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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

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

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