27

Straight Line 27

Chapter

CONTENTS
2.1 Definition
2.2 Slope (gradient) of a line
2.3 Equations of straight line in different forms
2.4 Equation of parallel and perpendicular lines to
a given line
2.5 General equation of a straight line and its
transformation in standard forms
2.6 Selection of co-ordinates of a point on a
straight line
2.7 Point of intersection of two lines
2.8 General equation of lines through the
intersection of two given lines Monge
2.9 Angle between two non-parallel lines
2.10 Equation of straight line through a given point
making a given angle with a given line A straight line is the simplest geometric curve.
2.11 A line equally inclined with two lines Monge (1781 A.D.) gave the modern 'Point Slope'
2.12 Equations of the bisectors of the angles form of equation of a line as y - y' = m (x - x') and
between two straight lines condition of perpendicularity of two lines as mm'
+1=0
2.13 Length of perpendicular
S.f. Lacroix (1765-1843 A.D.) was a prolific text
2.14 Position of a point with respect to a line book writer; but his contributions to analytic
2.15 Position of two points with respect to a line geometry are found scattered. He gave the 'two
2.16 Concurrent lines point' form of equation of a line as
β  β
2.17 Reflection on the surface y β  (x – ). He also gave the formula
α'α
2.18 Image of point in different cases for finding angles between two lines.
2.19 Some important results
Assignment (Basic and Advance Level)
Answer Sheet of Assignment
 28 Straight Line

2.1 Definition.
The straight line is a curve such that every point on the line segment joining any two points on it lies on it. The
simplest locus of a point in a plane is a straight line. A line is determined uniquely by any one of the following:
(1) Two different points (because we know the axiom that one and only one straight line passes through two given
points)
(2) A point and a given direction.

Y
Y
Y

45o
O X 45º 45º 45º 45º
X O X
Unique line through two O
Unique line through a given
given points point with a given direction Infinite number of lines
Infinite number of lines
with a given direction
through a given point

Thus, to determine a line uniquely, two geometrical conditions are required.

2.2 Slope (Gradient) of a Line.
The trigonometrical tangent of the angle that a line makes with the positive direction of the x-axis in anticlockwise
sense is called the slope or gradient of the line.
Y Y
The slope of a line is generally denoted by m. Thus, m = tan 
B

(1) Slope of line parallel to x – axis is m  tan 0  0 . o B

 
(2) Slope of line parallel to y – axis is m  tan 90 o   . X X X O A X
A O

(3) Slope of the line equally inclined with the axes is 1 or –1. Y Y

y 2  y1
(4) Slope of the line through the points A( x 1 , y 1 ) and B(x 2 , y 2 ) is taken in the same order.
x 2  x1
a
(5) Slope of the line ax  by  c  0, b  0 is  .
b
(6) Slope of two parallel lines are equal.
(7) If m 1 and m 2 be the slopes of two perpendicular lines, then m1 .m 2  1 .

Note :  m can be defined as tan  for 0     and  
2
 If three points A, B, C are collinear, then
Slope of AB = Slope of BC = Slope of AC
Example: 1 The gradient of the line joining the points on the curve y  x 2  2 x , whose abscissae are 1 and 3, is [MP PET 1997]
 Straight Line 29

(a) 6 (b) 5 (c) 4 (d) 3

Solution: (a) The points are (1, 3) and (3, 15)
y 2  y1 12
Hence gradient is =  6
x 2  x1 2

Example: 2 Slope of a line which cuts intercepts of equal lengths on the axes is [MP PET 1986]

(a) – 1 (b) 0 (c) 2 (d) 3

x y
Solution: (a) Equation of line is  1
a a

 x y a  y  x  a . Hence slope of the line is – 1.

2.3 Equations of Straight line in Different forms .

(1) Slope form : Equation of a line through the origin and having slope m is y = mx.
Y
(2) One point form or Point slope form : Equation of a line through the B
point (x 1 , y1 ) and having slope m is y  y1  m (x  x 1 ) .
(3) Slope intercept form : Equation of a line (non-vertical) with slope m and  c
X' X
cutting off an intercept c on the y-axis is y = mx + c. A O

The equation of a line with slope m and the x-intercept d is y  m (x  d ) Y'

(4) Intercept form : If a straight line cuts x-axis at A and the y-axis at B then OA and OB are known as the
intercepts of the line on x-axis and y-axis respectively.
Y
The intercepts are positive or negative according as the line meets with positive
or negative directions of the coordinate axes. B

In the figure, OA = x-intercept, OB = y-intercept. b

A
Equation of a straight line cutting off intercepts a and b on x–axis and y–axis X' X
O a
x y
respectively is   1 . Y'
a b
Note :  If given line is parallel to X axis, then X-intercept is undefined.
 If given line is parallel to Y axis, then Y-intercept is undefined.
y 2  y1
(5)Two point form: Equation of the line through the points A (x 1 , y1 ) and B(x 2 , y 2 ) is (y  y1 )  (x  x 1 ) . In
x 2  x1
the determinant form it is gives as:
Y
x y 1 L
B
x1 y1 1 = 0 is the equation of line. (x2, y2)

x2 y2 1
A O X
(x1,y1)
 30 Straight Line
(6) Normal or perpendicular form : The equation of the straight line upon which the length of the
perpendicular from the origin is p and this perpendicular makes an angle  with x- Y
axis is x cos   y sin   p .

B
P
p
Y'  E X
O A

Y'

(7) Symmetrical or parametric or distance form of the line : Equation of

a line passing through ( x 1 , y1 ) and making an angle  with the positive direction of x-
x  x1 y  y1 Y
axis is  r, (x1,y1) r
cos  sin  A  P(x, y)
where r is the distance between the point P (x, y) and A(x 1 , y1 ) .
X' 
The coordinates of any point on this line may be taken as O
X

(x 1  r cos  , y1  r sin  ) , known as parametric co-ordinates, ‘r’ is called the parameter.

Y'
Note :  Equation of x-axis  y = 0
Equation a line parallel to x-axis (or perpendicular to y-axis) at a distance ‘b’ from it  y  b
Y

b
X' X
O

Y

 Equation of y-axis  x = 0
Equation of a line parallel to y-axis (or perpendicular to x-axis) at a distance ‘a’ from it  x  a
Y

a
X' X
O

Y’

Example: 3 Equation to the straight line cutting off an intercept 2 from the negative direction of the axis of y and inclined at 30° to the
positive direction of x, is [MP PET 2003]

(a) y  x  3  0 (b) yx20 (c) y  3x  2  0 (d) 3y  x  2 3  0

Solution: (d) Let the equation of the straight line is y  mx  c .
1
Here m  tan 30   and c = – 2
3
1
Hence, the required equation is y  x 2  3y  x  2 3  0 .
3
Example: 4 The equation of a straight line passing through (– 3, 2) and cutting an intercept equal in magnitude but opposite in sign from the
axes is given by [Rajasthan PET 1984; MP PET 1993]
 Straight Line 31

(a) x  y  5  0 (b) x y 5  0 (c) x y 5  0 (d) x y 5 0

x y
Solution: (a) Let the equation be  1  x y a
a a
But it passes through (–3, 2), hence a  3  2  5 . Hence the equation of straight line is x  y  5  0 .
Example: 5 The equation of the straight line passing through the point (4, 3) and making intercept on the co-ordinates axes whose sum is – 1,
is [AIEEE 2004]
x y x y x y x y
(a)   1 and   1 (b)   1 and   1
2 3 2 1 2 3 2 1
x y x y x y x y
(c)   1 and   1 (d)   1 and   1
2 3 2 1 2 3 2 1
x y 4 3
Solution: (d) Let the equation of line is   1 , which passes through (4, 3). Then   1  a  2
a 1  a a 1  a
x y x y
Hence equation is   1 and  1 .
2 3 2 1
Example: 6 Let PS be the median of the triangle with vertices P(2, 2), Q(6,  1) and R(7, 3). The equation of the line passing through (1, – 1)
and parallel to PS is [IIT Screening 2000]
(a) 2 x  9 y  7  0 (b) 2 x  9 y  11  0 (c) 2 x  9 y  11  0 (d) 2x  9y  7  0
 6  7 1  3   13 
Solution: (d) S = mid point of QR   ,    ,1 
 2 2   2 
2 1 2 2
 Slope (m) of PS =  ;  The required equation is y  1  (x  1)  2 x  9 y  7  0
13 9 9
2
2
2.4 Equation of Parallel and Perpendicular lines to a given Line .
(1) Equation of a line which is parallel to ax  by  c  0 is ax  by    0
(2) Equation of a line which is perpendicular to ax  by  c  0 is bx  ay    0
The value of  in both cases is obtained with the help of additional information given in the problem.
x y
Example: 7 The equation of the line passes through (a, b) and parallel to the line   1 , is [Rajasthan PET 1986, 1995]
a b
x y x y x y x y
(a)  3 (b)  2 (c)  0 (d)  20
a b a b a b a b
x y
Solution: (b) The equation of parallel line to given line is   .
a b
This line passes through point (a, b).
a b
    2
a b
x y
Hence, required line is  2.
a b
Example: 8 A line passes through (2, 2) and is perpendicular to the line 3 x  y  3 . Its y-intercept is [IIT 1992]

1 2 4
(a) (b) (c) 1 (d)
3 3 3
Solution: (d) The equation of a line passing through (2, 2) and perpendicular to 3 x  y  3 is
1 4
y2 (x  2) or x  3 y  4  0 . Putting x  0 in this equation, we obtain y  .
3 3
4
So y-intercept = .
3
  4
Example: 9 The equation of line passing through   1,  and perpendicular to 3 sin   2 cos   is [EAMCET 2003]
 2 r
 32 Straight Line

(a) 2  3 r cos   2r sin  (b) 5  2 3 r sin  4 r cos 

(c) 2  3 r cos   2r sin  (d) 5  2 3 r sin   4 r cos 

4     k
Solution: (a) Equation of a line, perpendicular to 3 sin   2 cos   is 3 sin     2 cos     
r  2   2  r
 
It is passing through   1,  . Hence, 3 sin   2 cos   k /  1  k  2
 2
2
 3 cos   2 sin    2  3 r cos   2r sin  .
r
Example: 10 The equation of the line bisecting perpendicularly the segment joining the points (– 4, 6) and (8, 8) is [Karnataka CET 2003]
(a) 6 x  y  19  0 (b) y 7 (c) 6 x  2 y  19  0 (d) x  2y  7  0
Solution: (a) Equation of the line passing through (–4, 6) and (8, 8) is
8 6 2
y 6  (x  4 )  y  6  (x  4 )  6 y  x  40 ......(i)
84 12
Now equation of any line  to it is 6 x  y   = 0 .....(ii)
This line passes through the midpoint of (–4, 6) and (8, 8) i.e., (2, 7)
 From (ii) 12 + 7 +   0    19 ,  Equation of line is 6 x  y  19  0

2.5 General equation of a Straight line and its Transformation in Standard forms .
General form of equation of a line is ax  by  c  0 , its
a c a c
(1) Slope intercept form: y   x  , slope m   and intercept on y-axis is, C  
b b b b
x y  c  c
(2) Intercept form :   1 , x intercept is =    and y intercept is =   
c/a c/b  a  b
(3) Normal form : To change the general form of a line into normal form, first take c to right hand side and make it
positive, then divide the whole equation by a 2  b 2 like
ax by c a b c
   , where cos    , sin    and p 
a b 2 2
a b
2 2
a b2 2
a b
2 2
a b 2 2
a  b2
2

2.6 Selection of Co-ordinate of a Point on a Straight line.

(1) If the equation of the straight line be ax  by  c  0 , in order to select a point on it, take the x co-ordinate
a  c
according to your sweet will. Let x   ; then a  by  c  0 or y   ;
b
 a  c   c
   ,  is a point on the line for any real value of  . If   0 is taken then the point will be  0,  .
 b   b 
 c 
Similarly a suitable point can be taken as   ,0  .
 a 
(2) If the equation of the line be x  c then a point on it can be taken as (c,  ) where  has any real value.
In particular (c, 0) is a convenient point on it when   0 .
(3) If the equation of the line be y  c then a point on it can be taken as (, c) where  has any real value.
In particular (0, c) is a convenient point on it when   0 .
Example: 11 If we reduce 3 x  3 y  7  0 to the form x cos   y sin   p , then the value of p is [MP PET 2001]
 Straight Line 33

7 7 3 7 7
(a) (b) (c) (d)
2 3 3 2 3 2

Solution: (d) Given equation is 3 x  3 y  7  0 , Dividing both sides by 32  32

3x 3y 7 3 3 7 7 7
   0  x y ,  p  .
3 3
2 2
3 3
2 2
3 3
2 2 3 2 3 2 3 2 3 2 3 2

2.7 Point of Intersection of Two lines.

Let a1 x  b1 y  c1 =0 and a2 x  b 2 y  c 2  0 be two non-parallel lines. If ( x , y ) be the co-ordinates of their point of
intersection, then a1 x   b1 y   c1  0 and a 2 x   b 2 y   c 2  0

 b1 b2 c1 c2 
 
 b c  b 2 c1 c1 a 2  c 2 a1   c1 c2 a1 a2 
Solving these equation, we get (x , y )   1 2  
a 2 
, ,
 a1 b 2  a 2 b1 a1 b 2  a 2 b1   a1 a2 a1
 b1 b2 b1 b 2 


Note :  Here lines are not parallel, they have unequal slopes, then a1b 2  a2 b1  0 .

 In solving numerical questions, we should not remember the co-ordinates (x , y ) given above, but we
solve the equations directly.
2.8 General equation of Lines through the Intersection of Two given Lines .
If equation of two lines P  a1 x  b1 y  c1  0 and Q  a2 x  b 2 y  c 2  0 , then the equation of the lines passing
through the point of intersection of these lines is P   Q  0 or a1 x  b1 y  c  (a 2 x  b 2 y  c 2 )  0 ; Value of  is
obtained with the help of the additional information given in the problem.
Example: 12 Equation of a line passing through the point of intersection of lines 2 x  3 y  4  0 , 3 x  4 y  5  0 and perpendicular to
6 x  7 y  3  0 , then its equation is [Rajasthan PET 2000]
(a) 119 x  102 y  125  0 (b) 119 x  102 y  125 (c) 119 x  102 y  125 (d) None of these
 1 22 
Solution: (b) The point of intersection of the lines 2 x  3 y  4  0 and 3 x  4 y  5  0 is  , 
 17 17 
7
The slope of required line = .
6
22 7  2 
Hence, Equation of required line is, y   x    119 x  102 y  125 .
7 6  34 
Example: 13 The equation of straight line passing through point of intersection of the straight lines 3 x  y  2  0 and 5 x  2 y  7  0 and
having infinite slope is [UPSEAT 2001]
(a) x  2 (b) x  y  3 (c) x  3 (d) x  4
Solution: (c) Required line should be, (3 x  y  2)  (5 x  2 y  7)  0 .....(i)
 3  5  2  7
 (3  5  )x  (2  1)y  (2  7  )  0  y   x  ......(ii)
 2  1  2  1
1
As the equation (ii) has infinite slope, 2  1  0   
2
1  1 
Putting   in equation (i), We have (3 x  y  2)    (5 x  2 y  7 )  0  x  3
2  2 

2.9 Angle between Two non-parallel Lines.

 34 Straight Line
Let  be the angle between the lines y  m 1 x  c1 and y  m 2 x  c 2 .
and intersecting at A.
Y
where, m 1 = tan  and m 2 = tan  y=m1x + c1

y=m2x + c2
          A

tan   tan    
tan  
X
 O C B
1  tan  tan 

m1  m 2
   tan 1 .
1  m1m 2

(1) Angle between two straight lines when their equations are given : The angle  between the lines
a2 b1  a1b 2
a1 x  b1 y  c1  0 and a2 x  b 2 y  c 2  0 is given by, tan   .
a1a2  b1b 2

(i) Condition for the lines to be parallel : If the lines a1 x  b1 y  c1  0 and a2 x  b 2 y  c 2  0 are parallel then,
a1 a2 a b
m1  m 2    1  1 .
b1 b2 a2 b2

(ii) Condition for the lines to be perpendicular : If the lines a1 x  b1 y  c1  0 and a2 x  b 2 y  c 2  0 are
a1 a2
perpendicular then, m1m 2  1    1  a1a2  b1b2  0 .
b1 b2
(iii) Conditions for two lines to be coincident, parallel, perpendicular and intersecting : Two lines
a1 x  b1 y  c1  0 and a2 x  b 2 y  c 2  0 are,

a1 b1 c a1 b1 c a1 b
(a) Coincident, if   1 (b) Parallel, if   1 (c) Intersecting, if  1
a2 b 2 c 2 a2 b 2 c 2 a2 b 2

(d) Perpendicular, if a1a2  b1b 2  0

Example: 14 Angle between the lines 2 x  y  15  0 and 3 x  y  4  0 is [Rajasthan PET 2003]

(a) 90  (b) 45  (c) 180  (d) 60 

a2b1  a1b2 (3) (1)  (2) (1) 3  2 5

Solution: (b) tan     tan    |  1 |
a1a2  b1b2 (3) (2)  (1) (1) 6 1 5

  tan 1 | 1 |  tan 1 1  45 o .

Example: 15 To which of the following types the straight lines represented by 2 x  3 y  7  0 and 2 x  3 y  5  0 belongs [MP PET 1982]

(a) Parallel to each other (b) Perpendicular to each other

(c) Inclined at 45° to each other (d) Coincident pair of straight lines
a1 b c 2 2 7
Solution: (a) Here,  1  1 ;   . Hence, lines are parallel to each other.
a2 b 2 c 2 3 3 5

2.10 Equation of Straight line through a given point making a given Angle with a given Line.
 Straight Line 35

Since straight line L makes an angle (   ) with x-axis, then equation of line L is y  y1  tan(    )(x  x 1 ) and
straight line L makes an angle (   ) with x-axis, then equation of line L is
L
 y  y1  tan(    )(x  x 1 )
where m  tan  
y = mx+c

Hence, the equation of the straight lines which pass through a given point ( x 1 , y1 ) L

 ( – )
and make a given angle  with given straight line y  mx  c are O X

m  tan 
y  y1  (x  x 1 )
1  m tan 

Example: 16 The equation of the lines which passes through the point (3, – 2) and are inclined at 60 o to the line 3x  y  1
[IIT 1974; MP PET 1996]

(a) y  2  0, 3x  y  2  3 3  0 (b) x  2  0, 3x  y  2  3 3  0

(c) 3x  y  2  3 3  0 (d) None of these

Solution: (a) The equation of lines passing through (3, –2) is (y  2)  m (x  3) .....(i)

The slope of the given line is  3 .

m  ( 3 )
So, tan 60 o   . On solving, we get m  0 or 3
1  m( 3 )

Putting the values of m in (i), the required equation is y  2  0 and 3x  y  2  3 3  0 .

Example: 17 In an isosceles triangle ABC, the coordinates of the point B and C on the base BC are respectively (1, 2) and (2, 1). If the equation
of the line AB is y  2 x , then the equation of the line AC is [Roorkee 2000]

1 x
(a) y  (x  1) (b) y (c) y  x  1 (d) 2y  x  3
2 2
12
Solution: (b) Slope of BC =  1 A
2 1
 AB  AC ,  ABC  ACB
y = 2x
2 1 m 1 m 1 m 1 1
   | 3 |    3  m  2, .
1  2 (1) 1  m (1) 1m 1m 2

1
But slope of AB is 2;  m  (Here m is the gradient of the line AC) B C
2 (1, 2) (2, 1)
1 x
Equation of the line AC is y  1  ( x  2)  x  2 y  0 or y  .
2 2

2.11 A Line equally inclined with Two lines.

Let the two lines with slopes m 1 and m 2 be equally inclined to a line with slope m
 m m   m m 
then ,  1    2  m2
m

 1  m1m   1  m 2m 

Note :  Sign of m in both brackets is same. m1


Example: 18 If the lines y  3 x  1 and 2 y  x  3 are equally inclined to the line y  mx  4 , then m = [ISM Dhanbad 1976]
 36 Straight Line

13 2 13 2 13 2 15 2

(a) (b) (c) (d)
7 7 7 7
 1 
 3 
Solution: (d)
1  3 m 
If line y  mx  4 are equally inclined to lines with slope m1  3 and m 2  ,then   2   m  15 2
2  1  3 m   1  7
1  m 
 2 
2.12 Equations of the bisectors of the Angles between two Straight lines.
The equation of the bisectors of the angles between the lines a1 x  b1 y  c1  0 and a2 x  b 2 y  c 2  0 are given
a1 x  b1 y  c1 a2 x  b2 y  c2
by,  .....(i)
a12  b12 a22  b22
Algorithm to find the bisector of the angle containing the origin :
Let the equations of the two lines a1 x  b1 y  c1  0 and a 2 x  b 2 y  c 2  0 . To find the bisector of the angle
containing the origin, we proceed as follows:
Step I : See whether the constant terms c 1 and c 2 in the equations of two lines positive or not. If not, then multiply
both the sides of the equation by –1 to make the constant term positive.
a x  b1 y  c1 a 2 x  b 2 y  c 2
Step II : Now obtain the bisector corresponding to the positive sign i.e., 1  .
a12  b12 a 22  b 22
This is the required bisector of the angle containing the origin.
Note :  The bisector of the angle containing the origin means the bisector of the angle between the lines which
contains the origin within it.
(1) To find the acute and obtuse angle bisectors
Let  be the angle between one of the lines and one of the bisectors given by (i). Find tan  . If | tan  |  1 , then
this bisector is the bisector of acute angle and the other one is the bisector of the obtuse angle.
If | tan  | > 1, then this bisector is the bisector of obtuse angle and other one is the bisector of the acute angle.
(2) Method to find acute angle bisector and obtuse angle bisector
(i) Make the constant term positive, if not. (ii) Now determine the sign of the expression a1 a 2  b1 b 2 .
(iii) If a1 a 2  b1 b 2  0 , then the bisector corresponding to “+” sign gives the obtuse angle bisector and the bisector
corresponding to “–” sign is the bisector of acute angle between the lines. L1
(iv) If a1 a 2  b1 b 2  0 , then the bisector corresponding to “+” and “–” sign given
the acute and obtuse angle bisectors respectively. Acute bisector

Note :  Bisectors are perpendicular to each other. P(x, y)

 If a1 a 2  b1 b 2  0 , then the origin lies in obtuse angle and if

L2
a1 a 2  b1 b 2  0 , then the origin lies in acute angle. Obtuse bisector

Example: 19 The equation of the bisectors of the angles between the lines | x | | y | are [Orissa JEE 2002]
1 1
(a) y   x and x  0 (b) x  and y  (c) y  0 and x  0 (d) None of these
2 2
Solution: (c) The equation of lines are x  y  0 and x  y  0 .
x y x y
 The equation of bisectors of the angles between these lines are   x  y  (x  y )
1 1 1 1
Taking +ve sign, we get y  0 ; Taking –ve sign, we get x  0 . Hence, the equation of bisectors are x  0 , y  0 .
Example: 20 The equation of the bisector of the acute angle between the lines 3 x  4 y  7  0 and 12 x  5 y  2  0 is
[IIT 1975, 1983; Rajasthan PET 2003]
 Straight Line 37

(a) 21 x  77 y  101  0 (b) 11 x  3 y  9  0 (c) 31 x  77 y  101  0 (d) 11 x  3 y  9  0

3x  4y  7 12 x  5 y  2
Solution: (b) Bisector of the angles is given by 
5 13
 11 x  3 y  9  0 .....(i) and 21 x  77 y  101  0 ......(ii)
3 11

m1  m 2 4 3 35
Let the angle between the line 3 x  4 y  7  0 and (i) is  , then tan      1    45 o
1  m1m 2 3 11 45
1 
4 3
Hence 11 x  3 y  9  0 is the bisector of the acute angle between the given lines.

2.13 Length of Perpendicular.

(1) Distance of a point from a line : The length p of the perpendicular from the point (x 1 , y1 ) to the line
| ax1  by1  c |
ax  by  c  0 is given by p  .
a2  b 2
c
Note :  Length of perpendicular from origin to the line ax  by  c  0 is .
a2  b 2
 Length of perpendicular from the point (x 1 , y1 ) to the line x cos   y sin   p is
x 1 cos   y 1 sin   p .
(2) Distance between two parallel lines : Let the two parallel lines be ax  by  c1  0 and ax  by  c 2  0 .
| c1  c 2 |
First Method : The distance between the lines is d  . ax + by + c1 = 0

(a  b )
2 2

ax + by + c2 = 0


Second Method : The distance between the lines is d  , where ax + by + c1 = 0
(a  b 2 )
2

(i)  | c1  c 2 | if they be on the same side of origin.

(ii)  | c1 |  | c 2 | if the origin O lies between them. ax + by + c2 = 0

.O (0, 0)

Third method : Find the coordinates of any point on one of the given line, preferably putting x  0 or y  0 . Then
the perpendicular distance of this point from the other line is the required distance ax + by + c1 = 0
between the lines.
.O (0, 0)
Note::  Distance between two parallel lines ax  by  c1  0 and
c2
c1  ax + by + c2 = 0
k
kax  kby  c 2  0 is
a2  b 2
 Distance between two non parallel lines is always zero.
2.14 Position of a Point with respect to a Line.
Let the given line be ax  by  c  0 and observing point is (x 1 , y1 ) , then
 38 Straight Line
(i) If the same sign is found by putting in equation of line x  x 1 , y  y1 and x  0 , y  0 then the point (x 1 , y1 ) is
situated on the side of origin.
(ii) If the opposite sign is found by putting in equation of line x  x 1 , y  y 1 and x  0 , y  0 then the point
(x 1 , y1 ) is situated opposite side to origin.
2.15 Position of Two points with respect to a Line.
Two points ( x 1 , y 1 ) and (x 2 , y 2 ) are on the same side or on the opposite side of the straight line ax  by  c  0
according as the values of ax 1  by 1  c and ax 2  by 2  c are of the same sign or opposite sign.

Example: 21 The distance of the point (–2, 3) from the line x  y  5 is [MP PET 2001]

(a) 5 2 (b) 2 5 (c) 3 5 (d) 5 3

x1  y1  5 235  10
Solution: (a) p   5 2
1 1
2 2
1 1
2 2 2

Example: 22 The distance between the lines 4 x  3 y  11 and 8 x  6 y  15 is [AMU 1979; MNR 1987; UPSEAT 2000; DCE 1999]

7 7
(a) (b) 4 (c) (d) None of these
2 10

 11  15 11 15
Solution: (c) Given lines 4 x  3 y  11 and 8 x  6 y  15 , distance from the origin to both the lines are and  ,
25 100 5 10

Clearly both lines are on the same side of the origin.

11 15 7
Hence, distance between both the lines are,   .
5 10 10
Example: 23 If the length of the perpendicular drawn from origin to the line whose intercepts on the axes are a and b be p, then
[Karnataka CET 2003]
1 1 1 2 1 1 1
(a) a 2  b 2  p 2 (b) a 2  b 2  (c)   2 (d)   2
p2 a2 b 2 p a2 b 2 p

x y
Solution: (d) Equation of line is   1  bx  ay  ab  0
a b

 ab a2  b 2 1 a2  b 2 1 1 1 1
Perpendicular distance from origin to given line is p     2 2
 2  2  2  2
a2  b 2 ab p ab p a b p

x y
Example: 24 The point on the x-axis whose perpendicular distance from the line   1 is a, is[Rajasthan PET 2001; MP PET 2003]
a b

a  b  a 
(a)  (b  a 2  b 2 ), 0  (b)  a (b  a  b ), 0 
2 2
(c)  b (a  a  b ), 0 
2 2
(d) None of these
b     
bh  0  ab a
Solution: (a) Let the point be (h,0 ) then a    bh  a a2  b 2  ab  h  (b  a 2  b 2 )
a2  b 2 b

a 
Hence the point is  (b  a 2  b 2 ), 0 
 b 

Example: 25 The vertex of an equilateral triangle is (2, –1) and the equation of its base is x  2 y  1 . The length of its sides is [UPSEAT 2003]

4 2
(a) (b) A (2, –1)
15 15

60o
 Straight Line 39

4
(c) (d) None of these
3 3

2  2 1 1
Solution: (b) | AD |  
1 22 2 5

AD 1/ 5 1 2
 tan 60 o   3   BD   BC  2 BD 
BD BD 15 15

2.16 Concurrent Lines.

Three or more lines are said to be concurrent lines if they meet at a point.
First method : Find the point of intersection of any two lines by solving them simultaneously. If the point satisfies
the third equation also, then the given lines are concurrent.
Second method : The three lines a1 x  b1 y  c1  0 , a2 x  b 2 y  c 2  0 and a3 x  b3 y  c3  0 are concurrent if,
a1 b1 c1
a2 b 2 c 2  0
a3 b 3 c 3
Third method : The condition for the lines P  0 , Q  0 and R  0 to be concurrent is that three constants a, b, c
(not all zero at the same time) can be obtained such that aP  bQ  cR  0 .
Example: 26 If the lines ax  by  c  0 , bx  cy  a  0 and cx  ay  b  0 be concurrent, then [IIT 1985; DCE 2002]
(a) a  b  c  3 abc  0 (b) a  b  c  abc  0 (c) a  b  c  3 abc  0 (d) None of these
3 3 3 3 3 3 3 3 3

Solution: (c) Here the given lines are, ax  by  c  0 , bx  cy  a  0 , cx  ay  b  0

a b c
The lines will be concurrent, iff b c a  0  a 3  b 3  c 3  3 abc  0
c a b
Example: 27 If the lines 4 x  3 y  1, y  x  5 and 5 y  bx  3 are concurrent, then b equals
[Rajasthan PET 1996; MP PET 1997; EAMCET 2003]
(a) 1 (b) 3 (c) 6 (d) 0
Solution: (c) If these lines are concurrent then the intersection point of the lines 4 x  3 y  1 and y  x  5 , is (–2, 3), which lies on the third line.
Hence,  5  3  2b  3  15  2b  3  2b  12  b  6
Example: 28 The straight lines 4 ax  3 by  c  0 where a  b  c  0 , will be concurrent, if point is [Rajasthan PET 2002]
1 1 1 1
(a) (4, 3) (b)  ,  (c)  ,  (d) None of these
4 3 2 3
Solution: (b) The set of lines is 4 ax  3 by  c  0 , where a  b  c  0
Eliminating c, we get 4 ax  3by  (a  b)  0  a(4 x  1)  b(3 y  1)  0
1 1 1 1
They pass through the intersection of the lines 4 x  1  0 and 3 y  1  0 i.e., x  ,y  i.e.,  , 
4 3 4 3
2.17 Reflection on the Surface.
Here IP = Incident Ray
N R
PN = Normal to the surface I
PR = Reflected Ray
IPN  NPR
Normal

Then,
Angle of incidence = Angle of reflection Incident ray
Reflected ray

 
 
Tangent
P
2.18 Image of a Point in Different cases. Surface

(1) The image of a point with respect to the line mirror : The image of A (x1, y1)
A( x 1 , y1 ) with respect to the line mirror ax  by  c  0 be B (h, k) is given by, ax+by+c = 0

B (h, k)
 40 Straight Line
h  x1 k  y1  2(ax1  by1  c)
 
a b a2  b 2

(2) The image of a point with respect to x-axis : Let P( x , y ) be any point and P  ( x , y ) its image after
reflection in the x-axis, then Y

x = x (  O  is the mid point of P and P  )

y = – y
P(x, y)

X
O O

P(x, y)

(3) The image of a point with respect to y-axis : Let P( x , y ) be any point and P(x , y ) its image after
reflection in the y-axis Y

then x  x ( O  is the mid point of P and P  )

y  y P (x,y) P(x, y)
O

X X
O

Y

(4) The image of a point with respect to the origin : Let P( x , y ) be any Y

point and P(x , y ) be its image after reflection through the origin, then P(x, y)

x   x (  O  is the mid point of P and P  ) N

X
X O M
y  y
P(x,y)

Y

(5) The image of a point with respect to the line y = x : Let P( x , y ) be any point and P(x , y ) be its image
after reflection in the line y  x , then Y
P(x, y)
x  y (  O  is the mid point of P and P  )
y=x O
y  x 45º P(x, y)
X X
O

Y
(6) The image of a point with respect to the line y = x tan  : Y
P(x, y)
Let P( x , y ) be any point and P(x , y ) be its image after reflection in the line
y  x tan  then y=x tan 
O
P(x, y)
X 
x   x cos 2  y sin 2 ( O  is the mid point of P and P  ) O
X

Y
 Straight Line 41

y   x sin 2  y cos 2

Example: 29 The reflection of the point (4, –13) in the line 5 x  y  6  0 is [EAMCET 1994]
(a) (–1, –14) (b) (3, 4) (c) (1, 2) (d) (–4, 13)
 a  4 b  13 
Solution: (a) Let Q(a, b) be the reflection of P(4,  13) in the line 5 x  y  6  0 . Then the point R  ,  lies on 5 x  y  6  0
 2 2 
 a  4   b  13 
.  5     6  0  5a  b  19  0 .....(i)
 2   2 
 b  13   5 
Also PQ is perpendicular to 5 x  y  6  0 . Therefore      a  5b  69  0 .....(ii)
 a4   1 
Solving (i) and (ii), we get a  1, b  14 .
Example: 30 The image of a point A(3, 8) in the line x  3 y  7  0 , is [Rajasthan PET 1991]
(a) (–1, –4) (b) (–3, –8) (c) (1, –4) (d) (3, 8)
Solution: (a) Equation of the line passing through (3, 8) and perpendicular to x  3 y  7  0 is 3 x  y  1  0 . The intersection point of both
the lines is (1, 2). Now let the image of A(3, 8) be A(x1 , y1 ) .
x1  3 y 8
The point (1, 2) will be the midpoint of AA .  1  x1  1 and 1  2  y1  4 . Hence the image is (–1, –4).
2 2

2.19 Some Important Results.

1 (c1  c 2 )2
(1) Area of the triangle formed by the lines y  m 1 x  c1 , y  m 2 x  c 2 , y  m 3 x  c 3 is  .
2 m1  m 2
c2
(2) Area of the triangle made by the line ax  by  c  0 with the co-ordinate axes is .
2 | ab |
2c 2
(3) Area of the rhombus formed by the lines ax  by  c  0 is
ab
(4) Area of the parallelogram formed by the lines a1 x  b1 y  c1  0 ; a2 x  b 2 y  c 2  0 , a1 x  b 1 y  d 1 and
(d1  c1 )(d 2  c 2 )
a2 x  b 2 y  d 2  0 is .
a1b 2  a2 b1
(5) The foot of the perpendicular (h, k ) from (x 1 , y1 ) to the line ax  by  c  0 is given by
h  x 1 k  y1  (ax1  by1  c)
  . Hence, the coordinates of the foot of perpendicular is
a b a2  b 2
 b 2 x 1  aby1  ac a 2 y1  abx1  bc 
 , 
 a2  b 2 a2  b 2 
 
p p
(6) Area of parallelogram A  1 2 , where p 1 and p 2 are the distances between parallel sides and  is the angle
sin 
between two adjacent sides.
x y
(7) The equation of a line whose mid-point is (x 1 , y 1 ) in between the axes is  2
x 1 y1
x y
(8) The equation of a straight line which makes a triangle with the axes of centroid (x 1 , y 1 ) is  1.
3 x 1 3 y1
 42 Straight Line
Example: 31 The coordinates of the foot of perpendicular drawn from (2, 4) to the line x  y  1 is [Roorkee 1995]
1 3  1 3 4 1  3 1 
(a)  ,  (b)   ,  (c)  ,  (d)  , 
3 2  2 2 3 2 4 2 
 1 2  2  1  1  4  1 1 2  4  1  1  2  1   1 3 
Solution: (b) Applying the formula, the required co-ordinates is  ,  = , 
  2 2
 12  12 12  12 
Example: 32 The area enclosed within the curve | x |  | y |  1 is [Rajasthan PET 1990, 97; IIT 1981; UPSEAT 2003]

(a) 2 (b) 1 (c) 3 (d) 2

Solution: (d) The given lines are  x  y  1 i.e., x  y  1 , x  y  1 , x  y  1 and x  y  1 . These lines form a quadrilateral whose
vertices are A(1, 0) , B(0,  1) , C(1, 0) and D(0,1) . Obviously ABCD is a square. Length of each side of this square is

1 2  1 2  2 . Hence, area of square is 2  2  2 sq. units.

Example: 33 If x 1 , x 2 , x 3 and y1 , y 2 , y 3 are both in G.P. with the same common ratio, then the point (x 1 , y1 ) , (x 2 , y 2 ) and (x 3 , y 3 )
[AIEEE 2003]
(a) Lie on a straight line (b) Lie on an ellipse (c) Lie on a circle (d) Are vertices of a triangle
x y
Solution: (a) Taking co-ordinates as  ,  , (x, y) and (x r , y r ) . Above co-ordinates satisfy the relation y  mx ,  the three points lie on a
r r
straight line.
Example: 34 A square of side a lies above the x-axis and has one vertex at the origin. The side passing through the origin makes an angle
 
  0     with the positive direction of x-axis. The equation of its diagonal not passing through the origin is
 4
[AIEEE 2003]
(a) y(cos   sin  )  x (sin   cos  )  a (b) y(cos   sin  )  x (sin   cos  )  a
(c) y (cos   sin)  x(sin  cos )  a (d) y(cos   sin  )  x (sin   cos  )  a
 
Solution: (b) Co-ordinates of A  (a cos , a sin) ; Equation of OB y  tan     x Y
B
4 
 
 CA to OB ;  slope of CA   cot    
4  C
  A
Equation of CA, y  a sin    cot     (x  a cos  )
4  /4

 y (sin  cos )  x (cos a  sin)  a . O X

Example: 35 The number of integral points (integral point means both the coordinates should be integer) exactly in the interior of the triangle
with vertices (0, 0), (0, 21) and (21, 0) is [IIT Screening 2003]
(a) 133 (b) 190 (c) 233 (d) 105
Solution: (b) x  y  21
(0, 21)
The number of integral solution to the equation x  y  21 i.e., x  21  y B
19  20
Number of integral co-ordinates = 19  18  ..........  1   190 .
2

O A (21, 0)
(0, 0)

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