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