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Higher Linear Relationships Lesson 7

Higher Linear Relationships Lesson 7. Linear Relationships. 6. 0. Which of the following points does not lie on the line 2 y + 5x – 4 = 0?. (-0.8, 0) (-1,0.5) (0,2) (2,3) I don’t know. Linear Relationships. 6. 0.

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Higher Linear Relationships Lesson 7

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  1. Higher Linear Relationships Lesson 7

  2. Linear Relationships 6 0 Which of the following points does not lie on the line 2y + 5x – 4 = 0? • (-0.8, 0) • (-1,0.5) • (0,2) • (2,3) • I don’t know

  3. Linear Relationships 6 0 1 3y = 4x + 5 2 4y = 3x - 1 3 4y + 3x = 7 4 4x + 3y = 2 • Lines 1 and 2 are perpendicular • Lines 1 and 4 are parallel • Lines 2 and 4 are perpendicular • Lines 2 and 3 are parallel • I don’t know

  4. Linear Relationships 6 0 A straight line has equation 10y = 3x + 15. Which of the following is true? • The gradient is 0.3 and the y intercept is 1.5 • The gradient is 3 and the y intercept is 15 • The gradient is 15 and the y intercept is 3 • The gradient is 1.5 and the y intercept is 0.3 • I don’t know

  5. Higher Mathematics Linear Relationships Unit 1 Outcome 1 Intersecting Lines To find where two lines meet we can use simultaneous equations For this reason it is useful to write the equations in the form Ax + By = C Thursday, 13 November 2014 www.chmaths.wikispaces.com

  6. Example 1 Higher Mathematics Linear Relationships Unit 1 Outcome 1 Q(5,12) P(-3,6) K R(5,2) Intersecting Lines Example 1 Triangle PQR has vertices P(-3,6) , Q(5,12) & R(5,2). The median from Q meets the altitude from P at point K. Draw a diagram Find the equations of the median and altitude. Hence find the co-ordinates of K. Thursday, 13 November 2014 www.chmaths.wikispaces.com

  7. Higher Mathematics Linear Relationships Unit 1 Outcome 1 Q(5,12) P(-3,6) K R(5,2) Intersecting Lines Triangle PQR has vertices P(-3,6) , Q(5,12) & R(5,2). Find the equations of the median and altitude. To find the equation of the median QK y - b = m(x - a) y - 4 = 2(x - 1) Find the midpoint of PR Gradient of median QK = (12 - 4) (5-1) mQK = 2 y - 4 = 2x - 2 Using P = (-3, 6) R = (5, 2) (1,4). x -3 + 5 2 y 6 + 2 2 2x - y = -2 Midpoint of PR , K, is (1,4) To find the equation of the altitude PK Find gradient QR mQR = (12 - 2) / (5 - 5) = 10 / 0 = undefined y - b = m( x - a) y - 6 = 0(x + 3) mPK PK is zero y - 6 = 0 y = 6 www.chmaths.wikispaces.com

  8. Higher Mathematics Linear Relationships Unit 1 Outcome 1 Q(5,12) P(-3,6) K R(5,2) Intersecting Lines To find the point of intersection of K 2x - y = -2 (A) y = 6 (B) Substitute y = 6 in the top equation 2x - 6 = -2 2x = 4 x = 2 Hence K is the point (2,6) Thursday, 13 November 2014 www.chmaths.wikispaces.com

  9. Higher Mathematics Linear Relationships Unit 1 Outcome 1 F(3,3) G(6,2) E(-1,1) C Example 2 The points E(-1,1), F(3,3) and G(6,2) all lie on the circumference of a circle. Find the equations of the perpendicular bisectors of EF and FG. Hence find the co-ordinates of the centre of the circle, C. Diagram will be something like

  10. Higher Mathematics Linear Relationships Unit 1 Outcome 1 F(3,3) G(6,2) E(-1,1) C Find the equations of the perpendicular bisectors of EF and FG. Find Midpoint of EF Find Midpoint of FG Using E = (-1, 1) F = (3, 3) Using F = (3, 3) G = (6, 2) x 3 + 6 2 y 3 + 2 2 x -1 + 3 2 y 1 + 3 2 Midpoint of FG is B (4.5, 2.5) Midpoint of EF is A (1,2) mFG = (2-3)/(6-3) = -1/3 B mEF = (3-1)/(3+1) = 1/2 A Perpendicular m = -2 Perpendicular m = 3 y - b = m(x - a) y - b = m(x – a) y - 2 = -2( x - 1) y - 2.5 = 3( x - 4.5) y - 2 = -2x + 2 y - 2.5 = 3x - 13.5 2x + y = 4 3x - y = 11 A B

  11. Linear Relationships Unit 1 Outcome 1 F(3,3) G(6,2) E(-1,1) C Finding where the bisectors meet gives us the centre of the circle Solving 2x + y = 4 (A) 3x - y = 11 (B) 5x = 15 x = 3 Substituting x = 3 into A 2x + y = 4 6 + y = 4 y = -2 Hence centre of circle at (3,-2)

  12. Higher Mathematics Linear Relationships Unit 1 Outcome 1 The Equation of the straight line Intersecting Straight Lines using simultaneous equations Homework Page xx Exercise A Question xxxx Classwork Page 11 Exercise 6 Complete Thursday, 13 November 2014 www.chmaths.wikispaces.com

  13. Higher Mathematics Linear Relationships Unit 1 Outcome 1 Typical Exam Questions The Equation of the straight line y – b = m (x - a) Find the equation of the line which passes through the point (-1, 3) and is perpendicular to the line with equation Find gradient of given line: Find gradient of perpendicular: Find equation: Thursday, 13 November 2014 www.chmaths.wikispaces.com

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