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Unit 1
Chapter 5
Equation of Straight Line
Equation of the straight line
For every point P(x,y) to lie on a straight line, its coordinates x and y must
satisfy a certain relation. This relation is called the equation of the straight line.
For example, to find the equation of the line of slope 2 and through the point
A(2,1) is to find the connection such that every point P(x,y) on the line should satisfy,
i.e. slope of PA = given slope 2
y 1
2
x2
or 2x – y = 3 …………..(1)
Hence, in order that P(x,y) be a point on the line, its coordinates x and y must satisfy
(1), and (1) is called the equation of the line.
In fact, the equations of straight lines are of 1st degree, and equations in x, y of 1st
degree are called linear equations.
7.
Point-slope form
We have already found in the last section that the line through the point (x1,y1)
of slope m is:
y  y1
m
x  x1
Example 8: Find the equations of lines through the following given point(1,-2) and
with slopes –1.
Solution:
The equation of the line through (1,-2) with slope = -1 is :
y2
 1
x 1
or_________________________
8.
Two points form
From our last section, the line through (x1,y1) and of slope m is
y  y1
m .
x  x1
1
2
If the line also passes through (x2,y2), its slope is
y 2  y1
.
x 2  x1
Therefore, the equation of the line joining (x1,y1) and (x2,y2) is
y  y1
y  y1
 2
x  x1
x 2  x1
Example 9: Find the equation of the line joining (4,1) to (5,-2).
Soln: The equation is:
y 1  2 1

x4
54
or
____________________
Example 10: Find the equation of the line which makes x-intercept 5 and yintercept 2 on the axes.
Solution
y-axis
x-intercept = 5
y-intercept = 2
A
2
x-axis
O
5
B
As the line makes y-intercept 2 and x-intercept 5
From the figure, A = (0,2) and B = (5,0)
The equation of the line is
y2 02

x0 50
or
_______________
Example 11: Find the equation of the line passing through (2,3) and is parallel to the
line 3x + 4y =12.
Soln: When x=0, y=3
2
3
When y=0, x=4
The line 3x + 4y =12 passes through (0,3) and (4,0)
and its slope is m=
03 3

40
4
Equation of the line is:
Or
9.
y 3 3

x2
4
__________________
Slope-intercept form
The equation of the line with slope m and y-intercept b is the line through (0,b),
so its equation is:
y b
 m or
x0
y=mx+b.
By writing the equation of a line into this form, we can read the slope m of the line
easily.
Example 12: Find the slope of each of the line 3x + 4y = 10.
Soln: 4y = -3x + 10
y=
3
10
3
x
gives slope m =
4
4
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Example 13: Find the equation of the line through (5,7) and is perpendicular to
another line 3x + 2y – 7 = 0.
Soln: As 3x + 2y – 7 = 0
y=
3
7
x+
2
2
It’s slope m =
3
2
Hence equation of the line is
y7 2

x5 3
or ____________________
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Supplementary Problems
1) Write down the equation of the line through the origin and with gradient
1
1
(a) 2
(b)-1
(c)
(d) 
(e) 0
(f) 
3
4
Draw a sketch showing these lines on the same pair of axis.
Ans. a) y =2x
b) y = -x C) 3y = x d) 4 y  x  0 e) y = 0 f) x = 0.
2) Write down the equation of the line passing through the given point and with
the given gradient.
1
1
1
(a) (0,1),
(b) (0,0),
(c) ( 1,4),
2
2
2
Ans: a) 2 y  x  2
b) 2y = x
c) 2y = x-7
3) Write down the equation of the line passing through the given points.
(a) (0,0), (2,1) (b) (1,4), (3,0) (c) (2,0), (0,4) (d) (2,0), (0,4) (e) (1,3), (4,3)
Ans. a) 2y = x b) 2x  y  6  0 c) y  2x  4  0 d) y  2 x  5
4) Write down the inequality which defines the region :
(a) above the line through the origin with gradient 1.
(b) Below the line through (1,2) and (0,4)
(c) Above the line x  y  2  0
(d) Below the line 2x  y  4  0
Ans. a) y>x
b) y  4  2x c) y  2  x d) y  2 x  4
5) Write down the equation of the line passing through the origin and
perpendicular to :
(a) 3x  2 y  4  0 (b) x  2 y  3  0
Ans. a) 3 y  2 x  0
b) y  2x  0
6) Write down the equation of the line passing through the given point and
perpendicular to the given line.
(a) (2,1), 3x  y  2  0 (b) (-1,-2), 2x  3 y  6  0
Ans. (a) 3 y  x  1
(b) 2 y  3x  7  0
7) Write down the equation of the line passing through (3,-2) and parallel to:
(a) 5x  y  3  0 (b) x  7 y  5  0
Ans. a) 5x  y  17  0 b) x  7 y  11  0
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