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

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