1. 3.6-3.7 Parallel and Perpendicular Lines in the Coordinate Plane

2. Postulate 17 Slopes of Parallel Lines
In a coordinate plane, two non-vertical lines are parallel if and only if they have the same slope. Any two vertical lines are parallel.
Lines k1 and k2 have the same slope. k1 k2

3. Ex. 3 Deciding whether lines are parallel
Find the slope of each line. Is j1║j2?
M1 = 4 = 2 2
M2 = 2 = 2 1
Because the lines have the same slope, j1║j2.

4. Ex. 5: Writing an Equation of a Line
Write an equation of the line through the point (2, 3) that has a slope of 5.
y = mx + b
3 = 5(2) + b
3 = 10 + b
-7= b
Steps/Reasons why
Slope-Intercept form
Substitute 2 for x, 3 for y and 5 for m
Simplify
Subtract.

5. Write the equation Because m = - 1/3 and b = 3, an equation of n2 is
y = -1/3x + 3

6. Postulate 18: Slopes of Perpendicular Lines
In a coordinate plane, two non-vertical lines are perpendicular if and only if the product of their slopes is -1.
Vertical and horizontal lines are perpendicular

7. Ex. 1: Deciding whether lines are perpendicular
Find each slope.
Slope of j1
3-1 = - 2
Slope of j2
3-(-3) = 6 = 3
0-(-4)
Multiply the two slopes. The product of
-2 ∙ 3 = -1, so j1  j2
3 2

8. Ex.2 Deciding whether lines are perpendicular
Decide whether AC and DB are perpendicular.
Solution:
Slope of AC=
2-(-4) = 6 = 2
4 –
Slope of DB=
2-(-1) = 3 = 1
-1 –
The product of 2(-1/2) = -1; so ACDB

9. Ex.3: Deciding whether lines are perpendicular
Line h: y = ¾ x +2
The slope of line h is ¾.
Line j: y=-4/3 x – 3
The slope of line j is -4/3.
The product of ¾ and -4/3 is -1, so the lines are perpendicular.

10. Ex.4: Deciding whether lines are perpendicular
Line r: 4x+5y=2
4x + 5y = 2
5y = -4x + 2
y = -4/5 x + 2/5
Slope of line r is -4/5
Line s: 5x + 4y = 3
5x + 4y = 3
4y = -5x + 3
y = -5/4 x + 3/5
Slope of line s is -4/5
The product of the slopes is NOT -1; so r and s are NOT perpendicular.
-4 ∙ -5 = 1

11. Ex. 6: Writing the equation of a perpendicular line
The equation y = 3/2 x + 3 represents a mirror. A ray of light hits the mirror at (-2, 0). What is the equation of the line p that is perpendicular to the mirror at this point?
The mirror’s slope is 3/2, so the slope of p is -2/3. Use m =
-2/3 and (x, y) = (-2, 0) to find b.
0 = -2/3(-2) + b
0 = 4/3 + b
-4/3 = b
So, an equation for p is
y = -2/3 x – 4/3