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 L_{1} + l
L_{2} = 0 passes through a fixed point which is the point of
intersection of the lines L_{1} = 0 and L_{2} = 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.

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).

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).

If a linear expression L_{1}
contains an unknown coefficient, then the line L_{1} =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. .

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 .

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).

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).

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. .

Let n given points be (x_{i},
y_{i}) where i = 1, 2…. n and the variable straight line is ax + by + c
= 0. Given that

&⇒
a åx_{i} + båy_{i} + cn = 0

&⇒
.

Hence the variable straight line always passes through the fixed point .

If a family of straight lines can be written as L_{1}
+ lL_{2} = 0 where L_{1}, L_{2} 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 L_{1}
= 0 and L_{2} = 0.

Note: If L_{1} = 0 and
L_{2} = 0 are parallel lines , they will meet at infinity.

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.

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 L_{1} + l
L_{2} = 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. .

The condition for 3 lines a_{1}x
+ b_{1}y + c_{1} = 0, a_{2}x + b_{2}y + c_{2}
= 0, a_{3}x + b_{3}y + c_{3} = 0 to be concurrent is

(i) .

(ii) There exist 3 constants l,
m, n (not all zero at the same time) such that

l L_{1} + mL_{2} + nL_{3} = 0, where L_{1} = 0,
L_{2} = 0 and L_{3} = 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.