1. Analytical Geometry
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2. General Equation of a Line
General equation of first degree in two variables,
Ax + By + C= 0
where A, B and C are real constants such that A and B are
not zero simultaneously.
Graph of the equation
Ax + By + C= 0 is always a straight line.
Therefore, any equation of the form Ax + By + C= 0,
where A and B are not zero simultaneously is called
general linear equation or general equation of a line.
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3. Different forms of Ax + By + C = 0
The general equation of a line can be reduced into various
forms of the equation of a line, by the following procedures:
 Slope-intercept form
If B ≠ 0, then Ax + By + C= 0 can be written as
Where and
then from equation (1)
the slope of the st. line is and y intercept is
If B= 0 then x = which is a vertical line whose slope is
undefined and x-intercept is
B
C
x
B
A
y  or y = mx + c ------------- (1)
B
A
m 
B
C
c 
B
A

A
C

B
C

A
C

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4.  Intercept form
If C ≠0, then Ax + By + C = 0 can be written as
where
We know that equation (1) is intercept form of the equation
of a line whose
x-intercept is y- intercept is
If C = 0, then Ax + By + C = 0 can be written as
Ax + By = 0,
which is a line passing through the origin and,
therefore, has zero intercepts on the axes.
1



B
C
y
A
C
x
or 1
b
y
a
x
A
C
a 
B
C
b And
A
C

B
C

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5.  Normal form
Let x cos θ +y sin θ = p be the normal form of the line
represented by the equation
Ax + By + C = 0 or Ax + By = – C.
Thus, both the equations are same and therefore
Which gives,
and
Now,
Or
p
CBA

 sincos
C
Ap
cos
C
Bp
sin
1sincos
2
22













C
Bp
C
Ap

22
2
2
BA
C
p

 or 22
BA
C
p


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6. Therefore
and
Thus, the normal form of the equation Ax + By + C = 0 is
x cos θ + y sin θ = p,
where
,
and
Proper choice of signs is made so that p should be positive.
22
cos
BA
A

 22
sin
BA
B


22
cos
BA
A

 22
sin
BA
B


22
BA
C
p


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7. 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 : Ax + By + C = 0 be a line, whose distance from the
point P (x1, y1) is d.
Draw a perpendicular PM
from the point P to the line L.
If the line meets the x-and y-axes
at the points Q and R, respectively.
Then, coordinates of the points are
L : Ax + By + C = 0
X
Y
P(x1, y1)






B
C
R ,0






 0,
A
C
Q







B
C
R ,0





 0,
A
C
Q and
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8. Thus, the area of the triangle PQ is given by
area (  PQR)
, which gives
Also,
area (  PQR)
Or 2 area (ΔPQR)
QRPM.
2
1
 






QR
PQR)(trianglearea2
PM
 000
2
1
111 

















 yy
B
C
A
C
B
C
x
AB
C
A
C
y
B
C
x
2
11
2
1

,. 11 CBxAx
AB
C

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9. 22
22
00 BA
AB
C
B
C
A
C
QR 












Substituting the values of area (ΔPQR) and QR in (1), we get
or
Thus, the perpendicular distance (d) of a line Ax + By+ C = 0
from a point (x1, y1) is given by
22
11
BA
CByAx
PM



22
11
BA
CByAx
d



22
11
BA
CByAx
d



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10. Distance between two parallel lines
We know that slopes of two parallel lines are equal.
Therefore,
two parallel lines can be taken in the form
y = mx + c1 ... (1)
and y = mx + c2 ... (2)
Line (1) will intersect x-axis at the point
Distance between two lines is
equal to the length of
the perpendicular
from point A to line (2). X
Y
d
o






0,1
m
c
A
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11. Therefore, distance between the lines (1) and (2) is
or
Thus,
the distance d between two parallel lines y = mx + c1 and
y = mx + c2 is given by
If lines are given in general form, i.e., Ax + By + C1 = 0 and
Ax + By + C2 = 0,
then above formula will take the form
   2
1
c
m
c
m 






2
21
1 m
cc
d



2
21
1 m
cc
d



22
21
BA
cc
d



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12. The End
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