meters in twenty seconds, whereas we could walk that distance easily in a couple of minutes.

ü        These are some common sense notions of work and energy which can however be precisely defined and measured in physics.

ü        These definitions are measurements can be used consistently to describe and predict the behaviour of bodies and thus can form very powerful tool for analysis of physical systems.

OBJECTIVE

ü        In the previous chapter we studied the Newton’s laws of motion to understand how objects move under the influence of force acting on them or the relation between the force and the acceleration produced by this force.

ü        In this chapter however, we will study the situation where one is not interested in such an exhaustive study of object’s motion rather desires to relate the final velocity of an object to the forces acting on it without going into the deeper details how the object acquired that velocity. The work-energy theorem solves this purpose.

PRE-REQUISITE

Velocity

ü        Rate of change of position of an object is known as its velocity.

V = dx /dt

Acceleration

ü        Rate of change of velocity of an object is known as its acceleration.

a= dv / dt = d²x / dt²

Force

ü        Force is a push or pull which tends to change the position of the object on which it is applied.

Newton’s first law of motion

ü        It states that every object persists in its natural state of motion i.e. continues to be at rest or moves in a straight line with uniform (constant) velocity. (This is what is meant by natural state of motion); In the absence of a net external force acting (impressed) on it.

Newton’s Second law of motion

Newton’s Third Law of Motion

ü        When ever a body exerts a force on a second body, the second body always exerts a force of equal magnitude but opposite in direction on the first one.

Conservation of Energy

ü        Whenever one form of energy is transformed into other forms, the total amount of energy before transformation is always equal to the amount of energy after transformation, i.e., the total amount of energy remains conserved.

Heat energy is involved for changing the physical state of a chemical substance. For example in the conversion of water into steam, heat is absorbed and heat is evolved when steam is condensed.

Example

Consider the following two reactions –

It is observed that there is difference in the value of ΔH if water is obtained in gaseous or liquid state. ΔH value in second case is higher because heat is evolved when steam condenses. Hence, physical state always affects the heat of reaction.

Allotropic forms of the elements

Heat energy is also involved when one allotropic form of an element is converted into another. Thus, the value of ΔH depends on the allotropic form used in the reaction.

Example:

The  value of ΔH is different when carbon in the form of diamond or graphite is used.

The difference between two values is equal to the heat absorbed when 12 g of diamond is converted into 12 g of graphite

Reaction carried out at constant pressure or constant volume

When a chemical reaction  occurs at constant volume, the heat change is called the internal energy. However, most of the reactions are carried out at constant pressure and the enthalpy change is termed as the energy of reaction at constant pressure.

The relation between ΔH (Enthalpy change) and ΔE (Internal energy change) is given as follows:

ΔE + Δn_gRT = ΔH

Dn_g =             (Total number of moles of products) – (total number of moles of  reactants).

R = Gas Constant

T = Temperature (in Kelvin)

The difference between ΔH and ΔE value is negligible when solids and liquids are involved in a chemical change. But, in reactions which involve gases, the difference in two values is considerable.

Centre of mass of system of the discrete particles

The centre of mass of an object is a point that represents the entire body and move in the same way as a point mass having mass equal to that of the object, when subjected to the same external forces that act on the object. That is, if the resultant force acting on an object (or system of objects) of mass m is F, the acceleration of the centre of mass of the object (or system) is given by

Where the sums extend over all masses composing the object. In a uniform gravitational field, the centre of mass and the centre of gravity coincide.

Centre of mass of continuous distribution of particles

Centre of mass of a body having continuous distribution of particles (mass) is given by

Velocity & Acceleration of centre of mass

Position vector of the centre of mass of a system of particle is given by

Conservation of Linear Momentum

System of particles

Conservation of motion of centre of mass

In absence of a net external resultant force along a certain direction on a particulars system will retain its state of motion along that direction. If it had been at rest, it will remain at rest. If it had been moving with uniform velocity then it will continue moving with uniform velocity

Characteristic of linear momentum

If depends on the frame of reference, e.g., the linear momentum of a body at rest in a moving train, is zero relative to a person sitting in the train while it is not zero for a person standing on the ground.

Two bodies of same mass and moving with same speed will have different momenta unless their directions of motion are same.

Relation between kinetic energy and momentum

We are familiar with the representation of real numbers on a line, which we call a real line. In this representation we fix a point O (called origin) and represent a real number by a point A on this line such that its distance OA is equal to the value of real number. In the left side of O we represent negative real numbers and in the right side of O we represent positive real numbers. Thus, not only the magnitude of OA but the direction of the line OA is also considered for representation.

Hence OA′ = – AO′

Similarly ordered pairs are represented in a plane. To represent an ordered pair (a, b) we take two reference lines which are mutually perpendicular. The ordered pair (a, b) represents in such a plane, by a point                P(a, b) such that (see figure given below).

OA = a and OB = b.

This system is called Cartesian co-ordinate system. Since elements of an ordered pair are not inter changeable (i.e. (a, b) ≠ (b, a) unless a = b) so they are represented in particular order, the first element ‘a’ is represented on horizontal line called abscissa and the second element ‘b’ on a vertical line called ordinate. Like the real number notation the positive side of the x–axis is the right side of O and positive side of y–axis is upper side of O.

So, the two lines divide the region in 4 parts (See figure). These are called quadrants.

These quadrants are characterized as

I               quadrant                x > 0, y  > 0

II              quadrant                x < 0, y > 0

III             quadrant                x < 0, y < 0

IV             quadrant                x > 0, y < 0

Here the point ‘O’ represents x = 0 and y = 0, hence ordered pair becomes (0,0).

There is a second type of representation called the polar co-ordinate system. In this system a reference is fixed to a line (Called the initial line), and a point called the origin in the system. Any point P is represented by ordered pair (r, θ).

Such that

OP = r; The distance of point from origin and ∟POX  = 0  The angular displacement of line OP from fixed line i.e. the initial line. (in the anticlockwise direction)

Clearly ‘a’ = r cos θ  and ‘b’ = r sin θ (see figure given below)

We can find the distance between two points, then from given three points we should be able to find three sides of a triangle formed by these points. The area of this triangle.

Consider three points P₁, P₂ and P₃ in a plane. Let their co-ordinates be (x₁, y₁), (x₂, y₂) and (x₃, y₃) respectively (see figure)

Area of  ΔP₁P₂P₃ = Area of trapezium AC P₃P₁ – Area of trapezium AB P₂P₁ – Area of trapezium BC P₃P₂

Thus we observe that the area of a triangle is positive when vertices are taken in the anticlockwise direction and negative when the vertices are taken in the clockwise direction.

Important

This form is important. It can be used to find area of a quadrilateral, pentagon, hexagon and polygons.

Þ      If three points P₁, P₂ and P₃ are collinear then the determinant must vanish i.e. the area of triangle formed must be zero.

Several methods have been developed by mathematicians to uniquely locate the position of a point in space. The easiest and most widely used one is the Cartesian Coordinate System, which is based on mutually perpendicular axes. In this chapter you will learn about this systems: locating a point in these systems and finding the equation of line passing through these points.

Note that when we say a line passes through two given points, we are imposing two conditions on this line. The conditions can be imposed in several other manners also. e.g. we may say that a line passes through one point and is perpendicular to another line. These two conditions are sufficient to uniquely know our line. In this chapter you will also learn various manners of imposing conditions and finding the equation of line under those conditions.

Having known a line uniquely in space we shall try to a lot of timings with it e.g. divide it in a given ratio, find distances of other points or lines from it, find the angle, which this line makes with some straight line. We shall also try to see how to find the equation of line in a new coordinate system if the relationship between the new coordinate system with the old is known.

OBJECTIVE

Having known a line uniquely in space we shall try to a lot of timings with it e.g. divide it in a given ratio, find distances of other points or lines from it, find the angle, which this line makes with some straight line. We shall also try to see how to find the equation of line in a new coordinate system if the relationship between the new coordinate system with the old is known. Having known all this we will then try to see how these concepts can be used to solve some problems of conventional Geometry.

PRE-REQUISITE

This chapter requires the following Pre-requisites

Þ       High school knowledge of Geometrical Axioms & Identities.

Þ       Basic knowledge of Lines, Circles, Planes, Triangles etc.

Þ    Elementary algebra