7.3 Equation of a line:
Introduction:
In section 7.1 we
had studied that the linear equation of the generic form y=mx+c
is a straight line. We have learnt that every straight line can be represented
as a linear equation. We have also observed that any point on the line
satisfies the linear equation and conversely, any point which satisfies the
linear equation lies on that line
The ‘inclination’ of a line is the angle _{} which the line makes
with x axis.


Figure
1 
Figure
2 
Figure 3 
If
inclination angle is measured in anti clockwise direction as in Figure 1, _{} is positive 
If
inclination angle is measured in clockwise direction as in Figure 2, _{} is negative. 
The
‘slope’ of any inclined plane (see
Figure 3) is the ratio between the length of vertical line to the horizontal
line 
If
_{} is the angle made by
the inclined plane to the horizontal line, then Slope
= Length of Vertical line/Length of horizontal line = BC/AB=tan_{} (Refer section 8.1) The
‘slope’ of a straight line is the
tangent (tan) of inclination and is denoted by letter m. m=tan_{}. 
Observations:
1. Slope of x axis
= 0 (tan_{}=0 as _{}=0^{0})
2. Slope of y axis
= undefined (tan_{}= _{} as _{}=90^{0})
3. Slope of a line
is positive (tan_{}), if it makes an acute angle in the anticlockwise direction
with x–axis (Figure 1)
4. Slope of a line
is negative ( tan_{}= tan(_{}), if it makes an obtuse angle in the anticlockwise
direction with x–axis or an acute angle in the clockwise direction with
x–axis (Figure 2)
Slope
of a straight line passing through fixed points
Let
P (x_{1},y_{1}) and Q (x_{2},y_{2})
be the two given points. We
are required to find the slope of the line PQ. Extend
QP to meet xaxis to form the angle _{} with xaxis Since
CR is  xaxis _{}= _{}QPR _{}m = tan _{} = QR/PR= Vertical
distance/Horizontal Distance = (y_{2} y_{1})/ (x_{2}
x_{1}) = (y_{1} y_{2})/ (x_{1} x_{2}) =Opposite
Side/Adjacent Side 

1. Note that slope of two parallel lines are same. (_{}Parallel lines make same inclination with xaxis) Conversely, if the slopes of two lines are same then they are
parallel. 2. From above, it follows that slopes of lines joining points
on a straight line are same. 

Relationship
between slopes of two perpendicular lines: Let
BA_{}AC. Let _{} be the angle of
inclination of AB with x axis. Let
_{} be
the angle of inclination of AC with x axis (acute angle measured in
clockwise direction) Slope
of AB = tan_{} = AC/AB Since
slope is negative if the angle of inclination is obtuse in anticlockwise
direction, Slope
of CA =  tan_{} =  (AB/AC) _{} Slope of AB * Slope
of CA = (AC/AB)*(AB/AC) = 1. 3. Thus the product of slopes of two perpendicular lines is
1; conversely, if the product of slopes of two lines is 1 then the lines
are mutually perpendicular 

7.3 Problem 1: A(5,4),B(3,2) and C(1,8) are the vertices of a triangle
ABC. Find
(I) The slope of altitude of AB,
(II) The slope of
the median AD
(III) The slope of
the line parallel to AC.
Solution:
(I) By formula, the
slope of AB = (4(2))/(5(3)) = (6/8)
= 3/4 Since the product
of slopes of two perpendicular lines is 1 The slope of CP
which is perpendicular to AB = 4/3 (II) In
order to find the slope of AD we need to find the point D(x,y) Since
D is mid point of BC x=
(3+1)/2 = 1 : y = (2+(8))/2 = 5 Thus
D is D(1,5). Slope
of AD = ( 4(5))/(5(1)) = 9/6 = 3/2 (III) Slope
of AC = (4(8))/(51) = 12/4 = 3 Thus
any line parallel to AC has same slope of 3 

Intercepts:
In
section 7.1 we have learnt what intercepts are. To
recollect, xintercept is
the distance from O to the point where the line cuts xaxis(x coordinate). yintercept is
the distance from O to the point where the line cuts yaxis(y coordinate). 

Forming
equation of a line:
We have studied
that y=mx+c is the general format for the equation of
a line and we have also studied what slope is.
Let us formulate an
equation for a given line.
1.
Slopeintercept form
Let
P be any point on the given line with coordinates (x,y). Let
AB be the line and _{} be it’s angle of inclination. Let
its yintercept be c, hence _{} PQ = PRQR = yc From
the figure it is clear that _{}PBQ = _{} _{}m = tan _{} = PQ/BQ = (yc)/x _{} mx
= yc i.e.
y = mx+c is the equation of the given line AB. 

7.3 Problem 2: Find
the slope and y intercept of
the line 2x5y+4=0
Solution:
Slope of
the given line 2x5y+4=0 is 2/5
(_{}y = (2/5)x+(4/5)
and is of the form y=mx+c) and its y intercept is 4/5
7.3
Summary of learning
No 
Points to
remember 
1 
If
P (x_{1},y_{1}) and Q (x_{2},y_{2}) then m =
(y_{1} y_{2})/(x_{1} x_{2}) 
2 
If
m is the slope and c is the yintercept then y = mx+c 
Forming
equation of a line(continued):
2.
SlopePoint form
Let
AP be the line and _{} be it’s angle of inclination. Let
Q(x_{1}.y_{1}) be any point on AP. From
the figure it is clear that R is R(x_{1},y) Hence
PR=(x_{1}x) and RQ = (y_{1}y) Also
_{}QPR = _{} _{}m = tan _{} = QR/PR = (y_{1}y)/(x_{1}x) _{} m(x_{1}x) =
(y_{1}y) (Multiply both sides by 1 to get) _{} yy_{1} =
m(xx_{1}) Note y = mx+(y_{1} mx_{1})
is the equation of the given line AP which is of the form y=mx+c 

3.
Two Point form
Let
R(x_{2},y_{2}),Q(x_{1},y_{1})
be two points on the given line AB. We
are required to find the equation of this line. As
learnt earlier in this section, the slope of a line through two points is: Slope
of AB = m = (y_{2}y_{1})/(x_{2}x_{1}).
Let P (x,y) be another point on this line It’s
slope = m = (yy_{1})/(xx_{1}) _{} yy_{1} = m(xx_{1}) i.e. y = mx+(y_{1}mx_{1}) where m = (y_{2}y_{1})/(x_{2}x_{1})
which is of the form y=mx+c. This is the equation of the given line
AB passing through 2 points. 

4.
Intercepts form:
Let
P(a,0) and Q(0,b) be the x and y intercept
respectively of the line AB. _{}m = Slope of AB = tan _{} = OB/OA = b/(a) =
b/a From
Slope – intercept form which was discussed earlier in this section The
equation of a line AB is y = mx + (yintercept) =
(b/a)x+b (Note Y intercept = b) i.e.
y = (bx+ab)/a i.e.
ay = bx+ab i.e.
ay+bx = ab i.e. (y/b) + (x/a) = 1 

7.3 Problem 3: A(1,4),B(3,2) and C(7,5) are the vertices of a triangle ABC.
Find
(i) The coordinates of the centroid of triangle ABC
(ii) The equation
of the line through the centroid of the triangle and parallel to AB
Solution:
(i) In section 7.2
we have learnt that the coordinates of centroid G(x,y) of any triangle is: _{}x = (1+3+7)/3 =11/3 _{}y = (4+2+5)/3 = 11/3 _{} G(x,y)
= G(11/3,11/3) (ii)
Slope of the line passing through G and  to AB, is same as the slope
of AB. Slope
of AB = m = (y_{2}y_{1})/ (x_{2}x_{1}) =
(24)/(31) = 1 The
equation of the line passing through G(x,y) and
having a slope of 1 is (slope–point form) y
= mx+(y_{1}mx_{1}) Note:
Here x_{1} and y_{1} are coordinates of G. y
= x + [11/3(1)(11/3)] =
x+22/3 i.e.
3x+3y = 22 is the
equation to the line, parallel to AB and passing through the centroid G of
the triangle ABC. 

7.3 Problem 4: The mid
points of three sides of a triangle are (5,3), (6,6)
and (5,3).
Find the equation
of the sides of the triangle.
Solution:
Let
the triangle be ABC and D, E and F be the mid points of BC, AB and AC
respectively, so that the
co–ordinates of D, E and F are (5,3), (6,6) and (5,3) respectively. We
are required to find the equation to the lines BC, AB and AC. Hint: Note
that BCEF and hence the slope of BC = Slope of EF = 3/11 Hence
the equation to BC is 3x11y = 48 (use slopepoint form) DEAC,
Slope of AC = slope of DE = 9 and hence the equation to AC is 9xy+48=0 DFAB,
Slope of AB = slope of DF = 3/5 and hence the equation to AB is 3x+5y=48 

Equation of lines parallel and
perpendicular to a given line.
Let
L be a line passing through the point (x_{1},y_{1})
with slope m. Then
the equation of L is y = mx+(y_{1}mx_{1})
which is of the form y=mx+c_{1} where c_{1} = (y_{1}mx_{1}). Let L^{1} be a line parallel to L and passing through the point (x_{2},y_{2}) with slope m_{1.} Then the equation
of L^{1} is y = m_{1}x+(y_{2}m_{1}x_{2}) Since L and L^{1} are parallel, m=m_{1.} Hence the
equation of L^{1} is y = mx+(y_{2}mx_{2}) = mx+c_{2} ,where
c_{2}= (y_{2}mx_{2}). 1.
Thus equations of two parallel lines differ only by constants. (_{} For L it is mx+(y_{1}mx_{1})
and for L^{1} it is mx+(y_{2}mx_{2})) 

Let
L be a line passing through the point (x_{1},y_{1})
with slope m = tan_{}. Then
the equation of L is y = mx+(y_{1}mx_{1})
Let L^{1} be a line perpendicular to L and passing through the point (x_{2},y_{2}) with slope m_{1}._{} Then the equation
of
L^{1} is y = m_{1}x+(y_{2}m_{1}x_{2}) Earlier in this
section, We have learnt
that product of slopes of two perpendicular line is 1. _{} m_{1}= 1/m 2. Thus the equation of L^{1} which is perpendicular
to L is y = –(1/m)x+(y_{2}+(1/m)x_{2}) 

Let ax+by+c = 0 be an equation in first degree.
(i)
If a = 0 then y =c/b which represents a line parallel to xaxis
(ii)
If b = 0 then x = c/a which represents a
line parallel to yaxis
(iii) If b is
non zero then y = (a/b)x(c/b) = mx+c
which represents an equation of a straight line.
3.
Thus ax+by+c = 0 represents an equation to a straight
line.
The general form of
an equation to a line is ax+by+c = 0 with a,b,c _{}R. Properties of this line depends on the values of a, b, c and are as follows:
Value of 
Equation : 
Line : 
Slope : 
xintercept: 
yintercept: 

a 
b 
c 

=
0 
_{} 0 
= 0 
y=0 
xaxis 
0 
 
0 
_{}0 
y=c/b 
to
xaxis 
0 
 
c/b 

_{} 0 
_{} 0 
= 0 
y=(a/b)x 
Line
through O(0,0) 
(a/b) 
0 
0 
_{}0 
ax+by+c=0 
Neither
 to xaxis nor  to yaxis 
(a/b) 
(c/a) 
(c/b) 

=
0 
_{}0 
x=c/a 

yaxis 
_{} 
(c/a) 
 

= 0 
x=0

yaxis 
_{} 
0 
 
7.3 Problem 5: Find
the equation of a line passing through (4,3) and perpendicular to the line
2x5y+4=0
Solution:
Slope of
the given line 2x5y+4=0 is 2/5 (y = (2/5)x+(4/5))
Slope of
the line perpendicular to the above line is –(5/2) (_{}the product of the slopes of two perpendicular lines =1)
Since
this perpendicular line passes through (4,3) its equation is
y= mx+(y_{1}mx_{1})
= –(5/2)x + (3) + (5/2)4
= –(5/2)x+ 7
i.e. 2y
= 5x+14 Or 5x+2y14=0
Testing concurrency of straight lines when
their equations are given:
Steps:
1. Solve
any two equations to get their point of intersection
2.
Substitute the coordinates obtained from step 1 in the third equation
3. If
the third equation is satisfied then the point is on the third line and hence
the three lines are concurrent