View
4
Download
0
Category
Preview:
Citation preview
MOD4 L2 1
GEOMETRY
MODULE 4 LESSON 2
PARALLEL AND PERPENDICULAR LINES
OPENING EXERCISE
Plot the points 𝑂 0,0 , 𝑃 3,−1 , 𝑄 (2,3) on the coordinate plane. Determine whether 𝑂𝑃 and 𝑂𝑄
are perpendicular. Explain.
If we can show that the triangle formed be the segments is a
right triangle, we can say that 𝑂𝑃 and 𝑂𝑄 are perpendicular.
Using the distance formula, 𝑂𝑃 = 10, 𝑂𝑄 = 13, 𝑃𝑄 =
17.
10!+ 13
!≠ 17 !
So, 𝑂𝑃 and 𝑂𝑄 are not perpendicular.
CRITERIA FOR PARALLEL AND PERPENDICULAR LINES
It’s all about the slope!
• The slopes of parallel lines are equal. 𝑚! = 𝑚!
• The slopes of perpendicular lines are negative reciprocal of each other. 𝑚! = − !!!
What type of angle do perpendicular lines form? The perpendicular lines form a right (90°) angle.
TESTING FOR PERPENDICULARITY
• Find and compare the slopes of each segment/line created by the given points.
• Determine if the condition 𝑥!𝑥! + 𝑦!𝑦! = 0 is satisfied.
Note: Segments may need to be translated so that one endpoint is on the origin (0,0).
MOD4 L2 2
Apply each test to the points from the opening exercise.
• Find and compare slopes.
𝑠𝑙𝑜𝑝𝑒!" =𝑦! − 𝑦!𝑥! − 𝑥!
=3− 02− 0 =
32
𝑠𝑙𝑜𝑝𝑒!" =𝑦! − 𝑦!𝑥! − 𝑥!
=−1− 03− 0 =
−13
• Test 𝑥!𝑥! + 𝑦!𝑦! = 0
2 ∙ 3 + 3 ∙−1 = 6+ −3 ≠ 0
PRACTICE
• Are the pairs of lines parallel, perpendicular, or neither?
3𝑥 + 2𝑦 = 74 𝑎𝑛𝑑 9𝑥 − 6𝑦 = 15
Convert each equation to slope-intercept form and compare slopes.
3𝑥 + 2𝑦 = 74 → 𝑦 = −32 𝑥 + 37
9𝑥 − 6𝑦 = 15 → 𝑦 =32 𝑥 −
52
NEITHER
4𝑥 − 9𝑦 = 8 𝑎𝑛𝑑 18𝑥 + 8𝑦 = 7
4𝑥 − 9𝑦 = 8 → 𝑦 =49 𝑥 −
89
18𝑥 + 8𝑦 = 7 → 𝑦 = −94 𝑥 +
78
PERPENDICULAR
ON YOUR OWN
−4𝑥 + 5𝑦 = −35 𝑎𝑛𝑑 − 8𝑥 + 10𝑦 = 200
−4𝑥 + 5𝑦 = −35 → 𝑦 =45 𝑥 − 7
−8𝑥 + 10𝑦 = 200 → 𝑦 =45 𝑥 + 20
PARALLEL
MOD4 L2 3
PRACTICE
• Write the equation of the line passing through (−3, 4) and is perpendicular to−2𝑥 + 7𝑦 = −3.
• Write the equation of the line passing through (−3, 4) and is parallel to−2𝑥 + 7𝑦 = −3.
Convert the given equation to slope-intercept form.
𝑦 =27 𝑥 −
37
The slope of this line is !!, thus the perpendicular slope is − !
! and the parallel slope is !
!.
𝑦 − 𝑦! = 𝑚(𝑥 − 𝑥!)
𝑦 − 4 = −72 (𝑥 − (−3))
𝑦 − 4 = −72 𝑥 −
212
𝑦 = −72 𝑥 −
212 +
82
𝑦 = −72 𝑥 −
132
𝑦 − 𝑦! = 𝑚(𝑥 − 𝑥!)
𝑦 − 4 =27 (𝑥 − (−3))
𝑦 − 4 =27 𝑥 +
67
𝑦 =27 𝑥 +
67+
287
𝑦 =27 𝑥 +
347
VERTICAL AND HORIZONTAL LINES
The equation 𝑥 = 2 refers to a position on the x-axis. Note that the y-variable is not present in the
equation. Thus, y can be any value.
• The equation 𝑥 = 2 represents a vertical line.
• Its slope is undefined.
• Find an equation of a line that goes through the point
(−4, 7) and is parallel to 𝑥 = 2.
𝑥 = −4
MOD4 L2 4
The equation 𝑦 = 3 refers to a position on the y-axis. Note that the x-variable is not present in the
equation. Thus, x can be any value.
• The equation 𝑦 = 3 represents a horizontal line.
• Its slope is zero.
• Find an equation of a line that goes through the point
(−4, 7) and is parallel to 𝑦 = 3.
𝑦 = 7
• How would the x-axis be represented in an equation? 𝑦 = 0
• How would the y-axis be represented in an equation? 𝑥 = 0
• Find an equation of a line that goes through (−4, 7) and is perpendicular to 𝑥 = 2.
𝑦 = 7
Recommended