濟 記 程 式 版
Equations of straight lines
Please read the following and complete the Basic Level Exercises accordingly.
1.
Distance between two points
AB =
(a  c) 2  (b  d ) 2
B(c , d)
A(a , b)
bd
ac
2.
Slope of AB =
3.
Angle of inclination = 
L
tan = slope of L

x
Remarks
 is measured from the +ve x-axis,
anticlockwise = => positive 
clockwise = =>
negative 
e.g. 3.1 Find the angle of inclination of a straight line which passes through (1,3) and (4,2).
Solution: tan =
tan =
3  (2)
1 4
5
= 1
5
  = 135
Do Basic Level Exercise Q.1,2
程式版 (Equations of lines) P.
1
4.
Angle between two straight lines
Slope of L1 = m1
Slope of L2 = m2
m1
L2
Acute angle between L1 and L2 = 

m2
m1  m2
1  m1m2
tan =
L1
e.g. 4.1 L1 and L2 are two straight lines. The slope of L1 is
1
while L2 passes through (1,3) and
2
(4,2). Find the acute angle between L1 and L2.
1
2
Solution: m1 =
3  (2)
= 1
1 4
m2 =
1
 (1)
tan = 2
=3
1
1  ( )( 1)
2
 = 71.6 (3 sign. fig.)
e.g. 4.2 L1 and L2 are two straight lines. Given that the slope of L1 is
1
and the acute angle
2
between L1 and L2 is 60. Find the slope of L2.
1
, let m be the slope of L2
2
Solution: m1 =
1
m
tan60 = 2
1
1  ( )m
2
3 =
1  2m
2m
1  2m
=  3
2m
1  2m = 3 (2 + m) or 1  2m =  3 (2 + m)
( 3 +2)m = 1  2 3 or (2  3 )m = 1 + 2 3
m=
1 2 3
2 3
or
m=
1 2 3
2 3
Do Basic Level Exercise Q.3,4
程式版 (Equations of lines) P.
2
5.
Section formula
Given AP : PB = m : n
nx1  mx2
x=
nm
y=
m :
n
B(x2 , y2)
P(x , y)
A(x1 , y1)
ny1  my2
nm
e.g. 5.1 Let A(1,2) and B(3,6). P is a point lies on the line segment AB such that
AP : PB = 4 : 5. Find the coordinates of P.
Solution: x1 = 1
x2 = 3
y1 = 2
y2 = 6
m=4
n=5
By section formula,
x=
5( 1)  4(3)
7
=
54
9
y=
5(2)  4(6)
14
= 
54
9
 7 14 
 coordinates of P are  , 
9 9 
e.g. 5.2 Let A(1,2) and B(3,6). P divides the line segment AB externally such that
AP : PB = 4 : 5. Find the coordinates of P.
Solution: AP : PB = 4 : 5 and P cuts AB externally (as shown)
PA : AB = 4 : 1
Let P(x , y)
By section formula,
1( x)  4(3)
1 =
=> x + 12 = 5 => x = 17
1 4
2=
4
:
1
P(x , y)
A(1, 2)
B(3, 6)
1( y )  4(6)
=> y  24 = 10 => y = 34
1 4
 coordinates of P are (17,34)
Do Basic Level Exercise Q.5,6
程式版 (Equations of lines) P.
3
6.
Let A(x1 , y1), B(x2 , y2) and C(x3 , y3)
Area of ABC =
x1
y1
1 x2
2 x3
x1
y2
y3
y1
e.g. 6.1 Let A(1,2), B(3,6) and C(4,2). Find the area of ABC.
1
Solution: Area =
=
2
1 3 6
2
2 4
1 2
1
(1)(6)  (3)(2)  (4)(2)  (3)(2)  (4)(6)  (1)(2)
2
= 22 units square
e.g. 6.2 Let A(1,2), B(3,6) and C(t,t). Given that A,B and C are collinear (在同一直線上).
Find the value of t.
Solution: If A,B and C are collinear, the area of ABC will be zero.
1
2
1 3 6
=0
t
2 t
1 2
1
(1)(6)  (3)(t )  (t )(2)  (3)(2)  (t )(6)  (1)(t ) = 0
2
t=0
Do Basic Level Exercise Q.7,8
7. Let A1(x1 , y1), A2(x2 , y2),… An(xn , yn) be vertices of polygon A1A2A3…An.
x1
y1
x2
1
...
Area of polygon A1A2A3…An =
2
xn
y2
x1
... .
yn
y1
Note: Vertices should be arranged either in clockwise or anti-clockwise.
Do Basic Level Exercise Q.9
程式版 (Equations of lines) P.
4
8.
Equations of straight lines
“Equation” means “Relation”
What kind of relation?
Relation between x and y.
In details,
Relation between x-coordinate and y-coordinate of any point on the line.
“Find the equation of a line L”
means
“Find the relation between the x-coordinate and y-coordinate of any point on the line L”
The general method of finding equation of straight line is finding slope.
e.g. 8.1 Let A(1,2) and B(3,6). Find the equation of the line AB.
2  ( 6 )
Solution: slope of AB =
= 2
 1  (3)
suppose P(x,y) is a point on the line AB, then
slope of PA = slope of AB
y2
= 2
x  (1)
It is already an EQUATION of AB, because it tells us the RELATION between x and y.
Simplifying,
y  2 = 2(x + 1)
2x + y = 0 is the equation of AB
Do Basic Level Exercise Q.10-12
程式版 (Equations of lines) P.
5
9.
Equations of straight lines : Slope-intercept form
y = mx + c
slope = m
y-intercept = c
e.g. 9.1 Find the slope and the y-intercept of y = 2x  3.
Solution:
slope = 2
y-intercept = 3
e.g. 9.2 Find the slope and the y-intercept of 4x + 5y  6 = 0
Solution:
4x + 5y  6 = 0
5y = 4x + 6
y= 
4
6
x
5
5
slope = 
y-intercept =
4
5
6
5
e.g. 9.3 Find the x-intercept of 4x + 5y  6 = 0
Solution:
Put y = 0 into the equation,
4x + 5(0)  6 = 0
x=
6
3
=
4
2
 x-intercept =
3
2
Do Basic Level Exercise Q.13
10. Equations of straight lines : General form
Ax + By + C = 0
程式版 (Equations of lines) P.
6
11. Distance between a point and a line
(x1 , y1)
d=
d
Ax1  By 1  C
Ax + By + C = 0
A2  B 2
Remark : before using it, the equation of line must be expressed in General Form first.
e.g. 11.1 Find the distance between (3,4) and the line y = 2x + 1.
Solution: y = 2x + 1
2x  y + 1 = 0
d=
2(3)  (4)  1
2 2  (1) 2
hence the distance required =
3
5
Do Basic Level Exercise Q.14,15
12. Distance between two parallel lines
d=
Ax + By + C1 = 0
d
C1  C 2
A2  B 2
Ax + By + C2 = 0
Remark : before using it, the equation of lines must have the same corresponding coefficients
of x and y.
e.g. 12.1 Find the distance between 3x + 4y + 7 = 0 and 6x + 8y + 29 = 0
Solution: First, make the corresponding coefficients of x and y being the same
3x + 4y + 7 = 0
2(3x + 4y + 7) = 2(0)
6x + 8y + 14 = 0
d=
14  29
62  82
= 1.5
Do Basic Level Exercise Q.16,17
程式版 (Equations of lines) P.
7
13. Family of line
(Type I)
“L is parallel to 2x + 3y + 4 = 0” 
Let L be 2x + 3y + k = 0
i.e. 2x + 3y + k = 0 is a family of parallel lines
In general, Ax
+ By + k = 0 is a family of parallel lines
e.g. 13.1 Write down the family of line parallel to 2x  3y + 4 = 0
Solution: 2x  3y + k = 0
e.g. 13.2 Find the equation of a line, parallel to 2x  3y + 4 = 0, passes through (1,4).
Solution: Let 2x  3y + k = 0 be the line,
∵ it passes through (1,2)
 2(1)  3(4) + k = 0
k = 10
Hence the equation is 2x  3y + 10
(Type II)
“L passes through (1,2)”
i.e.

Let L be
y2
=k
x 1
y2
= k is a family of lines passing through (1,2)
x 1
In general,
y b
k
xa
is a family of lines passing through (a,b).
Do Basic Level Exercise Q. 18,19
e.g. 13.3 Write down the family of line passing through (1,2).
Solution:
y2
=k
x 1
e.g. 13.4 Find the equation of a line passing through (1,2) with slope 3.
Solution: The required equation is
y2
=3
x 1
y  2 = 3x + 3
3x  y + 5 = 0
Do Basic Level Exercise Q. 20,21
程式版 (Equations of lines) P.
8
(Type III)
“L passes through the point of intersection of

2 x  3 y  4  0
”

3x  4 y  5  0
Let L be (2x + 3y + 4) + k(3x  4y  5) = 0
or let L be k(2x + 3y + 4) + (3x  4y  5) = 0
(2x + 3y + 4) + k(3x  4y  5) = 0 is a family of lines passing through the point of intersection
2 x  3 y  4  0
of 
.
3x  4 y  5  0
In general, L1
+ kL2 = 0 is a family of lines passing through the point of intersection of
L1 = 0 and L2 = 0
e.g. 13.5 Write down the family of straight line passing through the point of intersection of
 5 x  3 y  4  0

3x  4 y  5  0
Solution: 5x  3y  4 + k(3x + 4y + 5) = 0
e.g. 13.6 Find the equation of a line passing through the point of intersection of
 5 x  3 y  4  0
with slope 2

3x  4 y  5  0
Solution: Let the equation of the line be 5x  3y  4 + k(3x + 4y + 5) = 0
(5 + 3k)x + (3 + 4k)y + 5k  4 = 0 --- (1)
y= 
 5  3k
5k  4
x
 3  4k
 3  4k
Hence, slope = 
 5  3k
 3  4k
= 2 (given)


 5  3k
=2
 3  4k
5 + 3k = 2(3 + 4k)
11k = 11
k=1
Put into (1), the required equation is
(5 + 3(1)) + (3 + 4(1))y + 5(1)  4 = 0
2x + y + 1 = 0 or 2x  y  1 = 0.
Do Basic Level Exercise Q. 22,23,24
程式版 (Equations of lines) P.
9
14. Parallel and perpendicular
If the slope of L1 = slope of L2, then L1 // L2.
If L1 // L2 , the slope of L1 = slope of L2.
If slope of L1  slope of L2 = 1, then L1  L2.
If L1  L2, then either
slope of L1  slope of L2 = 1
or
one of L1 and L2 is parallel to the x-axis and the other is parallel to the y-axis.
e.g. 14.1 Find the equation of a line passing through (1,2) and perpendicular to 2x + 3y  4 = 0
Solution: 2x + 3y  4 = 0
y=
2
x+2
3
 slope = 
2
3
Let m be the slope of the required line,
m(
m=
2
) = 1
3
3
2
Besides, the required line passes through (1,2), hence we can let the required equation be
y2
3
=
x 1
2
3x  3 = 2y  4
3x  2y + 1= 0 is the required line
For me, I’ll let the equation be 3x  2y + k = 0 as my first step, and everything is easy.
Do Basic Level Exercise Q. 25,26,27,28
15. Intercept form of straight line
x y
 1
a b
y
b
a
x
Remark: a ,b are non-zero x and y intercepts
Do Basic Level Exercise Q. 29
程式版 (Equations of lines) P. 10