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Investigation. Find the distance between two points A(1, 2) and B(3, 6) A(1,2) B(3,6) 1 3 2 6 Form a triangle and use Pythagoras to find the distance between.

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Presentation on theme: "Investigation. Find the distance between two points A(1, 2) and B(3, 6) A(1,2) B(3,6) 1 3 2 6 Form a triangle and use Pythagoras to find the distance between."— Presentation transcript:

1 Investigation. Find the distance between two points A(1, 2) and B(3, 6) A(1,2) B(3,6) Form a triangle and use Pythagoras to find the distance between the points (3 – 1 ) ( 6 – 2 ) y-length = 6 – 2 = 4 x-length = 3 – 1 = Length = √( = √20

2 Co-ordinate Geometry Chapter 7

3 Let A (x 1, y 1 ) and B (x 2, y 2 ) be two points. A (x 1, y 1 )   B (x 2, y 2 ) We often need to find d, the distance between A and B. This is found using PYTHAGORAS y 2 – y 1 x 2 – x 1 D Note 1 Note 1 : Distance between two poi nts

4 Example: Find the distance between the points (–3,2) and (3,-6) substitute into the formula d = 10 units (x 1, y 1 )(x 2, y 2 )

5 Applications Prove that the vertices A(1, 5), B(2, 9) and C(6, 10) are those of an isosceles triangle. AB = √17 BC = √17 AC = √50 Because there is two sides with the same length the triangle is isosceles.

6 Theata Page 128 Exercise 16.2

7 Investigation Consider the two numbers 6 and 10. The number exactly halfway between them, their MIDPOINT, is 8. The midpoint can easily be worked out by counting inwards from 6 and 10, but you can also find the midpoint by averaging the two numbers. This averaging is a really useful process when the numbers are not as easy to work with as 6 and 10

8 This concept can now be used to find the midpoint of two points on an x-y graph. Example.Find the midpoint of (– 3, – 4) and (1, 2) (–3,–4) (1,2) Step 1Average the x’s Step 2Average the y’s Step 3M = (–1,–1) (–1,–1)

9 A general formula that finds the midpoint of any two points is: Note 2 Note 2: Finding a Midpoint Let A (x 1, y 1 ) and B (x 2, y 2 ) be two points.  A (x 1, y 1 )  B (x 2, y 2 )  M (?, ?)

10 Example: Find the midpoint of the line segment joining E = (10, -3) and F = (6, 0) (6, 0) (x 1, y 1 )   (10,-3) (x 2, y 2 ) M = (8, -1.5)

11 Prove that the points A(-1,-2), B(1, 1), C(8,-1) and D(6,-4) are the vertices of a parallelogram. (HINT: find the lengths of all four sides and the diagonals) AB = 3.6 unitsBC = 7.3 units DC = 3.6 unitsAD = 7.3 units AC = 9.1 unitsBD = 7.1 units Length of AB = length of CD and length of BC = length of AD. The lengths of the diagonals are different, so therefore the shape is a parallelogram

12 Theata Page 127 Exercise 16.1

13 Starter Find the midpoints of the line segments joining (6, -2) and (1, 2) (3, -2) and (1, -6) midpoint = (3½, 0) midpoint = (2, -4)

14 Note 3: Gradient (slope) measures the steepness of a line is positive if the line leans to the right is negative if the line leans to the left is zero if the line is horizontal is not defined if the line is vertical is defined as RISE RUN

15 Examples: 4 cm 2 cm km 3 km Leans left, so m is negative!

16 Let A (x 1, y 1 ) and B (x 2, y 2 ) be any two points  B (x 2, y 2 )  A (x 1, y 1 ) C The GRADIENT of AB is given by x2 x2 x1 x1 y1 y1 y2 y2 (x 2 – x 1 ) (y 2 - y 1 ) rise is y 2 – y 1 run is x 2 – x 1 m =

17 A(4,3) B(1, –3) Find the gradient of the line joining points A(4, 3) and B(1, –3) M = C Example: (x 2, y 2 ) (x 1, y 1 )

18 Parallel lines have the same gradient Perpendicular lines gradients are the negative reciprocals of each other m 1 x m 2 = -1 Example: If line AB has a gradient of 2 / 3, and line CD is perpendicular to line AB, what is the gradient of CD? Gradient of CD = - 3 / 2 Points are collinear if they lie on the same line – their gradients are equal Parallel and Perpendicular lines

19 Page 137 Exercise 7A

20 The gradient/intercept form for the equation of a line is: y = mx + c gradient y-intercept Note 4: Revision of Equations of Lines

21 Example: Plot y = -2 / 3 x + 2 using the gradient and intercept method y-intercept (c) = 2 Gradient (m) = -2 3 Plot the y-intercept Plot the next point by making a triangle, whose fall is 2 and run is 3 Plot another point in the same manner Connect the points with a straight line Fall is 2 Run is 3 Negative sign means it leans to LEFT

22 Another option: up (+2) then left (-3)  y = - 2 / 3 x + 2 Don’t forget arrows & label One option: down (-2) then right (+3)  Now join the dots 

23 Exercises: Plot the following graphs; y = 2x – 3 Y = - ⅖ x + 2 y = x

24 The general form of the equation of a line is: ax + by + c = 0 Equations can be rearranged from gradient- intercept form to the general form by performing operations on both sides of the equation. (a is usually positive)

25 Example: Write the following equation in the general form y = -2 / 3 x + 2 3y = -2x + 6 2x + 3y – 6 = 0 Multiply equation by 3 Move everything to the LHS

26 Exercises: Write the following equations in the general form: y = 4x – 3 Y = - ⅖ x + 2 y = ⅙ x – ⅚ 4x – y - 3 = 0 2x + 5y - 10 = 0 x - 6y - 5 = 0

27 Note 5: Finding Equations of Lines If you know the gradient, mand any point, (x 1, y 1 ) then the equation of the line can be worked out using the formula y – y 1 = m (x – x 1 )

28 Example 1 Find the equation of the line passing through (– 3, 5) and with gradient 4 Step 1 Write m, x 1 and y 1 m = 4x 1 = –3 Step 2 y 1 = 5 Put these values into the formula y – y 1 = m(x – x 1 ) y – 5 = 4(x – – 3) Step 3 Remove brackets. Write in general form y – 5 = 4(x + 3) y – 5 = 4x + 12 y = 4x = 4x - y + 17

29 Example 2 Find the equation of the line passing through (– 2, -13) and (3, 2) Step 1 Find m m = Step 2 Put these values into the formula y – y 1 = m(x – x 1 ) y – -13 = 3(x – – 2) Step 3 Remove brackets. Write in general form Y + 13 = 3(x + 2) Y + 13 = 3x + 6 y = 3x - 7 (x 1, y 1 )(x 2, y 2 ) = 3 0 = 3x - y - 7

30 Note 5a: Finding Equations of Lines Quickly For slope the form of the line is Ax – By = … For slope the form of the line is Ax + By = … Example: Find the equation of the line which passes through (2,-5) with a gradient of ⅜ The equation is: 3x – 8y = 3(2) – 8(-5) 3x – 8y = 46

31 Exercise 7B and C

32 Starter A car club has an annual hill climb competition. A cross section of part of the hill they race on is drawn below. The beginning and end points of the section, in metres from the start of the hill are given as co-ordinate pairs.

33 Find:  The gradient over this section of the hill climb  What is the distance of this section of the hill climb in metres to the nearest metre?  What are the co-ordinates of the half-way point of this section of the hill climb? m = 1 / 5 153m (215, 85)

34 STARTERS: Graph the following lines: 4x + 12y = 24 y = - 2 / 3 x + 4

35 Note 6: Perpendicular Bisector The perpendicular bisector of AB is the set of all points which are the same distance from A and B. The perpendicular bisector (or mediator) is a line which is perpendicular to AB and passes through the midpoint of AB.

36 Example:Find the equation of the perpendicular bisector between the points A(-2, 5) and B(4, 9) Find the midpoint of AB: Find the gradient of AB: M = The perpendicular gradient of AB = - 3 / 2 Equation of perpendicular bisector: M = (1, 7) y – 7 = - 3 / 2 (x – 1) 2y – 14 = - 3 (x - 1) 2y – 14 = - 3x + 3 3x + 2y – 17 = 0 = ⅔

37 Page 146 Exercise 7D.1 and 7D.2

38 Starter Find the equation of the mediator of AB for the triangle A(-2,3), B(4,0) C(-2,-3) Find the midpoint of AB Find the gradient of AB Find the perpendicular gradient Solution Midpoint = (1, 1.5) Gradient = -0.5 Perpendicular gradient = 2 Equation: y – 1.5 = 2( x – 1 ) = 2x – 2 y = 2x – 0.5 A B C midpoint

39 Starter 3 darts are thrown at a dart board. The first land in the bulls eye, the 2 nd lands 2cm to the left and 1 cm above the bulls eye, the 3 rd lands 2cm to the right and 9cm above the bulls eye. Calculate the distance between the points of the 2 nd and 3 rd darts The equation of the line joining these two darts Find the intersection of this line with its perpendicular bisector Find the equation on this perpendicular bisector

40 Other Geometrical Terms Equidistant – find the midpoint Bisects – cuts into two equal parts Perpendicular – at right angles Vertex – corner of angle Concurrent – pass through the same point Collinear – lies on the same line

41 Note 7: Triangles – Altitude of Triangles The perpendicular distance from a vertex to the opposite side of a triangle is called the altitude (or height) Example: In the triangle A(-4,4), B(2,2) and C(-2,-1), calculate the length of the altitude of the triangle ABC through vertex C. Find the intersection of CI and AB Find equation of AB Find equation of CI Solve a and b simultaneously

42 Example: In the triangle A(-4,4), B(2,2) and C(-2,-1), calculate the length of the altitude of the triangle ABC through vertex C. Find the intersection of CI and AB Find equation of AB Find equation of CI Solve a and b simultaneously

43 Note 7: Triangles – Medians of Triangles A line drawn from a vertex of a triangle to a midpoint of the opposite side is called the median

44 Note 7: Triangles – Medians of Triangles


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