Solving Problems Involving Lines and Points

1. Solving Problems Involving Lines and Points

2. The Distance Between Two Points • Coordinate geometry describes shapes using the coordinates of points. It’s the study of geometric problems using algebraic methods. • A sketch will often help you understand the problem, but always show your working

3. The Distance Between Two Points • The distance between A (3,2) and B (7,2) is the distance between the x coordinates. Length AB = 7 – 3 = 4 This is the change in x.

4. The Distance Between Two Points • The distance between C (4,2) and D (4,5) is the distance between the y coordinates. Distance from C to D = 5 – 2 = 3 This is the change in y.

5. To find the distance between points on other straight lines we use the theorem of Pythagoras Find the distance between A (2,1) and B (6,4) • Step 1: Draw a diagram • Mark A and B and form a right angled triangle ABT • Step 2: Find the lengths of AT and BT • Length AT = change in x = 6 – 2 = 4 • Length BT = change in y = 4 – 1 = 3 • Step 3: Apply the theorem of Pythagoras = AT2 + BT2 = 42 + 32 = 25 so AB = 5 • So the distance between A and B is 5 units. B (6,4) A (2,1) T

6. What is the length of the line segment joining P (-3,4) and Q (2,-5)? P • Step 1: Draw a diagram • Step 2: Find the lengths of the sides • Length QT = 2 – -3 = 5 • Length PT = 4 – -5 = 9 • Step 3: Apply the theorem of Pythagoras PQ2 = QT2 + PT2 = 25 + 81 = 106 so PQ = 10.29563 • The length of PQ is 10.3 units (1dp). Q T

7. Find the distance between the following pairs of points • A (0,5) and B (0,2) • C (-1,1) and D (-3,6) • E (-2,6) and F (8,6) • G (1,6) and H (5,3) • I (-6,-4) and J (1,3) • M (-4,5) and N (3,2) 3 5 10 5 7√2 or 9.9 √58 or 7.62

8. The Formula for the distance between two points Two general points A (x1, y1) and B (x2, y2) • Step 1: The diagram • Step 2: The lengths of the sides length AT = change in x = x2 – x1 length BT = change in y = y2 – y1 • Step 3: by Pythagoras AB2 = AT2 + BT2 = (x2 – x1)2 + (y2 – y1)2 So AB = √ (x2 – x1)2 + (y2 – y1)2 B (x2, y2) T A (x1, y1)

9. Find the distance between A (3,-2) and B (-4,-1) • Let A (3,-2) be (x1,y1) then x1=3 & y1=-2 • If B (-4,-1) is (x2,y2) then x2=-4 & y2=-1 • Substitute into the formula AB = √ (x2 – x1)2 + (y2 – y1)2 = √ (-4 – 3)2 + (-1 – -2)2 = √ (-7)2 + 12 = √ 50 = 7.0710678 = 7.07 (3 sf)

10. Find the distance between • (1,2) and (5,7) • (-2,-3) and (-6,-8) • By finding the lengths of each side of the triangle joining the points A (-3,4), B (3,5) and C (2,-1), decide whether triangle ABC is scalene, isosceles or equilateral. • Use the theorem of Pythagoras to show that the triangle joining the points P (2,7), Q (4,-1) and R (-2,6) is a right angled triangle.

11. 1. • = √ (5 – 1)2 + (7 – 2)2 = √ 16 + 25 = √ 41 = 6.403 • = √ (-6 – -2)2 + (-8 – -3)2 = √ 16 + 25 = √ 41 = 6.403

12. 2. A (-3,4) B (3,5) Length AB = √ (3 – -3)2 + (5 – 4)2 = √ 37 = 6.083 Length BC = √ (3 – 2)2 + (5 – -1)2 = √ 37 = 6.083 Length AC = √ (-3 – 2)2 + (4 – -1)2 = √50 = 7.071 Since length AB = length BC, ABC is isosceles C (2,-1)

13. P (2,7) R (-2,6) Length PQ = √ (4 – 2)2 + (-1 – 7)2 = √ 68 Length RP = √ (2 – -2)2 + (7 – 6)2 = √ 17 Length RQ = √ (4 – -2)2 + (-1 – 6)2 = √85 If RP2 + PQ2 = RQ2, then PRQ is right angled. RP2 + PQ2 = (√17)2 + (√68)2 = 17 + 68 = 85 = RQ2 So PQR is a right angled triangle Q (4,-1)

14. Gradients All the lines in the graph have different slopes The gradient of the line is defined as: Gradient = change in y change in x = y2 – y1 x2 – x1

15. Find the gradient of the line joining P(1,2) and Q(6,4) Let P (1,2) be the point (x1,y1) then x1 = 1 and y1 = 2. If Q (6,4) is (x2,y2) then x2 = 6 and y2 = 4. So gradient of PQ = y2 – y1 x2 – x1 = 4 – 2 6 – 1 = 2 5 Q P

16. Find the gradient of the line joining S(-2,3) and T(4,-5) Let S(-2,3) be the point (x1,y2) then x1 = -2 and y1 = 3. If T(4,-5) is (x2,y2) then x2 = 4 and y2 = -5. So gradient of ST = y2 – y1 x2 – x1 = -5 – 3 4 – -2 = -8 6 S T The gradient of ST is -4/3.

17. Remember • Gradient = y2 – y1 x2 – x1 • The gradient of any horizontal line is zero. gradient = 2 – 2 5 – 1 = 0 4 • The gradient of any vertical line is undefined. gradient = 4 – 1 2 – 2 = 3 0

18. Find the gradients of the line segments joining the following pairs of points • A(2,3) B(6,7) • C(0,4) D(3,-2) • E(-5,6) F(2,-1) • G(3,0) H(-4,0) • I(-3,4) J(-6,8) • K(5,-1) L(5,2) 1 -2 -1 0 -4/3 The gradient is undefined

19. Gradients and Angles • If you know the gradient of a straight line, you can find the angle the line makes with the positive direction of the x axis.

20. Find tan θ, and then θ in the triangle shown. C (7,4) θ B (7,1) A (2,1) Tan θ = BC AB = 3 5 = 0.6 θ = 30.963756530 So θ = 310 (2 sf)

21. Remember • The gradient of a line is the tangent of the angle the line makes with the positive direction of the x axis.

22. What is the gradient of the line which meets the x axis at 500? • Gradient = tan θ = tan 500 = 1.1917536 So the gradient of the line is 1.19 (2dp) 500

23. At what angle does the line with gradient 4 meet the x axis? • Gradient = tan θ 4 = tan θ θ = 75.963757 So the line meets the x axis at 760 (2 sf)

24. Find the gradient of the line which meets the x axis at 350 • The gradient of PQ is 0.5. At what angle does PQ meet the x axis? • Find the angle which the line joining A (0,-8) and • B (3,-7) makes with the positive direction of the x axis. Gradient = tan 350 = 0.7002 (4sf) Tan θ = 0.5 θ = 26.60 (1dp) The line PQ meets the x axis at 26.60 x Q P x

25. Find the angle which the line joining A (0,8) and B (3,-7) makes with the positive direction of the x axis. Gradient AB = -7 - -8 3 – 0 = 1 3 If angle between AB and x axis is θ then tan θ = 0.3333 θ = 18.40

26. Finding the Mid-point of a Line Segment • The coordinates of the midpoint M of the line segment joining the points (x1,y1) and (x2,y2) are found using the formula: M = ( x1 + x2 ,y1 + y2 ) 2 2 x (x1,y1) midpoint x (x2,y2)

27. Find the midpoint of the line segment joining P(-3,4) and Q(-1,-5) Midpoint = , = ( , ) = (-2 , )

28. Find the midpoints of the line segments joining the following points: • a) (1,5) and (7,3) b) (-2,-4) and (4,-10) • 2. A (-4,1) , B (2,5) and C (4,-3) are the vertices of a triangle. • P is the midpoint of AB. • Q is the midpoint of BC. • Find the coordinates of P. • Find the coordinates of Q. • Find the length of PQ. • Find the length of AC. (4,4) (1,-7) B(2,5) P A(-4,1) Q C(4,-3)

29. P is (-1,3) b) Q is (3,1) • c) Length PQ = √(3 - -1)2 + (1 – 3)2 • = √ 20 • = 4.47 (2dp) • d) Length AC = √(4 - -4)2 + (-3 – 1)2 • = √ 80 • = 8.94 (2dp) B(2,5) P A(-4,1) Q C(4,-3)

30. Collinear Points If points lie on the same straight line we say they are collinear. It’s possible to decide whether points are collinear or not by finding the gradients of the line segments connecting the points x A, B and C are collinear x A x B C Q x P x R x P, Q and R are not collinear

31. Test for collinear points If A, B and C are collinear, then the gradient AB = gradient BC or gradient AB = gradient AC x C x x B A

32. Do the points A (-4, -17), B (-1 ,-7) and C (6, 13) lie on the same straight line? Gradient AB = yb – ya Gradient BC = yc - yb xb – xa ` xc – xb = -7 - -17 = 13 - -7 -1 - -4 6 - -1 = 10 = 20 3 7 Gradient AB ≠ gradient BC, A, B and C do not lie on the same straight line. y 10 -10 x -5 5

33. If A (-3, 11), B (1,3) and C (6, m) are collinear, find m. Gradient AB = yb – yaso -2 = yc - yb xb – xa ` xc – xb = 3 - 11 -2 = m - 3 1 - -3 5 = -8 -10 = m - 3 4 m = -7 = -2 Since A, B and C are C is the point (6, -7) collinear, gradient BC = -2 A B C

34. Show the points P (2, -7), Q (-1, -5) and R (-4, -3) are collinear. • 2. If K (3, -7), L (-1, a) and M (1, -1) are collinear, find a. • 3. Does the midpoint of M of the line segment joining A (6, 0) and B (-2, -2) lie on the same line as the points P (4, -2) and Q (-3, -1)?

35. 1. Show the points P (2, -7), Q (-1, -5) and R (-4, -3) are collinear. Gradient PQ = -7 - -5 Gradient QR = -5 - -3 2 - -1 -1 - -4 = -2 = -2 3 3 Since gradient PQ equals gradient QR, P, Q and R are collinear. 2. If K (3, -7), L (-1, a) and M (1, -1) are collinear, find a. Gradient KM = -7 - -1 Gradient LM = a - -1 3 - 1 -1 - 1 = -3 = a + 1 -2 If K, L and M are collinear, a + 1 = -3 -2 a = 5

36. Finding equations of straight lines • You may be given • A graph, so you can find the y intercept and gradient • The gradient of the line and one point on the line • Two points on the line

37. Finding the equation of a straight line from a graph • Use the formule y = mx + c • Find the y intercept. This is ‘c’ • Find the gradient. This is ‘m’ • Substitute these values into the equation y= mx + c

38. The point gradient equation of a line (x1 , y1) for the known point (x , y)to represent any point on the line m for the gradient y x (x,y) x (x1 ,y1) Gradient = change in y , substituting gives m = y – y1 change in x x – x1 Giving y – y1 = m (x – x1)

39. Find the equation of the line with gradient -3 which passes through (4, -5) (x1 , y1) = (4, -5) and m = -3 The equation is y – y1 = m (x – x1) Substituting gives y – -5 = -3 (x – 4) y + 5 = -3x + 12 y = -3x + 7 The equation of the line is y = -3x + 7

40. The two point equation of a line • You are given two points P(x1,y1) and Q(x2,y2) • m = y2 – y1 x2 – x1 The equation is y – y1 = y2 – y1 (x – x1) x2 – x1

41. Find the equation of a line passing through (1,-2) and (4,5) The equation is y – y1 = y2 – y1 (x – x1) x2 – x1 y – -2 = 5 – -2(x – 1) 4 – 1 y + 2 = 7/3 (x-1) 3(y+2) = 7(x-1) 3y + 6 = 7x – 7 0 = 7x – 3y – 13 The equation of the line is 7x – 3y – 13 = 0

42. Perpendicular lines • The product of the gradients of perpendicular lines is -1. m1m2 = -1 or m1 = -1 m2

43. Show that the following pair of lines are perpendicular 5x – y + 8 = 0 3x + 15y -7 = 0 5x – y + 8 = 0  y = 5x + 8 gradient is 5 3x + 15y – 7 = 0  y = - 1 x + 7 gradient is -1 5 15 5 Gradient i x gradient ii = 5 x -1 = -1 5 Since m1m2 = -1, the lines are perpendicular

44. Find the gradient of the line which is perpendicular to y = x + 1 The gradientof the line is Then m2 = -1 = The gradient for the perpendicular line is 5 2