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Chabot Mathematics. §2.5 Line Eqn Point-Slope. Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu. MTH 55. 2.4. Review §. Any QUESTIONS About §’s2.4 → Slope-Intercept Eqn, Modeling Any QUESTIONS About HomeWork §’s2.4 → HW-06. The Point-Slope Equation.
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Chabot Mathematics §2.5 Line EqnPoint-Slope Bruce Mayer, PE Licensed Electrical & Mechanical EngineerBMayer@ChabotCollege.edu
MTH 55 2.4 Review § • Any QUESTIONS About • §’s2.4 → Slope-Intercept Eqn, Modeling • Any QUESTIONS About HomeWork • §’s2.4 → HW-06
The Point-Slope Equation • The equation y−y1 = m(x−x1) is called the point-slope equation for the line with slope m that contains the point (x1,y1). • Note that (x1,y1) is a KNOWN point • e.g.; (x1,y1) = (−7,11) • Sometimes (x1,y1) is called the ANCHOR Point
Point-Slope Derivation • Suppose that a line through (x1, y1) has slope m. Every other point (x, y) on the line must satisfy the equation • Because any two points can be used to find the slope. Multiply both sides by (x −x1) yielding: • which is the point-slope form of the equation of the line.
Find m & b for the line through (−1, −1) and (3, 4). Find m by y-chg divided by x-chg Use Pt-Slope Eqn and Solve for y to reveal b Example Point-Slope Eqn • Last Line to shows both m & b m b
Example Point-Slope • Write a point-slope equation for the line with slope 2/3 that contains the point (4, 9) • SOLUTION: Substitute 2/3 for m, and 4 for x1, and 9 for y1 in the Pt-Slope Eqn:
Example Pt-Slope → Slp-Inter • Write the slope-intercept equation for the line with slope 3 and point (4, 3) • SOLUTION: There are two parts to this solution. First, write an equation in point-slope form:
Example Pt-Slope → Pt-Inter • slope 3 & point (4, 3) → y = mx + b • SOLUTION: Next, we find an equivalent equation of the form y = mx + b: By Distributive Law Add 3 to Both Sides to yield Slope-Intercept Line:y = mx + b = 3x + (−9)
Graphing and Point-Slope Form When we know a line’s slope and a point that is on the line (i.e., an ANCHOR Point), we can draw the graph.
SOLUTION: Since y− 3 = 2(x − 1) is in point-slope form, we know that the line has slope 2 and passes through the point (1, 3). We plot (1, 3) and then find a second point by moving up 2 units and to the right 1 unit. Example Graph y− 3 = 2(x− 1) right 1 up 3 (1, 3)AnchorPoint • Connect the Dots to Draw the Line
SOLUTION: Find an equivalent equation: Example Graph y+3 = (−4/3)(x+2) (2, 3) down 4 • The line passes through Anchor-Pt (−2, −3) and has slope of −4/3 right 3
Parallel and Perpendicular Lines • Two lines are parallel (||) if they lie in the same plane and do not intersect no matter how far they are extended. • Two lines are perpendicular (┴) if they intersect at a right angle (i.e., 90°). E.g., if one line is vertical and another is horizontal, then they are perpendicular.
Para & Perp Lines Described • Let L1 and L2 be two distinct lines with slopes m1 and m2, respectively. Then • L1 is parallel to L2 if and only if m1 = m2and b1≠b2 • If m1 = m2. and b1 = b2 then the Lines are CoIncident • L1 is perpendicular L2 to if and only if m1•m2 = −1. • Any two Vertical or Horizontal lines are parallel • ANY horizontal line is perpendicular to ANY vertical line
Parallel Lines by Slope-Intercept • Slope-intercept form allows us to quickly determine the slope of a line by simply inspecting, or looking at, its equation. • This can be especially helpful when attempting to decide whether two lines are parallel • These Lines All Have the SAME Slope
Example Parallel Lines • Determine whether the graphs of the lines y = −2x− 3 and 8x + 4y = −6 are parallel. • SOLUTION • Solve General Equation for y • Thus the Eqns are • y = −2x− 3 • y = −2x− 3/2
Example Parallel Lines • The Eqns y = −2x− 3 & y = −2x− 3/2 show that • m1=m2 = −2 • −3 = b1≠b2 = −3/2 • Thus the LinesARE Parallel • The Graph confirmsthe Parallelism
Example ║& ┴ Lines • Find equations in general form for the lines that pass through the point (4, 5) and are (a) parallel to & (b) perpendicular to the line 2x − 3y + 4 = 0 • SOLUTION • Find the Slope by ReStating the Line Eqn in Slope-Intercept Form
Example ║& ┴ Lines • SOLUTION cont. • Thus Any line parallelto the given line must have a slope of 2/3 • Now use the GivenPoint, (4,5) in thePt-Slope Line Eqn • Thus ║- Line Eqn
Example ║& ┴ Lines • SOLUTION cont. • Any line perpendicularto the given line must have a slope of −3/2 • Now use the GivenPoint, (4,5) in thePt-Slope Line Eqn • Thus ┴ Line Eqn
Example ║& ┴ Lines • SOLUTION Graphically
Estimates & Predictions • It is possible to use line graphs to estimate real-life quantities that are not already known. To do so, we calculate the coordinates of an unknown point by using two points with known coordinates. • When the unknown point is located BETWEEN the two points, this process is called interpolation. • Sometimes a graph passing through the known points is EXTENDED to predict future values. Making predictions in this manner is called extrapolation
Example Aerobic Exercise • A person’s target heart rate is the number of beats per minute that bring the most aerobic benefit to his or her heart. The target heart rate for a 20-year-old is 150 beats per minute (bpm) and for a 60-year-old, 120 bpm • Graph the given data and calculate the target heart rate for a 46-year-old • Calculate the target heart rate for a 70-year-old
Solution a) We draw the axes and label, using a scale that will permit us to view both the given and the desired data. The given information allows us to then plot (20, 150) and (60, 120) Example Aerobic Exercise
Example Aerobic Exercise • Find the Slope of the Line • Use One Point to Write Line Equation
Solution a) To calculate the target heart rate for a 46-year-old, we sub 46 for x in the slope-intercept equation: Example Aerobic Exercise 130 46 • The graph confirms the target heart rate INterpolation
Solution b) To calculate the target heart rate for a 70-year-old, we substitute 70 for x in the slope-intercept equation: Example Aerobic Exercise 112 70 • The graph confirms the target heart rate EXtrapolation
WhiteBoard Work • Problems From §2.5 Exercise Set • PPT-example, 14, 22, 26, 40, 52, 62 • The LineEquations
Given Column Chart DropOut Rates Scatter Plot • Read Chart to Construct T-table • Use T-table to Make Scatter Plot on the next Slide
Intercept 15.2% (x1,y1) = (8yr, 14%) “Best” Line(EyeBalled)
Calc Slope from Scatter Plot Measurements DropOut Rates Scatter Plot • Thus the Linear Model for the Data in SLOPE-INTER Form • To Find Pt-Slp Form use Known-Pt from Scatter Plot • (x1,y1) = (8yr, 14%) • Read Intercept from Measurement
Thus the Linear Model for the Data in PT-SLOPE Form DropOut Rates Scatter Plot • X for 2010 → x = 2010 − 1970 = 40 • In Equation • Now use Slp-Inter Eqn to Extrapolate to DropOut-% in 2010 • The model Predicts a DropOut Rate of 9.2% in 2010
All Done for Today CommunityCollegeEnrollmentRates
Chabot Mathematics Appendix Bruce Mayer, PE Licensed Electrical & Mechanical EngineerBMayer@ChabotCollege.edu –