3-6 Perpendiculars and Distance.pdf

Chapter 3, Section 6 Perpendiculars and distance

1. Splash Screen

2. Vocabulary
•equidistant

3. Concept

4. Concept

5. Example 1
Construct Distance From Point to a Line
CONSTRUCTION A certain roof
truss is designed so that the
center post extends from the
peak of the roof (point A) to the
main beam. Construct and
name the segment whose
length represents the shortest
length of wood that will be
needed to connect the peak
of the roof to the main beam.
The distance from a line to a point not on the line is the
length of the segment perpendicular to the line from
the point. Locate points R and S on the main beam
equidistant from point A.

6. Example 1
Construct Distance From Point to a Line
Locate a second point not on
the beam equidistant from R
and S. Construct AB so that
AB is perpendicular to the beam.
Answer: The measure of AB represents the shortest
length of wood needed to connect the peak of the roof
to the main beam.
___

7. Example 1
A.AD
B.AB
C.CX
D.AX
KITES Which segment represents the
shortest distance from point A to DB?

8. Step 1Find the slope of line s.
Begin by finding the slope of the
line through points (0, 0) and
(–5, 5).
COORDINATE GEOMETRY
Line s contains points at (0, 0)
and (–5, 5). Find the distance
between line s and point
V(1, 5).
Example 2
Distance from a Point to a Line on Coordinate Plane
(–5, 5)
(0, 0)
V(1, 5)

9. Example 2
Distance from a Point to a Line on Coordinate Plane
Then write the equation of this line by using the point
(0, 0) on the line.
Slope-intercept form
m = –1, (x
1
, y
1
) = (0, 0)
Simplify.
The equation of line s is y = –x.
Simplify.y = -1x + 0
Simplify.y = -x

10. Example 2
Distance from a Point to a Line on Coordinate Plane
Step 2Write an equation of the line t perpendicular
to line s through V(1, 5).
Since the slope of line s is –1, the slope of line t is 1.
Write the equation for line t through V(1, 5) with a slope
of 1.
Slope-intercept form
m = 1, (x
1
, y
1
) = (1, 5)
Simplify.
The equation of line t is y = x + 4.
Subtract 1 from each side.

11. Example 2
Distance from a Point to a Line on Coordinate Plane
Step 3Solve the system of equations to determine
the point of intersection.
line s: y = –x
line t:(+) y = x + 4
2y = 4Add the two equations.
y = 2Divide each side by 2.
Solve for x.
2 =–xSubstitute 2 for y in the first
equation.
–2 =xDivide each side by –1.
The point of intersection is (–2, 2). Let this point be Z.

12. Example 2
Distance from a Point to a Line on Coordinate Plane
Step 4Use the Distance Formula to determine the
distance between Z(–2, 2) and V(1, 5).
Distance formula
Substitution
Simplify.
Answer:The distance between the point and the line is
or about 4.24 units.

13. Example 2
COORDINATE GEOMETRY
Line n contains points (2, 4)
and (–4, –2). Find the
distance between line n and
point B(3, 1).
A.
B.
C.
D.
B(3, 1)
(2, 4)
(–4, 2)

14. Concept

15. Concept

16. Example 3
Distance Between Parallel Lines
Find the distance between the parallel lines a and b
whose equations are y = 2x + 3 and y = 2x – 1,
respectively.
You will need to solve a system
of equations to find
the endpoints of a segment
that is perpendicular to both
a and b. From their equations,
we know that the slope of line
a and line b is 2.
Sketch line p through the
y-intercept of line b, (0, –1),
perpendicular to lines a and b.
a b
p

17. Example 3
Distance Between Parallel Lines
Step 1
Use the y-intercept of line b, (0, –1), as one of the
endpoints of the perpendicular segment.
Write an equation for line p. The slope of p is the
opposite reciprocal of
Point-slope form
Simplify.
Subtract 1 from each side.

18. Example 3
Distance Between Parallel Lines
Use a system of equations to determine the point of
intersection of the lines a and p.
Step 2
Substitute 2x + 3 for y in the
second equation.
Group like terms on
each side.

19. Example 3
Distance Between Parallel Lines
Simplify on each side.
Multiply each side by.
Substitutefor x in the
equation for p.

20. Example 3
Distance Between Parallel Lines
Simplify.
The point of intersection is or (–1.6, –0.2).

21. Example 3
Distance Between Parallel Lines
Use the Distance Formula to determine the distance
between (0, –1) and (–1.6, –0.2).
Step 3
Distance Formula
x
2
= –1.6, x
1
= 0,
y
2
= –0.2, y
1
= –1
Answer: The distance between the lines is about
1.79 units.

22. Example 3
A.2.13 units
B.3.16 units
C.2.85 units
D.3 units
Find the distance between the parallel lines a and b
whose equations are and,
respectively.

23. End of the Lesson