Straight Line
Any equation of first
degree of the form Ax + By + C = 0, where A, B, C are constants always
represents a straight line (at least one out of A and B is non zero).
Slope:
If q
is the angle at which a straight line is inclined to the positive direction of
x axis, slope of the line is m = tanq, 0 £ q
< 180° (q > 90°
).
Intercept of a straight
line on the axis:
If a line AB cuts the
xaxis and yaxis at A and B respectively and O be the origin then OA and OB
with proper sign are called the intercepts of the line AB on x and y axes
respectively.
Notes:
l If
we say that the length of intercept of a line on xaxis is 3, it means that
intercept on xaxis is either 3 or –3.
·
If intercept of a
line on xaxis –3, then length of intercept of axis is  –3  = 3.
Standard equations of
Straight Lines:
SlopeIntercept Form :
y = mx +
c,
where m =
slope of the line = tanq
c = y intercept


Intercept Form :
x/a + y/b =
1
x intercept = a
y intercept = b


Normal Form :
x
cosa + y sina = p, where a
is the angle which the perpendicular to the line makes with the axis of x
and p is the length of the perpendicular from the origin to the line. p is
always positive. .



Slope Point Form:
Equation: y  y_{1} = m(x 
x_{1}), where
(a) One point on the straight line is
(x_{1}, y_{1})
(b) The direction of the straight
line i.e., the slope of the line = m.
Two Points Form:
Equation: ,
where (x_{1}, y_{1})
and (x_{2}, y_{2}) are the two given points. Here .
Parametric Form:
To find
the equation of a straight line which passes through a given point A(h, k)
and makes a given angle q with the positive direction of the
xaxis.
P(x, y) is any point on the line LAL¢.
Let AP = r. x – h = r cosq
y – k = r sinq.


is the equation of the straight line
LAL¢.
Any point on the line
will be of the form (h+r cosq, k+ r sinq)
and for any value of r
this will represent a
point on the straight line. Here  r  gives the distance of the point P from
the fixed point (h, k).
Note:
If point P is taken relatively upward
to A then r is positive otherwise negative. If line is parallel to xaxis then
for the point right to A, r is positive and for left to A, r is negative. .
Reduction of the General
Equation to Different Standard Forms:
SlopeIntercept Form:
To
reduce the equation Ax + By + C = 0 to the form y = mx + c
Given
equation is Ax + By + C = 0
&⇒ y =
which
is of the form y = mx + c
where
m = ( B >
0)
Note:
Slope
of the line Ax + By + C = 0 is –,
i.e. –,
y intercept of the line = C/B
Intercept Form:
To
reduce the equation Ax + By + C = 0 to the form
This
reduction is possible only when C > 0
Given
equation is Ax + By = –C
&⇒ which is of the form
. where a =
Note:
Intercept
on the xaxis = ,
Intercept on the yaxis =
Thus intercept of a straight line on
the xaxis can be obtained by putting y = 0 in the equation of the line and
then finding the value of x. Similarly, intercept on the yaxis can be obtained
by putting x = 0 and solving for y.
Normal Form:
To reduce the equation Ax + By + C =
0 to the form xcosa + ysina = p
Given equation is Ax +
By + C = 0 Or, Ax + By = –C
Case I: When –C>0,
then normal form is
where cosa
= sina
= , p =
Case II: When –C < 0, then write
the equation as Ax –By = C
where cosa
= , sina
= p =
Note:
In the normal form xcosa
+ ysina = p, p is always taken as positive. .
Illustration  7:
Reduce the line 2x 
3y + 5 = 0, in slopeintercept, intercept and
normal forms.
Solution:
SlopeIntercept
Form:
Intercept Form:
Normal Form :
, ,
Example 1
Find the equations
of the lines which pass through the point (3, 4) and the sum of their
respective intercepts on the axes is 14. .
Solution:
Let the equation of the line be . . .
(i)
This passes through
(3, 4), therefore .
. . (ii)
It is given that a +
b = 14
&⇒ b = 14 –a . Putting b = 14  a in (ii),we get
&⇒
a^{2} 13 a + 42 = 0
&⇒
(a – 7)(a –6) = 0
&⇒ a = 7, 6
&⇒
two such lines are there.
For a = 7, b = 14 –7
= 7 and for a = 6 , b = 14 – 6 = 8. .
Putting the values of a
and b in (i), we get the equations of lines
and or x + y = 7 and
4x + 3y = 24.
Example 2
A
line joining two points A(2, 0) and B(3, 1) is rotated about A in the
anticlockwise direction through an angle of 15°. Find the equation of the line
in the new position. If B goes to C, in the new position, what will be the
coordinates of C.
Solution:
Slope of the line AB is m = 1
&⇒
tanq = 1 &⇒ q = 45°
AB = Ö2
&⇒ AC = Ö2
Slope of AC = tan(45° + 15°)
= tan(60°)
[because angle between AB
and AC=15°]
A is (2, 0) &⇒
h = 2, k = 0. .
Here, C is given by the formula


(h + r cosq,
k + r sinq), (r = Ö2)
C = (2 + Ö2
cos 60°, 0 + Ö2 sin 60°), C =
&⇒
Equation of the line AC is y = Ö3(x – 2).
Example 3
If the straight line
through the point P (3, 4) makes an angle p/6
with the xaxis and meets the line 12x + 5y + 10 = 0 at Q, find the
length of PQ.
Solution:
The
equation of a line passing through P (3, 4) and making an angle p/6
with the xaxis is
, where r represents the distance of
any point on this line from the given point P (3, 4). The coordinates of any
point Q on this line are
If Q lies on 12x + 5y + 10 = 0, then
&⇒
&⇒
length PQ = .
Example 4
A
canal is 4½ kms from a place and the shortest route from this place to the
canal is exactly northeast. A village is 3 kms north and 4 kms east from the
place. Does it lie on canal?.
Solution:
Let AB be the canal and O be the
given place. Let L be the foot of perpendicular from O to AB. Given, OL = 9/2.
And ∠AOL = 45°.
Now
equation of line AB will be
xcos45° +
ysin45° = 9/2
Or x + y =
9/Ö2 ……..(1).
Let S be
the given village, then
Sº (4, 3)
Putting x=4
and y = 3 in equation (1), we get 4 + 3 = 9/Ö2, which is
not true. Thus the coordinates of S .


doesn’t satisfy equation (1) and hence the given village
does not lie on the canal.
Example 5
Find
the equation of the straight line which makes an angles of 15°
with the positive direction of xaxis and which cuts an intercept of length 4
on the negative direction of yaxis. .
Solution:
Here
m = tan15° = tan(45°
– 30°)
=
= 2 –
and
c = intercept on yaxis = 4
Therefore,
equation of the required line will be
y
= (2 – ) x – 4
or
(2 –)x – y – 4 = 0.
Example 6
Find the equation of the straight line which passes
through the point
(3, 2) and whose gradient is .
Find the coordinates of the points on the lines that are 5 units away from the
point (3, 2).
Solution:
Let
P º (3, 2)and let the required line make an angle q
with the positive direction of xaxis . .
Given tanq
=
\
cos q =and
sinq =
Now
equation of the required line will be


or 3 (x – 3) = 4( y – 2) or 3x – 4y
–1 = 0 . . . (1).
Now the coordinates of the points
which are at a distance of 5 units from P are (3 ±
5 cosq, 2 ± 5 sinq) or (3 ±
4, 2 ± 3)
or (7, 5 ) and ( –1, –1) . .
Position of two points with
respect to a given line:
Let the given line be
ax + by + c = 0 and P(x_{1}, y_{1}), Q(x_{2}, y_{2})
be two points. If the quantities ax_{1} + by_{1} + c and ax_{2}
+ by_{2} + c have the same signs, then both the points P and Q lie on
the same side of the line ax + by + c = 0.
If the quantities ax_{1} + by_{1}
+ c and ax_{2} + by_{2} + c have opposite signs, then they lie
on the opposite sides of the line.
Example 7
Find the range of q in the interval (0, p) such that the points (3, 5) and
(sinq, cosq) lie on the same side of the line x
+ y  1 = 0.
Solution:
Here 3 + 5  1 = 7 > 0
Hence sinq
+ cosq  1 > 0
&⇒ sin(p/4 + q)
> 1/Ö2
&⇒
p/4
< p/4 + q < 3p/4
&⇒
0 < q < p/2.
Angle between two straight
lines:
If q
is the acute angle between two lines, then
tanq = where
m_{1} and m_{2} are the slopes of the two lines and are
finite.


Notes:
 If the two lines are perpendicular to
each other then m_{1}m_{2} = 1.
 Any line perpendicular to ax + by + c
= 0 is of the form bx – ay + k = 0.
 If the two lines are parallel or
coincident, then m_{1} = m_{2}.
 Any line parallel to ax + by + c = 0
is of the form ax + by + k = 0.
 If any of the two lines
is perpendicular to xaxis, then the slope of that line is infinite. Let m_{1}
= ¥. Then =
or q
= 90°  a
where tana = m_{2} i.e. angle q
is the complimentary to the angle which the oblique line makes with the xaxis.
Example 8
Find the equation to
the straight line which is perpendicular bisector of the line segment AB, where
A, B are (a, b) and (a¢, b¢) respectively.
Solution:
Equation of AB is y  b =
i.e. y(a¢
 a)  x(b¢  b) = a¢b
 ab¢.
Equation to the line perpendicular to
AB is of the form
(b¢  b)y + (a¢
 a)x + k = 0 ...(1).
Since the midpoint of AB lies on (1)
.
Hence the required equation of the
straight line is
2(b¢  b)y + 2(a¢
 a)x = (b¢^{2}  b^{2} + a¢^{2}  a^{2}).
Equation of Straight
Lines passing through a given point and equally inclined to a given
line:
Let the straight lines passing
through the point (x_{1}, y_{1}) and make equal angles with the
given straight line y = m_{1}x + c. .
If m is the slope of the required
line and a is the angle which this line makes
with the given line, then tana = ±.
1. The above
expression for tana, gives two values of m, say m_{A}
and m_{B}. .
The required
equations of the lines through the point (x_{1}, y_{1}) and
making equal angles a with the given line are y – y_{1}
= m_{A}(x  x_{1}), y  y_{1} = m_{B}(x  x_{1}).


Example 9
Find
the equation to the sides of an isosceles rightangled triangle, the equation
of whose hypotenuse is 3x + 4y = 4 and the opposite vertex is the point (2, 2).
Solution:
The problem can be restated as:
Find the equations to the straight
lines passing through the given point (2, 2) and making equal angles of 45°
with the given straight line 3x + 4y  4 = 0. Slope of the line 3x + 4y  4 =
0 is m_{1} = 3/4.
&⇒
tan 45° =±
m_{A}
= and m_{B}
= 7
Hence the required equations of the
two lines are


y  2 = m_{A}
(x  2) and y  2 = m_{B}(x  2)
&⇒
7y  x  12 = 0 and 7x + y = 16.
Example 10
A line 4x + y =1
through the point A(2, 7) meets the line BC whose equation is 3x – 4y +1 =
0 at the point B. Find the equation to the line AC so that AB = AC. .
Solution:
Clearly, AB
and AC both pass through A(2, 7) and are equally inclined to 3x – 4y +1
= 0. So, they form an isosceles triangle and their equations are given by .
(y + 7) = ,
where
m = slope of BC = 3/4 and


tana = =
&⇒ (y +7) =
&⇒ (y +7) = and y +7 =
&⇒ y+ 7 = 4(x
–2) and y + 7 = 
&⇒
4x + y = 1 and 52 x + 89y + 519 = 0
Clearly, 4x + y = 1 is the equation of AB. So, the
equation of AC is .
52x + 89 y + 512 = 0
Alternatively:
Slopes of the lines AB and BC are –4
and 3/4. Let the slope of the line AC be m. .
Then = ±
.
Hence equation of the line AC is
y + 7 = (x  2)
&⇒
52x + 89y + 519 = 0.
The distance between two
parallel lines:
The distance between
two parallel lines:
ax + by + c_{1} = 0 and ax +
by + c_{2} = 0 is .
Length of the Perpendicular
from a Point on a Line:
The length of the
perpendicular from P(x_{1}, y_{1}) on ax + by + c = 0 is .
The length of the
perpendicular from origin on ax + by + c = 0 is .
Example 11
Find all points on x
+ y = 4 that lie at a unit distance from the line
4x + 3y – 10 = 0. .
Solution:
Coordinates of an arbitrary point on
the given line are P(t, 4t). Let
P(t, 4t) be the required point. Then, distance of P from the line
4x+3y– 10 = 0 is unity i.e. =
1
&⇒  t + 2 =5
&⇒
t + 2 = ± 5
&⇒ t = 7 or t = 3. .
Hence, the required points are (7, 11)and (3, 1). .
Example 12
Three lines x + 2y + 3 = 0, x + 2y  7 =
0 and 2x  y  4 = 0 form 3 sides of two squares. Find the equation of
remaining sides of these squares.
Solution:
Distance between the two
parallel lines is . The
equations of sides A and C are of the form
2x – y + k = 0. Since distance between sides A and B = distance between sides


B and C
&⇒
&⇒
k = 6, 14.
Hence the fourth sides
of the two squares are
(i) 2x  y + 6 = 0,
(ii) 2x  y  14 = 0.
Notes:
(i)
In order to find
the foot of the perpendicular from a given point to a given line, the most
convenient method is to write the equation of the given line in such a way so
that the coefficient of y becomes negative and then write the equation of the
perpendicular in the parametric form. Then put r = the algebraic length
(without the modulus) of the perpendicular = p. In this way, we get the foot of
the perpendicular directly.
(ii)
To find the
mirror image of a given point in a given line, we use the same process, taking
r = 2p.
Example 13
Find the foot of the
perpendicular drawn from the point (2, 3) to the line 3x – 4y + 5 = 0. Also,
find the image of (2, 3) in the given line.
Solution:
Let
AB º 3x4y + 5 = 0, P º (2, 3) and
PM ^ AB. Slope of AB =
&⇒
slope of PM = = tanq
(say)
&⇒
,


Now, r = p =
which is the foot of
the perpendicular. .
&⇒
M = =
Let Q be the image of P
&⇒
Q = =