1. Details of Module and its structure

Module Detail

Subject Name Mathematics

Course Name Mathematics 02 (Class XI, Semester - 2)

Module Name/Title Straight Lines- Part 3

Module Id kemh_21003

Pre-requisites Basic knowledge of Straight Lines

Objectives After going through this lesson, the learners will be able to

understand the following :

● General Equation of a line

● Intercept form

● Normal form

● Distance of a point from a line

● Distance between two points

Keywords Equation of a line, Slope-intercept form, Intercept form,

Distance between two points

2. Development Team

Role Name Affiliation

National MOOC Coordinator

(NMC)

Prof. Amarendra P.

Behera

CIET, NCERT, New Delhi

Program Coordinator Dr. Mohd. Mamur Ali CIET, NCERT, New Delhi

Course Coordinator (CC) / PI Dr. Til Prasad Sarma DESM, NCERT, New Delhi

Course Co-Coordinator / Co-PI Dr. Mohd. Mamur Ali CIET, NCERT, New Delhi

Subject Matter Expert (SME) Dr. Sadhna Shrivastava KVS, Faridabad, Haryana

Review Team Prof. V.P. Singh (Retd.) DESM, NCERT, New Delhi

Table of contents:

1) Introduction

2) General Equation of a Line

3) Converting general equation of a line into different forms

(a) Slope-intercept form

(b) Intercept form

(c) Normal form

4) Distance of a Point From a Line

5) Distance between two parallel lines

6) Examples based on different forms of equation of a line

7) Summary

1. Introduction:

In our previous Modules on straight lines we have learnt that a straight line in a plane is defined

as set of points whose coordinates satisfy a given linear equation in two variables. We have also

discussed equation of a line under different conditions to get various forms of equation of line. In

all these forms we have two constants; in fact we need two conditions to determine equation of a

line. In this module we will study general equation of straight line and procedures to reduce it into

various forms of the equation of straight line.

2. General Equation of a Line:

We have seen that equation of a straight line is a linear relation between two variables, these

variables are coordinates of the points lying on the line. We know that general equation of first

degree in two variables is Ax + By + C = 0, where A, B and C are real constants such that A and

B both are not zero simultaneously. We also know from earlier classes that Graph of the equation

Ax + By + C = 0 is always a straight line. Therefore, any linear equation in two variables of the

form Ax + By + C = 0, where A and B both are not zero simultaneously, is a general equation of

a straight line.

Different forms of Ax + By + C = 0

We can reduce general equation of a straight line into various forms of the equation of a line, by

the following procedures:

(a) Slope-intercept form:

The slope intercept form is; y = m x + c,

The general equation of a straight line Ax + By + C = 0, can be re-written as,

, if B ≠ 0,

Comparing with slope intercept form, we get,

Thus general equation of straight line Ax + By + C = 0, represents a straight line whose slope is

and y-intercept is

Where B ≠ 0

If B = 0, then general equation reduces to;

Ax + C = 0,

Or,

It represents a line which is a vertical line and its slope is undefined i.e. the line is parallel to y-

axis and it cuts x-axis at,

Example:

Convert 4x − 2y − 5 = 0 to Slope-Intercept Form

Solution:

We know that slope intercept form is,

y = mx + c

given equation is, 4x – 2y – 5 = 0

we write it as, – 2y = – 4x + 5

or y = 2x –

this is the required slope intercept form, comparing with slope intercept form y = m x + c,

slope of the given line; m = 2 and y-intercept; c = –

(b) Intercept form:

If A, B and C all three are not equal to 0, then Ax + By + C = 0 can be written as Ax +By=–C

Further simplifying we get,

comparing it with intercept form,

we get,

Thus general equation Ax + By + C = 0, represents a straight line which cuts x-axis at the point (–

C/A , 0 ) and y-axis at (0, – C/B).

If C = 0, then Ax + By + C = 0 will become, Ax + By = 0, which represents a line passing through

the origin and makes zero intercepts on the axes.

Example:

Write equation of the line which passes through the points (1,2) and (3,1).

(i) What is its slope?

(ii) Write the equation of the line in intercept form.

(iii) What is its x-intercept?

(iv) What is itsy-intercept?

(v) Write the equation of the line in general form.

Solution:

Using two point form equation of the line is

we get,

simplifying and arranging in slope intercept form we get,

Thus, slope of the line m =

Simplifying the equation becomes,

2y =– x + 5

Re-arranging the intercept form is,

Therefore, x-intercept = 5, y-intercept =

further simplifying the equation, the general form of the

equation is, x + 2y – 5 = 0

(c) Normal form:

Let, xcosω + ysin ω = p, be the normal form of the line represented by the general equation Ax

+ By + C = 0

We can write it as, Ax + By=– C

comparing the two equations,we get,

We select signs such that p (the length of perpendicular from origin to the line) is always positive.

Example:

Reduce the line 4x + 3y - 19 = 0 to the normal form and find the perpendicular distance from origin

to the straight line; also find the angle that the perpendicular makes with the positive direction of

the x-axis.

Solution:

We will shift the constant term on the RHS and ensure that it is positive there,the equation

becomes,

4x+ 3y = 19

Comparing it, with standard equation, Ax + By = C,

We get, A (Coefficient of x)= 4, B (Coefficient of y)= 3

Dividing both sides of the equation by 5, we get

This is the required normal form of the given equation 4x + 3y = 19

Comparing it, with the equation, cosω + ysin ω = p

Perpendicular distance of the straight line from origin =

cosω = and sinω =

tanω =

In the above equation, ω is the angle made by the perpendicular(from origin to the line)with the

positive direction of the x-axis:

the angle made by the perpendicular to the line from origin, with the positive direction of x-

axis, is given by;

ω = tan-1

Example:

Find the equation of a line with slope 2 and length of the perpendicular from the origin equal to

.

Solution:

The slope intercept form is; y = mx + c …………….(i)

Slope of required line is, m = 2

equation of required line is, y = 2x + c,

or - 2x + y = c , where ‘c’ is the y-intercept.

to convert the equation in normal form, dividing both the sides by , we get,

x + y =

This is the normal form of the equation of straight line, hence RHS denotes the length of the

perpendicular from origin to the line. But the length of the perpendicular from origin is .

or c = ± 5

Putting value of c in equation (i), the required equation of the line is;

y = 2x ± 5

Distance of a Point from a Line:

The distance of a point from a line is the length of the perpendicular drawn from the point to the

line.

Let L be a line, whose distance from the point P (x1, y1) is d and general equation of the line be,

Ax + By + C = 0.

Let us draw perpendicular PM from the point P to the line L as shown in the figure below, the line

L meets the x-axis and y-axis at the points Q and R respectively.

The coordinates of points Q and R are ( , 0 ) and ( 0, ) respectively.

Area of triangle PQR is given by,

area (ΔPQR) = ( QR x PM ) 2 area(ΔPQR) = ( QR x PM )

….………………… (i)

Using formula,

area (ΔPQR) = [ x1 (0+ ) + ( )( - y1 ) + 0( y1 - 0)]

= [ x1 + + y1 ]

Taking , common in R.H.S. we get,

2 area (ΔPQR) = | |.|Ax1 + By1 + C|

……………………… (ii)

By distance formula we have,

Distance between points (x1, y1) and (x2, y2)

distance between points Q and R,

QR = = | |

......................... (iii)

From (i), (ii) and (iii) substituting values we get,

| |.|Ax1 + By1 + C| = (| | ) PM

Hence, perpendicular distance PM (= d) of the line Ax + By + C = 0 from point (x1, y1) is given

by,

Example:

Find distance of the point (2, 1), from the line, 12x – 5y + 9 = 0.

Solution:

We have just learnt that, length of perpendicular from point (x1 , y1) to the line Ax + By + C = 0,

is given by the formula,

From the given information we have,

x1= 2 y1 = 1, A = 12, B = – 5, C = 9

hence, required distance is,

Or

Or

units

Distance between two parallel lines:

We know that, if two lines are parallel then the distance between them is throughout same and

their slopes are equal,

let the lines be L1 and L2 with slope ‘m’ , then their equations are,

y = mx + c1

and y = mx + c2

these equations can be written as,

L1: – mx + y – c1 = 0 …………….(i)

L2: –mx + y – c2 = 0 ……………(ii)

Line L1 intersects x-axis at the point A as shown in figure.

Distance between lines L1 and L2 is equal to the length of the perpendicular from point A to line

L2. Therefore the distance between the lines L1 and L2 is;

Or

Thus, the distance d between two parallel lines, y = mx + c1 and y = mx + c2 is given by

…………… (iii)

If lines are given in general form, i.e., if equations of the lines are, L1: Ax + By + C1 = 0 and

L2: Ax + By + C2 = 0

then,

since, the lines are parallel, the coefficients of x, y will be same because their slopes are to be

equal.

Substituting the value of m in the equation (iii), the formula for distance between the parallel lines

is given by

Example:

Find the distance between the lines 3x – 4y +9 = 0 and 6x – 8y – 17 = 0

Solution:

Converting both the equations into slope-intercept form, we get

y = x + and y = x -

slopes of both the lines are equal, hence both the lines are parallel, comparing with the equations,

y = mx + c1 and y = mx + c2

we get, m = , c1 = , c2 =

substituting the values in the formula,

the distance between the given lines is,

Or

Thus the distance between the given parallel lines is 7/2 units.

Use of different forms of equation of straight line:

Let us study some examples to understand the use of different forms of equation of straight line.

Example:

Find area of the triangle formed by the lines,

y = m1x + c1, y = m2x + c2 and x = 0

Solution:

Given lines are;

y = m1x + c1 …………………..(i)

y = m2x + c2 ………………… (ii)

x = 0 …………………..(iii)

We know that line y = mx + c meets y-axis ( x = 0 ) at the point (0, c).

Therefore, two vertices of the triangle formed by lines (i),(ii) and (iii) are (0, c1) and (0, c2) shown

as points P and R in the figure. Third vertex is obtained by solving equations (i) and (ii).

Solving equations (i) and (ii), we get,

Therefore coordinates of third vertex Q are,

area of the Δ PQR is given by,

Considering coordinates of points P,Q,R as (x1, y1), (x2, y2) and (x3, y3) and substituting their

values in the formula, the required area of the ΔPQR is,

(area of any triangle is always positive)

Example:

Find the image of the point P (3, 8) with respect to the line

x + 3y = 7, assuming the given to be a plane mirror.

Solution:

Let AB be the given line whose equation is

x + 3y = 7 …………………….(i)

Slope-intercept form of the equation is,

y = - x + , slope of the given line is, m = -

Let P (3, 8) be the given point and P' be its image with respect to the line AB. Join of PQ intersects

AB at N

PN AB and PN = P'N

Let slope of PP' be m,

Then, m x (- ) = - 1, because PP' AB

m = 3

Line PP' passes through the point (3, 8) and its slope is 3, hence using point-slope form, y

– y1= m (x – x1)

Equation of line PP' is, y – 8 = 3 (x – 3)

y = 3x – 1 or 3x – y – 1 = 0 ……………(ii)

Point N is the point of intersection of lines AB and PP', hence solving equations (i) and (ii)

coordinates of point N are x = 1, y = 2

AB and PP' intersect at point N (1, 2), since P' is the image of point P with respect to line AB,

hence N is the mid-point of the line segment PP',

Let coordinates of point P' be (x1,y1), then by mid-point formula,

x1= – 1, y1 = – 4

Hence coordinates of image of point (3, 8), with respect to line

x + 3y = 7, is (-1, - 4).

Example:

Find the equation of median, passing through vertex A, of triangle ABC, whose vertices are, A (2,

5), B (- 4, 9) and C (-2, -1).

Solution:

Median through vertex A will pass through the mid-point of side BC. Let mid-point of side BC be

D, then coordinates of point D are;

x = , y =

Required median AD passes through two points A and D, hence using two point form,

The required equation of median AD is,

y – 5 =

y – 5 =

5y – 25 = x – 2

x – 5y + 23 = 0, is the required equation of the median AD.

Similarly equations of other two medians can also be found.

Summary:

1) Any equation of the form Ax + By + C = 0, where A and B both are not zero simultaneously,

is called the general equation of a line.

2) The general equation of a line can be reduced into various forms of the equation of a line,

like;

(a) Slope-intercept form

(b) Intercept form

(c) Normal form

3) The perpendicular distance (d) of a line Ax + By+ C = 0 from a point (x1, y1) is given by;

4) Distance between the parallel lines Ax + By + C1 = 0 and Ax + By+ C2 = 0, is given by

5) Distance d between two parallel lines,y = mx + c1 and y = mx + c2 , is given by

…………… (iii)