Unit 3A Notes PDF

## Unit 3A Notes ### Slope - rise / run - m = (y2 - y1)/(x2 - x1) **Example:** Find the slope of the line that goes through the points: (3, 4) and (-2, 7) - m = (7 - 4)/(-2 - 3) - m = 3/-5 - the slope is 3/5 - Graphically: ### Parallel Lines - Slopes are the same ### Perpendicular Lines - Slop...

## Unit 3A Notes ### Slope - rise / run - m = (y2 - y1)/(x2 - x1) **Example:** Find the slope of the line that goes through the points: (3, 4) and (-2, 7) - m = (7 - 4)/(-2 - 3) - m = 3/-5 - the slope is 3/5 - Graphically: ### Parallel Lines - Slopes are the same ### Perpendicular Lines - Slopes are opposite reciprocals - Example: 2/3 → -3/2 or -2/3 → 3/2 **Example:** AB has a slope of -3. What is the slope of a line parallel? -3, perpendicular? 1/3 ### Forms of Equations #### Slope-Intercept Form - y = mx + b - m = slope - b = y-intercept (0, b) #### Point-Slope Form - y - y1 = m(x - x1) - m = slope - point: (x1, y1) **Example:** Write an equation that is parallel to the line: y = -2x + 4 and goes through (0, -5) - m = -2 (same slope) - point: (0, -5) - also y int: -5 because x is 0 - Equation #1 - y = -2x - 5 or y - (-5) = -2(x - 0) ### Perpendicular Bisector - midpoint formula: (x1 + x2)/2, (y1 + y2)/2 - midpoint - New Slope: - 3/2 **Example:** Find the equation for the perpendicular bisector of JM where J(-6, 4) and M(8, 10) - Slope of JM: m = (10 - 4)/(8 - (-6)) = 6/14 = 3/7 - Midpoint of JM: (-6 + 8)/2, (4 + 10)/2 = (1, 7) - Equation of perp. bisector: y - 7 = -3/2 (x - 1)