Family of Lines
The general equation of the family of
lines through the point of intersection of two given lines is L + lL¢
= 0, where L = 0 and L¢ = 0 are the two given lines, and l
is a parameter. Conversely, any line of the form L1 + l
L2 = 0 passes through a fixed point which is the point of
intersection of the lines L1 = 0 and L2 = 0.
- The family of lines perpendicular to
a given line ax + by + c = 0 is given by
bx – ay + k = 0, where k is a parameter. - The family of lines parallel to a
given line ax + by + c = 0 is given by
ax + by + k = 0, where k is a parameter.
Example 1
A
variable line through the point of intersection of the lines x/a + y/b=1 and
x/b + y/a = 1 meets the coordinate axes in A and B. Show that the locus of the
midpoint of AB is the curve: 2xy(a + b) = ab(x + y).
Solution:
|
Let
(h, k) be the mid point of the variable line AB. .
The
equation of the variable line AB is
(bx
+ ay – ab) + l(ax + by - ab) = 0
Coordinates
of A are 
Coordinate
of B are 
|
|
Mid point of AB is 
&⇒
h = 
k = 
&⇒
&⇒
&⇒
(h + k)ab = 2hk (a + b).
Hence the locus of the mid-point of
AB is
(x + y) ab = 2xy (a + b).
One Parameter Family of
Straight Lines:
If a linear expression L1
contains an unknown coefficient, then the line L1 =0 can not
be a fixed line. Rather it represents a family of straight lines known as
one parameter family of straight lines. e.g. family of lines parallel to the
x-axis i.e. y = c and family of straight lines passing through the origin i.e.
y = mx.
Each member of the
family passes through a fixed point. We have two methods to find the fixed
point. .
Method (i):
Let the family of straight lines be
of the form ax + by + c = 0 where a, b, c are variable parameters satisfying
the condition al + bm + cn = 0, where l, m, n are given and n >
0. Rewriting the condition as 
,
and comparing with the given family of straight lines, we find that each
member of it passes through the fixed point 
.
Example 2
If a, b and c are
three consecutive odd integers then prove that the variable line ax + by + c =
0 always passes through (1, –2).
Solution:
Since a, b and c are
three consecutive odd integers, these must be in A.P.
&⇒
2b = a + c or a - 2b + c = 0. .
Hence, the variable line always
passes through (1, –2).
Example 3
If the
algebraic sum of perpendiculars from n given points on a variable straight line
is zero then prove that the variable straight line passes through a fixed
point. .
Solution:
Let n given points be (xi,
yi) where i = 1, 2…. n and the variable straight line is ax + by + c
= 0. Given that 
&⇒
a åxi + båyi + cn = 0
&⇒
.
Hence the variable
straight line always passes through the fixed point 
.
Method (ii):
If a family of straight lines can be written as L1
+ lL2 = 0 where L1, L2 are
two fixed lines and l is a parameter, then each member of
it will pass through a fixed point given by point of intersection of L1
= 0 and L2 = 0.
Note: If L1 = 0 and
L2 = 0 are parallel lines , they will meet at infinity.
Example 4
Prove that
each member of the family of straight lines
(3sinq + 4cosq)x + (2sinq
- 7cosq)y + (sinq + 2cosq) = 0 (q
is a parameter) passes through a fixed point.
Solution:
The given family of straight lines
can be rewritten as
(3x + 2y + 1) sinq
+ (4x – 7y + 2) cosq = 0
or, (4x – 7y + 2) + tanq
(3x + 2y + 1) = 0
which is of the form L1 + l
L2 = 0
Hence each member of it will pass
through a fixed point which is the intersection of 4x – 7y + 2 = 0 and 3x + 2y
+1 = 0 i.e. 
.
Concurrency of Straight
Lines:
The condition for 3 lines a1x
+ b1y + c1 = 0, a2x + b2y + c2
= 0, a3x + b3y + c3 = 0 to be concurrent is
(i) 
.
(ii) There exist 3 constants l,
m, n (not all zero at the same time) such that
l L1 + mL2 + nL3 = 0, where L1 = 0,
L2 = 0 and L3 = 0 are the three given straight lines.
(iii) The three
lines are concurrent if any one of the lines passes through the point of
intersection of the other two lines.
