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# Topic 8 (Writing Equations Of A Straight Lines)

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• 1. PROPERTIES OF STRAIGHT LINES A. WRITING EQUATION OF A LINE : 1. GIVEN TWO POINTS. 2. GIVEN THE SLOPE AND A POINT B. PARALLEL AND PERPENDICULAR LINES.
• 2. Writing Equations of Lines Equations of lines come in several different forms. Two of those are: 2. Slope-intercept form y = mx + b where m is the slope and b is the y-intercept 2. General form Ax + By + C = 0 Answers will be written in either of these two forms only
• 3. Writing Equations of Lines A. Given a Point and a Slope Find the equation of the line that goes through the point (4, 5) and has a slope of 2. Solution: m = 2 x1 = 4 y1 = 5 y − y1 = m( x − x1 ) Substitute the above given to point- slope form equation a of line. y − 5 = 2( x − 4) Simplify
• 4. Slope-intercept form General Form y − 5 = 2( x − 4) y − 5 = 2( x − 4) y = 2x − 8 + 5 y − 5 = 2x − 8 y = 2x − 3 −2 x + y + 3 = 0 2x − y − 3 = 0 FINAL ANSWER
• 5. Find the equation of the line that goes through the point (-3, 2) and has a slope of -4/5. Solution: m = -4/5 x1 = -3 y1 = 2 y − y1 = m( x − x1 ) Substitute the above given to point- slope form equation a of line. 4 y − 2 = − ( x − −3) 5
• 6. Slope-intercept form General Form 4 4 2 y − 2 = − [ x − (−3)] y = − x− 5 5 5 4 5 y = −4 x − 2 y − 2 = − ( x + 3) 5 4 12 4x + 5 y + 2 = 0 y−2= − x− 5 5 4 2 y = − x− 5 5 FINAL ANSWER
• 7. Writing Equations of Lines B. Given Two Points Find the equation of the line that passes through the points (-2, 3) and (1, -6). Solution: x1 = -2 y1 = 3 x2 = 1 y2 = -6 y2 − y1 Substitute the above given to slope m= formula to find the slope. x2 − x1 −6 − 3 −9 m= m= m = −3 1+ 2 3
• 8. Slope-intercept form General Form y − y1 = m( x − x1 ) y = −3x − 3 y − 3 = −3[ x − (−2)] y − 3 = −3( x + 2) 3x + y + 3 = 0 y − 3 = −3x − 6 y = −3x − 3 FINAL ANSWER
• 9. Definitions Perpendicular Two lines that makes a 90° angle. Lines The slopes of perpendicular lines are negative reciprocal of each other . Parallel Lines Lines that never meet . The slopes of parallel lines are the same.
• 10. Writing Equations of Lines • Given a point and equation of a line parallel to it. A Find the equation of the line that passes through (1, -5) and is parallel to 4x – 2y =3. Solution: x1 = 1 y1 = -5 4x - 2y =3. Rewrite the equation to slope- intercept form to get the slope. - 2y =-4x +3. -4 3 3 y= x+ . y =2x - . m=2 -2 -2 2
• 11. Slope-intercept form General Form y − y1 = m( x − x1 ) y = 2x − 7 y − (−5) = 2( x − 1) y + 5 = 2x − 2 2x − y − 7 = 0 y = 2x − 2 − 5 y = 2x − 7 FINAL ANSWER
• 12. Writing Equations of Lines • Given a point and equation of a line perpendicular to it. B Find the equation of the line that passes through (1, -5) and is perpendicular to 4x – 2y =3. Solution: x1 = 1 y1 = -5 4x - 2y =3. Rewrite the equation to slope- intercept form to get the slope. - 2y =-4x +3. -4 3 3 y= x+ . y =2x - . m=2 -2 -2 2 m = -½
• 13. Slope-intercept form General Form y − y1 = m( x − x1 ) 1 9 y = − x− 2 2 1 y − (−5) = − ( x − 1) 2 2y = −x − 9 1 1 x + 2y + 9 = 0 y+5= − x+ 2 2 1 9 y = − x− 2 2 FINAL ANSWER
• 14. LINEAR EQUATION WITH TWO VARIABLES SOLVING SYSTEM OF EQUATION BY: 2.Graph 3.Substitution 4.Elimination 5.Cramer’s Rule
• 15. Solving Systems of Linear Equations For two-variable systems, there are then three possible types of solutions: Properties A. Independent system: one solution and 1. two distinct non-parallel lines one intersection point 2. cross at exactly one point 3. "independent" system 4. one solution at (x,y )point.
• 16. Solving Systems of Linear Equations B. Inconsistent system: Properties no solution and no intersection point. 1. two distinct parallel lines 2. never cross 3. No point of intersection 4. "inconsistent" system 5. no solution.
• 17. Solving Systems of Linear Equations Properties C. Dependent system: infinitely many solution 1. only one line. 2. same line drawn twice. 3. "intersect" at every point 4. "dependent" system, 5. Infinitely many solutions.
• 18. Methods of Solving Systems of Linear Equations
• 19. A. Systems of Linear Equations: Solving by Graphing Solve the following system by graphing. 2x – 3y = –2 4x + y = 24 Solve for y for each equation Equation 1 Equation 2 2x – 3y = –2 2x + 2 = 3y 4x + y = 24 y = (2/3)x + (2/3) y = –4x + 24
• 20. Get the ( x, y) values for both equation to facilitate easy graphing. The table below shows it x y = (2/3)x + (2/3) y = –4x + 24 –4 –8/3 + 2/3 = –6/3 = –2 16 + 24 = 40 –1 –2/3 + 2/3 = 0 4 + 24 = 28 2 4/3 + 2/3 = 6/3 = 2 –8 + 24 = 16 5 10/3 + 2/3 = 12/3 = 4 –20 + 24 = 4 8 16/3 + 2/3 = 18/3 = 6 –32 + 24 = –8
• 21. Using the table of values we can now graph and look for the intersection: y= (2/3 )x + (2/3 24 x+ solution: (x, y) = (5, 4) ) –4 y=
• 22. B. Systems of Linear Equations: Solving by Substitution Solve the following system by substitution. 2x – 3y = –2 4x + y = 24 Solution: 4x + y = 24 solve the second equation for y: y = –4x + 24 2x – 3(–4x + 24) = –2 substitute it for "y" in 2x + 12x – 72 = –2 the first equation 14x = 70 x=5 solve for x
• 23. Equation 1 Equation 2 x=5 x=5 4x + y = 24 plug this x-value back 2x – 3y = –2 into either equation, 4( 5 ) + y = 24 and solve for y 2( 5 ) – 3y = –2 20 + y = 24 10 – 3y = –2 y = 24 - 20 - 3y = –2 - 10 - 3y = - 12 y=4 y=4 Then the solution is ( x, y ) = (5, 4).
• 24. C. Systems of Linear Equations: Solving by Elimination Solve the following system using elimination. 2x + y = 9 3x – y = 16 Solution: 2x + y = 9 add down, the y's will cancel out 3x – y = 16 5x = 25 x=5 divide through to solve for x
• 25. Equation 1 Equation 2 x=5 x=5 2x + y = 9 using either of the 3x – y = 16 original equations, to 2( 5 ) + y = 9 find the value 3( 5 ) – y = 16 10 + y = 9 of y 15 – y = 16 y = 9 - 10 - y = 16 - 15 -y= 1 y = -1 y = -1 Then the solution is ( x, y ) = (5, -1).
• 26. D. Systems of Linear Equations: Solving by Cramer’s Rule Solve the following system using cramer’s rule. 2x – 3y = –2 4x + y = 24 Solution: Nx Ny x= ,y= D D D - determinant of the coefficient of the variables Nx - determinant taken from D replacing the coefficient of x Ny - and y by their corresponding constant terms leaving all other terms unchanged
• 27. 2 −3 D= D = (2) −( −12) =14 4 1 −2 −3 N x = (−2) − (−72) = 70 Nx = N x 70 24 1 x= = =5 D 14 N y = (48) − (−8) = 56 2 −2 Ny = Ny 56 4 24 y= = =4 D 14 FINAL ( 5, 4 ) ANSWER
• 28. ANSWER THE FOLLOWING PROBLEMS 1. (a) Explain why the simultaneous equations 8x – 4y = 20 and y =2x – 3 have no solution . What can you say about the straight lines representing these two equations? They are parallel • The diagram shows the graph of 2y = x - 2. The values of a and b are respectively. 2 and -1
• 29. • The graphs of x - 2y - 3 = 0 and 6 + 4y - 2x = 0 are identical lines • Find the graph of y = -2x - 1? 5. The diagram shows the graph of y = ax + b. Find the values of a and b. a = 2, b = 2
• 30. Find the solution for each system of equation using any method: 1. 5x – 2y = 0 3. y=x+3 4x + y = 13 5y + 6x = 15 Solution : Solution : ( x , y) = ( 2, 5 ) ( x , y) = ( 0, 3 ) 2. 5y = 6x – 3 4. 1 1 x + y =1 2y = x – 4 3 3 1 Solution : y =1 − x ( x , y) = ( -2, -3 ) 2 Solution : ( x , y) = ( -1, 4 )