1. Chapter 3 Lesson 6 Objective: Objective: To relate slope and perpendicular lines.

2. If two nonvertical lines are perpendicular, the product of their slopes is -1. 2468 2 4 6 8 -2 -4 -6 Example 1: Checking for Perpendicular Lines Lines l 1 and l 2 are neither vertical nor horizontal. Are they perpendicular? Explain. (-2,3) (6,-3) (0,2) (-3,-2) Slope of l 1 Slope of l 2 Lines l 1 and l 2 are perpendicular because the product of their slopes is -1. (-2,3) (6,-3) x 1 y 1 x 2 y 2 (-3,-2) (0,2) x 1 y 1 x 2 y 2 Find the product of the slopes.

3. 2468 2 4 6 8 -2 -4 -6 Example 2: Checking for Perpendicular Lines Lines l 1 and l 2 are neither vertical nor horizontal. Are they perpendicular? Explain. (0,5) (3,-2) (5,5) (-4,1) Slope of l 1 Slope of l 2 Lines l 1 and l 2 are not perpendicular because the product of their slopes is not -1. (0,5) (3,-2) x 1 y 1 x 2 y 2 (-4,1) (5,5) x 1 y 1 x 2 y 2 Find the product of the slopes.

4. Example 3: Checking for Perpendicular Lines Line l 1 contains M(0,8) and N(4,-6). Line l 2 contains P(-2,9) and Q(5,7). Are they perpendicular? Explain. Slope of l 1 Slope of l 2 Lines l 1 and l 2 are not perpendicular because the product of their slopes is not -1. (0,8) (4,-6) x 1 y 1 x 2 y 2 (-2,9) (5,7) x 1 y 1 x 2 y 2 Find the product of the slopes.

5. Example 4: Writing Equations for Perpendicular Lines Write an equation for the line perpendicular to y=-3x-5 that contains (-3,7)). Slope Use point-slope form to write an equation for the new line. m y-y 1 =m(x-x 1 ) ( 1 / 3 ) y-7=( 1 / 3 )(x-(-3)) Y-7=( 1 / 3 )(x+3) x1x1x1x1 y1y1y1y1 Find the negative reciprocal for the slope. Slope=(-3) Negative Reciprocal = ( 1 / 3 )

6. Example 5: Writing Equations for Perpendicular Lines Write an equation for the line perpendicular to 5y-x=10 that contains (15,-4). Use point-slope form to write an equation for the new line. m y-y 1 =m(x-x 1 ) -5) y-(-4)=(-5)(x-15) y+4=(-5)(x-15) x1x1x1x1 y1y1y1y1 Find the negative reciprocal for the slope. Slope=( 1 / 5 ) Negative Reciprocal = -5 Get 5y-x=10 in slope-intercept form. 5y-x=10 5y=x+10 y=( 1 / 5 )x+2

7. Example 6: Writing Equations for Perpendicular Lines Write an equation for the line perpendicular to 5x+2y=1 that contains (10,0) Use point-slope form to write an equation for the new line. m y-y 1 =m(x-x 1 ) 2 / 5 ) y-(0)=( 2 / 5 )(x-10) y=( 2 / 5 )(x-10) x1x1x1x1 y1y1y1y1 Find the negative reciprocal for the slope. Slope=(- 5 / 2 ) Negative Reciprocal =( 2 / 5 ) Get 5x+2y=1 in slope-intercept form. 5x+2y=1 2y=-5x+1 y=(- 5 / 2 )x+( 1 / 2 )

8. Homework pg.161-163 #16-30;35;38