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

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

Self adjusting Properties

Then x = sin θ                                                  …(2)

From (1) putting the value of θ in (2), we get,

Illustrations

Orthocentre is the point of concurrence of the altitudes of a triangle. We denote it by H. The triangle obtained by joining the feet of the altitudes is called pedal triangle. The position vector of the orthocentre is given by

It is possible to find distance of orthocentre from vertices and sides of a triangle.

Let H be the orthocentre

In the figure, we observe that

HK = BK tan ∟HBK = BK tan ∟LBC

HK = BK tan (90°-C) (in ΔBHK)

= AB cos B cot C

=c cos B cos C/ sin C = (c/sin C) cos B cos C

= 2 R cos B cos C

Similarly, we can prove that

HL = 2R cos A cos C

and       HM = 2R cos A cos B

Now, AH = AL sec (90 – C)

= AB cos A cosec C

= (c/sin C) cos A = 2 R cos A

Similarly we get BH = 2R cos B, CH = 2R cos C

Thus, distances of orthocentre from sides a, b, c, are given by
2 R cos B cos C, 2R cos A cos C, 2 R cos A cos B respectively and from vertices A,B,C are given by 2R cos A, 2R cos B and 2R cos C  respectively.

ΔKLM is pedal triangle for ΔABC

In quadrilateral BKHM,

∟BMH +∟ BKH= 90° + 90° = 180°

Þ         BKHM is a cyclic quadrilateral

Þ         ∟BMK = ∟BHK = 90° – ∟HBK

= 90° – ∟LBC

= 90° – (90° – C) = C

=> ∟KMH = 90° – C

Similarly, we can prove that

∟HML = 90° – C

=> ∟KML = 180° – 2 C

Similarly

∟KLM = 180° – 2B and ∟MKL = 180° – 2A

i.e. angles of pedal triangle are supplement of double of opposite angles of the original triangle

Now, in Δ AML,

Similarly, it can be shown that KM = b cos B, KL = c cos C

i.e. sides of pedal triangle are a cos A, b cos B, c cos C.

Illustration

1. MKL is the pedal triangle of ABC; prove that its area is

2 S cos A cos B cos C, where S is area of ΔABC.

Circumcircle:

Circumcircle is a circle circumscribing the triangle i.e. passing through the three vertices of the triangle. Its center is naturally found to be at the point of intersection of perpendicular bisectors of the sides of the triangle. Its centre is usually denoted by O and it’s radius by R and called circumradius. It may happen that the circumcentre is lying out side the triangle as any angle may be out use. If O’ denotes the origin, then the position vector of circumcentre O is

Incircle

A circle touching all the three sides of a triangle is said to be incircle. It’s centre lies at the point of intersection of angle bisectors of the angles of the triangle. It’s centre is denoted by I and radius by r and called inradius. This is a circle inscribed in a triangle. Its centre can never lie outside the triangle. Its position vector is given by

Escribed circle

It is a circle touching on of the three sided externally and the rest two extended sides internally. Obviously, every triangle will lead to three escribed circles. The circle opposite to vertex A gets it’s centre denoted by I₁ and radius by r₁ called exradius. Similarly, circles opposite to vertex B and C get their centers and radii denoted by I₂, r₂ and I₃, r₃ respectively. Obviously its centre will lie on the point of concurrence of two external angle bisectors and one internal angle bisector. The position vector of the escribed centre I₁ is given by

As the name suggests an equation (or) Inequation In valuing one (or) more trigonometric ratios of unknown angles is called a trigonometric equation (or) inequation.

In this lesson we will learn how to affirm the general solution of trigonometric equations and inequations in valuing one (or) more of the six trigonometric ratios. We shall highlight the areas where a student must be careful while solving such equations and inequations we will also discuss the methods to solve equations / Inequations invaluing both trigonometric and non trigonometric functions.

(i) Basic Trigonometric equations:

Consider the following

cos q – sin q = 1                                                                   (1)

sin² q + cos² q = 1                                                                (3)

Equation (1) is satisfied if we put q = 0, 2Π, 4Π  …etc.

In it. Equation (2) is satisfied if we put q = 7Π/12 , 11Π/12, -Π/12

etc in it. But equation (3) is satisfied for any value of q. Equation (1) and (2) are called trigonometric equations while (3) is a trigonometric Identify. A trigonometric equation has three kinds of solutions.

(1)   Principal solution: Numerically smallest value of the equation unknown angle satisfying the given equation

(2)   Particular solution : Any value of angle satisfying the given equation.

(3)           General solution: Collection of all particular solution. For example the equation of (q – Π/6) = √3 is satisfied if q = -Π/6 (or) 7Π/6 (or) -5Π/6, etc

→q = Π/3 , 4Π/3 , -2Π/3 etc. Out of these the numerically smallest is q = Out of these the numerically smallest is q =/3.

This is the principal solution. It is a particular solution also.

Also the above values of q taken individually are particular solutions. For general solution we observe that since cotangent function is periodic with period p and it takes the value  only once in the Interval
[0, Π], (q , -Π/6) can be equal to any particular solution plus a multiple of Π. If we take the particular solution as the principal solution then q – Π/6 = nΠ + Π/6 , n ε l

q = nΠ+Π/3 , n ε l is the general solution. If in any equation we get two values of the unknown angle which are numerically equal but opposite in sign, the principal solution is assumed as the positive angle (e.g.) principal solution of sec q = 2 is q = Π/3, although q = –Π/3 also satisfies it.

Method to find principal value (numerically least angle)

(i)    First draw a trigonometric – circle and mark the quadrant in which the angle may be.

(ii)   Select anticlockwise direction for 1^st and 2^nd quadrants and select clockwise for 3^rd and 4^th quadrants

(iii)  Find the angle in the first rotation.

(iv)   Select the numerically least angle from these two values the angle thus found will be the principal value.

(v)    In case two angles (one with positive sign and the other with negative sign) qualify for the numerically least angle then we select the angle with positive sign as the principal value

Inverse of trigonometric ratios

We know that y = sin x means y is the value of sine of angle x if we consider domain and co-domain both as set R of real numbers. Sine ratio as seen from the fig. is many-one into function.

ü        But it is clear that if we restrict the domain to [-Π/2 , Π/2]  and range to [–1, 1], then. y = sin x is one-one onto and hence it is invertible.

So, y = sin x                             x ε [-Π/2 , Π/2] , y ε [–1, 1]

Þ  x = sin^–1 y                           y ε [–1, 1] , x ε [-Π/2 , Π/2]

ü        This value of x is called the principal value, i.e. belonging to [-Π/2 , Π/2]  and [-Π/2 , Π/2] range and it is called principal value range.

ü        The smallest numerical angle is called principal value.

ü        In general the inverse circular functions with their domain and range can be as given below:

Inverse Circular Function

Þ            sin^-1 x = θ        iff           sin θ = x, -Π/2 ≤ θ ≤ Π/2

Domain [-1,1]

Range [-Π/2 , Π/2]

Graph

Inverse Circular Function

Þ            cos^-1 x = θ   iff          cos θ = x, 0 ≤ θ ≤ Π

Domain [–1, 1]

Range [0, Π]

Graph

Inverse Circular Function

Þ            tan^-1 x = θ               iff   tan θ= x,  -Π/2< θ < Π/2

Domain (–∞, ∞)

Range (-Π/2 , Π/2)

Graph

GENERAL SOLUTION OF SOME SIMPLE EQUATION

sin Θ = 0                                   →  Θ = nΠ

cos Θ = 0                                  → Θ = (2n + 1) Π/2

tan Θ = 0                                  → Θ = nΠ

cot Θ = 0                                 → Θ = (2n + 1)Π/2

sin Θ = 1                                  → Θ = (4n + 1) Π/2

sin Θ = -1                                → Θ = (4n + 3) Π/2

cos Θ = 1                                → Θ = 2n Π

cos Θ = -1                              → Θ = (2n + 1) Π

tan Θ = not defined           → Θ = (2n + 1) Π/2

cot Θ = Not defined          → Θ = nΠ

cosec Θ = Not defined     → Θ = nΠ

sec Θ = not defined            → Θ= (2n + 1) Π/2

ü      There are some cautions to be taken while solving some Trigonometric Equations. They are listed down here.

Þ   Check the validity of the given equation

e.g., 2sin Θ – cos Θ = 4 can never be true for any Θ as the value (2 sin Θ – cos Θ) can never exceed √(2²+(-1)² = √5. So there is no solution to this equation.

Þ   Equation involving sec Θ and / or tan Θ can never have a solution of the form Θ = (2n + 1) Π/2. similarly, equation involving cosec Θ and / or cot Θ can never have solution of the form Θ = nΠ. The corresponding function are undefined at these values of Θ.

Þ   Avoid squaring the equations as far as possible because it leads to extraneous solution. If it has to be squared, check whether the solution (s) arrived at satisfy the original unsquared equation or not. e.g., given that x = 4 → x² = 16 → x = ± 4. But originally x = 4 only.

Þ   Do not cancel common factor involving the unknown angle on L.H.S. and R.H.S because it may delete some solution. e.g. In the equation sin Θ (2cos Θ – 1) = sin Θ cos²Θ if we cancel sin Θ on both sides we get cos² Θ – 2cos Θ + 1 = 0 → (cos Θ – 1)² = 0 → cos Θ = 1 → Θ = 2n Π.

Þ   But Θ = nΠ also satisfies the equation because it makes sin Θ = 0. So, the complete solution is Θ = nΠ, n Î Z.

Þ   Denominator terms of the equation if present should never become zero at any stage while solving for any value of Θ contained in the answer.

Þ        Something the equation has some limitation also. e.g., cot² Θ cosec² Θ = 1 can be true only if cot² Θ = 0 and cosec² Θ = 1 simultaneously as                 cosec² Θ ≥ 1. Hence the solution is Θ= (2n + 1) Π/2.

Illustrations

1. Obtain the general solution of secx + tanx =√3

Sol. Since secx and tanx. Both are undefined for x = (2n + 1)Π/2, the final solution should not include any odd multiple of Π/2.

Now,√3 cosx – sin x = 1 => cos(Θ + Π/6) = ½ = cosΠ/3

Þ Θ-+ Π/6 = 2nΠ + Π/3.

With ‘+’ sign, it given Θ = 2nΠ + Π/6

With ‘-‘ sign, it given Θ = 2nΠ – Π/2 = (4n – 1)Π/2

Since (4n -1) is always odd therefore we should delete this solution.

Thus, Θ= (12n + 1)Π/6, n Î Z.