Advanced Topic for 9th & 10th only
Chapter 3-6
Perpendiculars and
Distance
Distance between a Point and a Line:
The distance between a point and a line, is the
length of the segment perpendicular to the line
from the point.
C
A
Shortest
distance
B
Which segment in the diagram
represents the distance from R to
XY?
A.
B.
C.
D.
___
RY
A
___
RX
B
___
MX
C
___
RM
D
Equidistant: same distance.
Theorem:
In a plane if two lines are equidistant from
a third line, then the two lines are parallel
to each other.
d
d
a
b
c
If the distance
between line a
and b is d and
distance between
b and c is d
then a and c are
Parallel.
Find the distance between the parallel lines
1
1
1
y
x  3 and y 
x
3
3
3
Graph the original two equations.
1
1 1
y  x  3 and y  x 
3
3
3
9
8
7
6
5
4
3
2
1
-9 -8
-7 -6 -5 -4 -3 -2
-1
1
-1
-2
-3
-4
-5
-6
-7
-8
-9
2
3
4
5
6
7
8
9
Use y  y1  m( x  x1 ) to find the equation of
the line perpendicular to the original two equations.
Use one of the y intercepts of the original equations.
y  y1  m( x  x1 )
9
8
y  (3)  3( x  0)
7
6
5
y  3x  3
4
3
2
1
So the equation of the green
line is y  3 x  3
-9 -8
-7 -6 -5 -4 -3 -2
-1
1
-1
-2
-3
-4
-5
-6
-7
-8
-9
2
3
4
5
6
7
8
9
Use system of equations to determine where the
green line intersects the top blue equation.
y  3x  3
1
1
y
x
3
3
9
8
7
6
5
1
1
x
=
3
3
4
3x  3
1
1
 x  3 x  3 
3
3
10
10

x
3
3
x 1
3
2
1
-9 -8
-7 -6 -5 -4 -3 -2
-1
1
-1
-2
-3
-4
-5
-6
-7
-8
-9
2
3
4
5
6
7
8
9
Now you know that at x=1 the green graph
crosses the graph on top, plug in x=1 into the
equation of the green line.
y  3x  3
y  3(1)  3
y0
The intersection point is (1,0)
Now use the distance formula: d  ( x2  x1 )  ( y2  y1 )
Between points (0,-3) and (1,0).
2
d  ( x2  x1 )  ( y2  y1 )
2
2
9
8
7
d  (0  1)  (3  0)
2
6
2
5
4
3
d  10
2
1
-9 -8
-7 -6 -5 -4 -3 -2
-1
1
-1
-2
-3
-4
-5
-6
-7
-8
-9
2
3
4
5
6
7
8
9
2
Homework
• Textbook pages 185 – 187,
• problems 1, 4 – 7, 10 – 18 evens,
• and 36 – 42 evens.