Chapter 10 Topics in Analytical Geometry
10.1 Line
Objective: students will be able to find the inclination of a line, the angle between two lines, and the distance between a
point and a line.
I. Inclination of a line (p. 728-729)
The inclination of a non-horizontal line is the positive angle  (less than ) measured counterclockwise
from the x-axis to the line.
The following are examples of inclination of a line for  = 0, /2, and other angles (horizontal line, vertical
line, acute angle, and obtuse angle)
 = /2
 = 0 Horizontal Line
 = /2 Vertical Line
 is Acute Angle
 is Obtuse Angle
 = 0 Horizontal line
Inclination and slope – if a non-vertical line has inclination  and slope m, then 𝒎 =
; therefore
the
 = 0𝒕𝒂𝒏
Horizontal
line
inclination can be determined by using the slope: 𝜃 = tan−1 𝑚
slope = m =
e
y2
Rise
=
= tan 
Run
x 2 - x1
Example 1: find the inclination of the line given 2 x  y 5
Solution: The slope m = -2, then tan  = -2. Thus  = tan−1 (−2)  - 1.0171  2.0344 or 63.435
1
Now try this yourself: Find the inclination of the line y  x  5 ; answer is   26.565
2
II. The angle between two lines
If two non-perpendicular lines have slopes m1, and m2, the angle between the two lines is
tan   m
1
1
tan 
2
 m
2
By External Angle Theorem:  2     1 ( where  2   1 )
thus tan  
tan  2  tan 1
1 tan 1 tan  2
If we rewrite the equation by using slope and inclination relation,
m1  m2
1 m1m2
Example 2: Find the angle between the lines given by 3 x28 and 4 x 5 y 1
therefor tan  
Solution: m1 = -3/2, m2 = 4/5 thus tan  
4/5( 3/2)
1 (4/5)( 3/2)
a
(One easy way to find the slope of a given line ax by c 0 , one easy way is m )
b
III. The Distance Between a point and a line
The distance d between the point (x1, y1) and the line is Ax  By C 0 is defined as d =
(note that if (x1, y1) is on the line, then d = 0)
Example 3: Find the distance between the point (2, 3) and the line 𝑥 − 4𝑦 = 0
Solution: d 
(2) 4(3)
12 ( 4)2

10 10 17

17
17
Ax1  By1 C
A2  B 2
IV. Application of the distance formula between a point and a line
Example 4: Given a triangle with vertices A(-1, 2), B(2, 3), and C(1, -1), (a) find the altitude from B to line
segment formed by vertices A and C and (b) the area of the triangle ABC
Solution:
(a) The altitude of from B to the line segment AC is the distance
from B to the line AC (i.e. you draw a perpendicular line
from B to AC, that distance is altitude) we need to find the
equation for the line AC first.
The line segment AC has the equation : 3 x  2 y 10
Thus the altitude (distance from B to AC) is
h
32 231
32  22
 13
1
(b) The area of the triangle ABC is bh , the length (distance)
2
b  ( x1  x2 )2  ( y1  y2 )2
between point A and C is
 ( 11) 2  (2  ( 1) 2  13
1
13
Thus the area is
13 13 
2
2