Higher Linear Relationships Lesson 7

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

Linear Relationships 1 3y = 4x + 5 2 4y = 3x - 1 3 4y + 3x = 7 4 4x + 3y = 2 6 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 1 2 3 4 5 6

The gradient is 0.3 and the y intercept is 1.5 Linear Relationships A straight line has equation 10y = 3x + 15. Which of the following is true? 6 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 1 2 3 4 5 6

To find where two lines meet we can use simultaneous equations 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 Tuesday, 03 July 2018 www.chmaths.wikispaces.com

Triangle PQR has vertices P(-3,6) , Q(5,12) & R(5,2). Example 1 Higher Mathematics Linear Relationships Unit 1 Outcome 1 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 P(-3,6) Q(5,12) R(5,2) K Find the equations of the median and altitude. Hence find the co-ordinates of K. Tuesday, 03 July 2018 www.chmaths.wikispaces.com

Triangle PQR has vertices P(-3,6) , Q(5,12) & R(5,2). Higher Mathematics Linear Relationships Unit 1 Outcome 1 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 Q(5,12) K P(-3,6) R(5,2) 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

To find the point of intersection of K Higher Mathematics Linear Relationships Unit 1 Outcome 1 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 P(-3,6) Q(5,12) R(5,2) K 2x = 4 x = 2 Hence K is the point (2,6) Tuesday, 03 July 2018 www.chmaths.wikispaces.com

Higher Mathematics Linear Relationships Unit 1 Outcome 1 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. E(-1,1) F(3,3) G(6,2) C Hence find the co-ordinates of the centre of the circle, C. Diagram will be something like

A B Higher Mathematics Linear Relationships Unit 1 Outcome 1 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 E(-1,1) F(3,3) G(6,2) C 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

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

Intersecting Straight Lines using simultaneous equations 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 Tuesday, 03 July 2018 www.chmaths.wikispaces.com

Typical Exam Questions Higher Mathematics Linear Relationships Unit 1 Outcome 1 The Equation of the straight line y – b = m (x - a) Typical Exam Questions 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: Tuesday, 03 July 2018 www.chmaths.wikispaces.com