3.6-3.7 Parallel and Perpendicular Lines in the Coordinate Plane

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

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.

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.

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

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

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

Ex.2 Deciding whether lines are perpendicular Decide whether AC and DB are perpendicular. Solution: Slope of AC= 2-(-4) = 6 = 2 4 – 1 3 Slope of DB= 2-(-1) = 3 = 1 -1 – 5 -6 -2 The product of 2(-1/2) = -1; so ACDB

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.

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 5 4

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