2. HIGHER MATHEMATICS Unit 1 - Outcome 1 The Straight Line

3. THE DISTANCE FORMULA A is the point ( x 1, y 1 ) and B is the point ( x 2, y 2 ) By Pythagoras’ Theorem, AB 2 = AC 2 + CB 2 AB 2 = (x 2 – x 1 ) 2 + ( y 2 – y 1 ) 2 AB =  [(x 2 – x 1 ) 2 + ( y 2 – y 1 ) 2 ] A B C y x O

4. EXAMPLE 1 Calculate the length of the line joining the points A (3,-4) and B (-5,7). x y O A B AB =  ( 64 + 121 ) AB =  (185) AB = 13.6 units AB =  [( -5 – 3 ) 2 + ( 7 – (-4)) 2 ]

5. EXAMPLE 2 Prove that the triangle with vertices at A (-7,9),B (3,13) and C (2,1) is isosceles. ISOSCELES - TWO EQUAL SIDES x y A B C AC =  [(-7 - 2 ) 2 + ( 9 -1 ) 2 ] AC =  (81 + 64 ) AC =  145 BC =  [( 3 - 2 ) 2 + ( 13 -1) 2 ] BC =  ( 1 + 144) BC =  145 AC = BC,so triangle ABC is isosceles.

6. GRADIENT OF A STRAIGHT LINE Gradient = Change in y Change in x m AB = y 2 – y 1 x 2 – x 1 O x y A ( x 1, y 1 ) B ( x 2, y 2 )  Also,tan  = = m AB Opposite Adjacent y 2 – y 1 x 2 – x 1 =

7. SUMMARY m AB = y 2 – y 1 x 2 – x 1 m AB = tan  If lines are parallel, then their gradients will be equal. where  is the angle between the line and the positive direction of the x-axis.

8. x LINE SLOPES UP FROM LEFT TO RIGHT tan  is POSITIVE GRADIENT IS POSITIVE LINE IS HORIZONTAL  = 0 tan  = 0 GRADIENT IS ZERO LINE SLOPES DOWN FROM LEFT TO RIGHT tan  is NEGATIVE GRADIENT IS NEGATIVE LINE IS VERTICAL  = 90 tan  NOT DEFINED GRADIENT IS NOT DEFINED 

9. EXAMPLE 1 Calculate the gradient of the line joining the points P (-4,-7) and Q (6,8). Calculate the angle the line makes with the positive direction of the x-axis. O x y P Q  m PQ = y 2 – y 1 x 2 – x 1 8 – (-7 ) 6 – (–4 ) == 15 10 = 3 2 tan  = 3 2  = tan -1 3 2 = 56.3 o

10. EXAMPLE 2 Show that the points A (8,6), B (–1,3) and C (-4,2) are COLLINEAR. COLLINEAR Points lie on the same straight line m AB = y 2 – y 1 x 2 – x 1 = 3 - 6 -1 - 8 = -3 -9 = 1 3 m BC = y 2 – y 1 x 2 – x 1 = 2 - 3 -4 –(-1) = -3 = 1 3

11. m AB = m BC So AB and BC are PARALLEL But B is a common point So A, B and C are collinear. C B B A continuedcontinued

12. PERPENDICULAR LINES O x y A(a, b ) B(-b, a ) Line OA has been rotated 90 o anti-clockwise to OB. m OA = b - 0 a - 0 b a = m OB = a - 0 -b - 0 = a -b

13. m OA x m OB = b a x a -b = -1 SUMMARY If m 1 x m 2 =-1,then the lines are PERPENDICULAR. If the lines are PERPENDICULAR, then m 1 x m 2 =-1 continued

14. m1m1 m2m2 2-½ ¼-4 -¾ 4 3 m 1 and m 2 are the RECIPROCALS of each other ( with the sign changed ). You need only remember this simple rule : flip it and change the sign. continued

15. EXAMPLE 1 Write down the gradient of the lines PERPENDICULAR to the following lines. y = 3x + 5 3 m PERP. = 3 1 y = -4x + 5 -4 m PERP. = 4 1 6y = 5x + 2 m PERP. = 5 6 m = 5 6

16. EXAMPLE 2 Prove that the triangle with vertices at A (4,1), B (7,5) and C (0,4) is right - angled. x y O A B C m AC = 4 - 1 0 - 4 = 3 -4 m AB = 5 - 1 7 - 4 = 4 3

17. continued m AB xm AC = So AB and AC are perpendicular to each other Therefore triangle ABC is right- angled at A.

18. EQUATION OF A STRAIGHT LINE x y O A B A is the point ( 0, c ) and B is ( x, y ) m AB = y - c x - 0 y - c x CROSS MULTIPLY y - c = mx y = mx + c GRADIENT Y-INTERCEPT

19. For example, A line with gradient –5 which passes through the point ( 0, 4 ) has equation … y = -5x + 4 A line with gradient 3 which passes through the point ( 0, -2 ) has equation … y = 3x - 2 A line with gradient -1 which passes through the point ( 0, 0 ) has equation … y = -x continued

20. EXAMPLE Find the equation of the line which passes through the points (0,-5) and (-3,1 ). x y O m = 1 - (-5 ) -3 - 0 = 6 -3 = -2 Equation of line is y = -2x - 5

21. Equation Of A Straight Line Part 2 x y O A B A is the point ( a, b ) and B is the point ( x, y ) m AB = y - b x - a CROSS MULTIPLY y - b = m ( x - a ) Straight Line, Gradient m, passes through the point ( a, b )

22. To find the equation of the line now, we need only know the GRADIENT and ANY OTHER POINT - we DO NOT require the y - intercept. This version is much better than y = mx + c, because often it is difficult ( and unnecessary ) to write the equation in this form.

23. EXAMPLE 1 Find the equation of the line joining the points A ( 2, 3 ) and B ( 6, 6 ). m AB = 6 - 3 6 - 2 = 3 4 We now use y - b = m ( x - a ). We can use either point A or point B when we substitute numbers in place of a and b.

24. y - b = m ( x - a ) Use A ( 2, 3 ) y - 3 = ( x - 2 ) 3 4 x 4 4y - 12 = 3 ( x - 2 )4y - 12 = 3x - 64y – 3x = 6

25. EXAMPLE 2 Find the equation of the straight line parallel to the line 2x + y + 3 = 0, and which passes through the point ( 3, 4 ) 2x + y + 3 = 0 y = -2x - 3 m = -2 y - b = m ( x - a ) y - 4 = -2 ( x - 3 ) y - 4 = -2x + 6 y + 2x = 10

26. MEDIANS AND ALTITUDES MEDIAN Straight line from one vertex (corner) of a triangle to the MIDPOINT of the opposite side. ALTITUDEStraight line from one vertex (corner) of a triangle PERPENDICULAR to the opposite side.

27. O y x A B C P P is midpoint of BC Median AP CQ at right angles to AB Q Altitude CQ

28. EXAMPLE Triangle PQR has vertices at P (4,-4), Q (11,10) and R (-9,6). The median PS and the altitude RT intersect at K. Find the coordinates of K. K x y O P Q R S T

29. Median PS Midpoint of QR = S ( 1, 8 ) = m PS = 8 - (-4) 1 - 4 -4 y - b = m ( x - a ) y - 8 = -4 ( x - 1 ) y - 8 = -4x + 4 y + 4x = 12 Altitude RT = m PQ = 10 - (-4) 11 - 4 2 SO m RT = 1 2 - y - b = m ( x - a ) y - 6 =- ( x + 9 ) 1 2 2y - 12 = -x - 9 2y + x = 3

30. Coordinates of K 7x = 21 2y + x = 3 x 2 2y + 8x = 24 2y + x = 3 SUBTRACT y + 4x = 12 x = 3 y = 0 Therefore K is ( 3, 0 )

31. PERPENDICULAR BISECTOR A straight line which BISECTS ( cuts in half ) another straight line and which is at RIGHT ANGLES to the line. LINE BISECTOR Perpendicular Bisectors

32. EXAMPLE Find the equation of the perpendicular bisector of the line PQ, where P is ( 7, -12 ) and Q is ( -3, 4 ). P Q Midpoint of PQ= ( 2, -4 ) m PQ = -12 - 4 7 - (-3) = -16 10 = -8 5 SO m perp. = 5 8

33. Equation Of Perpendicular Bisector y - b = m ( x - a ) Use A ( 2, -4 ) y + 4 = ( x - 2 ) 5 8 x 8 8y + 32 = 5 ( x - 2 )8y + 32 = 5x - 108y – 5x = -42