Transcript 5_6 Parallel-Perpendicular Lines

5.6 Parallel and Perpendicular Lines:
Parallel lines ( || ): Lines in the same plane
with the same slope that never intersect.
Perpendicular lines ( ): Lines that
intersect to form right angles and have
opposite-inverse (opposite reciprocal)
slopes.
Opposite Reciprocals: Two numbers
whose product is -1.
GOAL:
Additional information we must know:
Parallel Lines:
Lines that have same slope and different y-intercept
y=
𝟏
x
𝟐
+1
and
y=
𝟏
x
𝟐
-2
Additional information we must know:
Perpendicular Lines:
Lines that have opposite and inverse slopes
(opposite reciprocal slopes)
𝟏
y = -2x + 1
and
y = x -2
𝟐
Checking:
Looking at their
slopes we see that:
-2(1/2) = -1
WRITING AN EQUATION OF A PARALLEL LINE:
Remember: Parallel lines we must have the
same slope and different y-intercepts.
Ex:
A line passes through (-3, -1) and is
parallel to the graph of y = 2x +3. What
equation represents the line in
slope-intercept form?
SOLUTION: Looking at the given info we have
the following:
A line passes through (-3, -1) and is parallel to
the graph of y = 2x +3. What equation
represents the line in
slope-intercept form?
Point: (-3, -1)
Slope: 2
Using point-slope {𝒚- 𝒚𝟏 = m(𝒙-𝒙𝟏)} form
equation we get:
𝒚-−𝟏 = 2(𝒙--3)
𝒚 + 𝟏 = 2(𝒙 +3)
𝒚 + 𝟏 = 2𝒙 + 𝟔)
𝒚= 2x+5
YOU TRY IT:
What equation represents the line in
slope-intercept form of a line passing
through (-2, 1) and parallel to 2x+3y = 6
SOLUTION: Looking at the given info we have
the following:
What equation represents the line in
slope-intercept form of a line passing
through (-2, 1) and parallel to 2x+3y = 6
Finding the slope:  y = Point: (-2, 1)
𝟐
x
𝟑
+2
Slope: -
𝟐
𝟑
Using point-slope {𝒚- 𝒚𝟏 = m(𝒙-𝒙𝟏)} form equation
we get:
𝟐
𝒚-𝟏 = - (𝒙--2)
𝒚-1 = -
𝟑
𝟐
(𝒙
𝟑
+ 𝟐) 
𝟐
𝟑
𝒚= - x -
𝟏
𝟑
WRITING AN EQUATION OF PERPENDICULAR
LINES:
Remember: Perpendicular lines must have
opposite and inverse slopes.
Ex:
Write an equation of the line that
passes through (2, 4) and is perpendicular
𝟏
to y = x – 1.
𝟑
SOLUTION: Looking at the given info we have
the following:
Write an equation of the line that passes
𝟏
through (2, 4) and is perpendicular to y = x – 1.
𝟑
Point: (2, 4)
𝟏
𝟑
Slope:  opposite inverse slope: -3
Using point-slope {𝒚- 𝒚𝟏 = m(𝒙-𝒙𝟏)} form
equation we get:
𝒚 - 𝟒 = -3(𝒙-2)
𝒚 - 4 = -3(𝒙 - 2) 
𝒚= -3x+10
YOU TRY IT:
What equation represents the line in
slope-intercept form of a line passing
through (-2, 1) and perpendicular
to 2x+3y = 6
SOLUTION: Looking at the given info we have
the following:
What equation represents the line in
slope-intercept form of a line passing
through (-2, 1) and perpendicular to 2x+3y = 6
Finding the slope:  y = 𝟐
𝟑
𝟐
x
𝟑
+2
Point: (-2, 1) Slope: - opposite inverse
𝟑
𝟐
Using point-slope {𝒚- 𝒚𝟏 = m(𝒙-𝒙𝟏)} form equation
we get:
𝟑
𝒚-𝟏 = (𝒙--2)
𝒚-1 =
𝟐
𝟑
(𝒙
𝟐
+ 𝟐) 
𝟑
𝟐
𝒚= x + 4
CLASSIFYING LINES:
To classify a pair of lines we must look at
their slope and y-intercepts thus we must
write the equations in slope-intercept form
(y=mx+b).
MUST ALWAYS REMEMBER:
* Parallel lines must have the same slope
and different y-intercepts.
*Perpendicular lines must have opposite and
inverse slopes.
CLASSIFYING LINES:
Ex:
Decide if the given equations are
parallel, perpendicular or neither? Explain.
4y = -5x + 12
and
5x + 4y = -8
SOLUTION:
Writing the equations in slope-intercept
form to compare we have:
1) 4y = -5x + 12
y=-
𝟓
x
𝟒
+3
2) 5x + 4y = -8
y=-
𝟓
x
𝟒
-2
Divide by 4
Slope: -
𝟓
,
𝟒
y-intercept: 3
Subtract 5x and divide by 4
Slope: -
𝟓
,
𝟒
y-intercept: - 2
Slopes are equal, and y-intercepts different,
thus we have PARALLEL LINES.
YOU TRY IT:
Decide if the given equations are
parallel, perpendicular or neither? Explain.
y=
𝟑
x
𝟒
+7
and
4x - 3y = 9
YOU TRY IT (SOLUTION):
Writing the equations in slope-intercept
form to compare we have:
y=
𝟑
x
𝟒
+7
Already in y=mx+b form
Slope:
2) 4x - 3y = 9
y=
𝟒
x
𝟑
-3
𝟑
,
𝟒
y-intercept: 7
Subtract 4x and divide by - 3
Slope:
𝟒
,
𝟑
y-intercept: - 3
Since the slopes are not equal or opposite
reciprocals, the lines are neither.
SOLVING REAL-WORLD PROBLEMS
In real-world situation we keep on using
linear equations to provide important
information that will aid us in making
important decisions.
EX:
An architect uses
software to design the ceiling
of a room. The architect needs
to enter and equation that
represents a new beam. The
new beam will be
perpendicular to the existing
beam, which is represented
the red line. The new beam
will pass through the corner
represented by the blue dot.
What is an equation of the
new beam?
SOLUTION:
1) Find the slope of the red line:
Using A(3,6)and B(6,4) we find slope:
𝟒−𝟔
𝟔−𝟑
2) Find the opposite reciprocal slope of
3) Use the slope =
𝟑
𝟐
=
𝟐
−
𝟑

and the point ( 12, 10)
In 𝒚- 𝒚𝟏 = m(𝒙-𝒙𝟏) to get:
𝟑
𝟐
𝒚- 𝟏𝟎 = (𝒙-𝟏𝟐)
𝟑
𝟐
𝒚- 𝟏𝟎 = 𝒙-𝟏𝟖 →
𝟑
𝟐
𝟐
−
𝟑
𝒚 = 𝒙-𝟖
𝟑
𝟐
VIDEOS:
Parallel/Perpendicular
https://www.khanacademy.org/math/algebra/line
ar-equations-and-inequalitie/more-analyticgeometry/v/equations-of-parallel-andperpendicular-lines
https://www.khanacademy.org/math/algebra/line
ar-equations-and-inequalitie/more-analyticgeometry/v/perpendicular-line-slope
CLASSWORK:
Page 330-332
Problems: As many as
needed to master
the concept