Coordinate Geometry powerpoint for high school
- 1. © Boardworks Ltd 2005 1 of 33 Coordinate Geometry
- 2. © Boardworks Ltd 2005 2 of 33 Contents © Boardworks Ltd 2005 2 of 33 The distance between two points The mid-point of a line segment Calculating the gradient of a straight line The equation of a straight line Parallel and perpendicular lines Examination-style questions The distance between two points
- 3. © Boardworks Ltd 2005 3 of 33 The Cartesian coordinate system The Cartesian coordinate system is named after the French mathematician René Descartes (1596 – 1650). Points in the (x, y) plane are defined by their perpendicular distance from the x- and y-axes relative to the origin, O. The x-coordinate, or abscissa, tells us the horizontal distance from the y-axis to the point. The y-coordinate, or ordinate, tells us the vertical distance from the x-axis to the point. The coordinates of a point P are written in the form P(x, y).
- 4. © Boardworks Ltd 2005 4 of 33 The distance between two points Given the coordinates of two points, A and B, we can find the distance between them by adding a third point, C, to form a right-angled triangle. We then use Pythagoras’ theorem.
- 5. © Boardworks Ltd 2005 5 of 33 Generalization for the distance between two points What is the distance between two general points with coordinates A(x1, y1) and B(x2, y2)? The horizontal distance between the points is . x2 – x1 The vertical distance between the points is . y2 – y1 Using Pythagoras’ Theorem, the square of the distance between the points A(x1, y1) and B(x2, y2) is The distance between the points A(x1, y1) and B(x2, y2) is ( ) ( ) x x y y 2 2 2 1 2 1 + ( ) ( ) x x y y 2 2 2 1 2 1 +
- 6. © Boardworks Ltd 2005 6 of 33 Worked example Given the coordinates of two points we can use the formula to directly find the distance between them. For example: What is the distance between the points A(5, –1) and B(–4, 5)? A(5, –1) B(–4, 5) x1 x2 y1 y2 ( ) ( ) x x y y 2 2 2 1 2 1 + 2 2 2 2 ( 4 5) +(5 1) = ( 9) +6 = 81+36 = 117 = 3 13
- 7. © Boardworks Ltd 2005 7 of 33 Contents © Boardworks Ltd 2005 7 of 33 The distance between two points The mid-point of a line segment Calculating the gradient of a straight line The equation of a straight line Parallel and perpendicular lines Examination-style questions The mid-point of a line segment
- 8. © Boardworks Ltd 2005 8 of 33 Finding the mid-point of a line segment
- 9. © Boardworks Ltd 2005 9 of 33 Generalization for the mid-point of a line In general, the coordinates of the mid-point of the line segment joining (x1, y1) and (x2, y2) are given by: , 1 2 1 2 + + 2 2 x x y y is the mean of the x-coordinates. x x 1 2 + 2 is the mean of the y-coordinates. y y 1 2 + 2 (x2, y2) (x1, y1) x y 0 , 1 2 1 2 + + 2 2 x x y y
- 10. © Boardworks Ltd 2005 10 of 33 Finding the mid-point of a line segment The mid-point of the line segment joining the point (–3, 4) to the point P is (1, –2). Find the coordinates of the point P. Let the coordinates of the points P be (a, b). We can then write (1, –2) Equating the x-coordinates: –3 + a = 2 a = 5 Equating the y-coordinates: 4 + b = –4 b = –8 The coordinates of the point P are (5, –8) , 3+ 4+ = 2 2 a b a 3+ =1 2 b 4+ = 2 2
- 11. © Boardworks Ltd 2005 11 of 33 Contents © Boardworks Ltd 2005 11 of 33 The distance between two points The mid-point of a line segment Calculating the gradient of a straight line The equation of a straight line Parallel and perpendicular lines Examination-style questions Calculating the gradient of a straight line
- 12. © Boardworks Ltd 2005 12 of 33 Calculating gradients
- 13. © Boardworks Ltd 2005 13 of 33 x y (x1, y1) (x2, y2) 0 Finding the gradient from two given points If we are given any two points (x1, y1) and (x2, y2) on a line we can calculate the gradient of the line as follows: the gradient = change in y change in x x2 – x1 y2 – y1 Draw a right-angled triangle between the two points on the line as follows: y y m x x 2 1 2 1 the gradient, =
- 14. © Boardworks Ltd 2005 14 of 33 Contents © Boardworks Ltd 2005 14 of 33 The distance between two points The mid-point of a line segment Calculating the gradient of a straight line The equation of a straight line Parallel and perpendicular lines Examination-style questions The equation of a straight line
- 15. © Boardworks Ltd 2005 15 of 33 The equation of a straight line The equation of a straight line can be written in several forms. You are probably most familiar with the equation written in the form y = mx + c. The value of m tells us the gradient of the line. The value of c tells us where the line cuts the y-axis. This is called the y-intercept and it has the coordinates (0, c). For example, the line y = 3x + 4 has a gradient of 3 and crosses the y-axis at the point (0, 4). x y 1 m c 0
- 16. © Boardworks Ltd 2005 16 of 33 The equation of a straight line A straight line can be defined by: one point on the line and the gradient of the line two points on the line If the point we are given is the y-intercept and we are also given the gradient of the line, we can write the equation of that line directly using y = mx + c. For example: Using y = mx + c with and c = –4 we can write the equation of the line as m 2 5 = 2 5 = 4 y x A line passes through the point (0, –4) and has a gradient of . What is the equation of the line? 2 5
- 17. © Boardworks Ltd 2005 17 of 33 Finding the equation of a line Finding the equation of a line given a point on the line and the gradient Suppose we are given the gradient of a line but that the point given is not the y-intercept. For example: A line passes through the point (2, 5) and has a gradient of 2. What is the equation of the line? Let P(x, y) be any point on the line. x y A(2, 5) 0 We can then write the gradient as But the gradient is 2 so y x 5 2 y x 5 = 2 2 x – 2 y – 5 P(x, y)
- 18. © Boardworks Ltd 2005 18 of 33 Finding the equation of a line Rearranging: y – 5 = 2(x – 2) y – 5 = 2x – 4 y = 2x + 1 So, the equation of the line passing through the point (2, 5) with a gradient of 2 is y = 2x + 1. Now let’s look at this for the general case. y x 5 = 2 2
- 19. © Boardworks Ltd 2005 19 of 33 Finding the equation of a line Suppose a line passes through A(x1, y1) with gradient m. Let P(x, y) be any other point on the line. x y A(x1, y1) 0 This can be rearranged to give y – y1 = m(x – x1). The equation of a line through A(x1, y1) with gradient m is y – y1 = m(x – x1) y y x x 1 1 The gradient of AP = So y y m x x 1 1 = x – x1 y – y1 P(x, y) In general:
- 20. © Boardworks Ltd 2005 20 of 33 Finding the equation of a line Finding the equation of a line given two points on the line x y B(5, 4) 0 A(3, –2) A line passes through the points A(3, –2) and B(5, 4). What is the equation of the line? Let P(x, y) be any other point on the line. The gradient of AP, mAP = ( 2) 3 y x The gradient of AB, mAB = 4 ( 2) 5 3 But AP and AB are parts of the same line so their gradients must be equal. P(x, y)
- 21. © Boardworks Ltd 2005 21 of 33 Finding the equation of a line ( 2) 3 y x 4 ( 2) 5 3 = + 2 3 y x = 3 y + 2 = 3(x – 3) y + 2 = 3x – 9 y = 3x – 11 So, the equation of the line passing through the points A(3, –2) and B(5, 4) is y = 3x – 11. Now let’s look at this for the general case. Putting mAP equal to mAB gives the equation
- 22. © Boardworks Ltd 2005 22 of 33 Finding the equation of a line Suppose a straight line passes through the points A(x1, y1) and B(x2, y2) with another point on the line P(x, y). The equation of a line through A(x1, y1) and B(x2, y2) is 1 1 2 1 2 1 = y y x x y y x x x y B(x2, y2) 0 A(x1, y1) P(x, y) The gradient of AP = the gradient of AB. So 1 2 1 1 2 1 = y y y y x x x x Or 1 1 2 1 2 1 = y y x x y y x x
- 23. © Boardworks Ltd 2005 23 of 33 The equation of a straight line One more way to give the equation of a straight line is in the form ax + by + c = 0. This form is often used when the required equation contains fractions. For example, the equation 3 1 4 2 = y x can be rewritten without fractions as 4y – 3x + 2 = 0. It is important to note that any straight line can be written in the form ax + by + c = 0. In particular, equations of the form x = c can be written in the form ax + by + c = 0 but cannot be written in the form y = mx + c.
- 24. © Boardworks Ltd 2005 24 of 33 Contents © Boardworks Ltd 2005 24 of 33 The distance between two points The mid-point of a line segment Calculating the gradient of a straight line The equation of a straight line Parallel and perpendicular lines Examination-style questions Parallel and perpendicular lines
- 25. © Boardworks Ltd 2005 25 of 33 Parallel lines If two lines have the same gradient they are parallel. Show that the lines 3y + 6x = 2 and y = –2x + 7 are parallel. We can show this by rearranging the first equation so that it is in the form y = mx + c. 3y = –6x + 2 y = –6x + 2 3 3y + 6x = 2 y = –2x + 2 /3 The gradient, m, is –2 for both lines and so they are parallel.
- 26. © Boardworks Ltd 2005 26 of 33 Exploring perpendicular lines
- 27. © Boardworks Ltd 2005 27 of 33 Perpendicular lines If the gradients of two lines have a product of –1 then they are perpendicular. Find the equation of the perpendicular bisector of the line joining the points A(–2, 2) and B(4, –1). The perpendicular bisector of the line AB has to pass through the mid-point of AB. In general, if the gradient of a line is m, then the gradient of the line perpendicular to it is . 1 m Let’s call the mid-point of AB point M, so M is the point 2+ 4 2+( 1) , = 2 2 1 2 (1, )
- 28. © Boardworks Ltd 2005 28 of 33 Perpendicular lines The gradient of the line joining A(–2, 2) and B(4, –1) is The gradient of the perpendicular bisector of AB is therefore 2. 1 2 = 2( 1) y x mAB = 1 2 = 4 ( 2) 3 = 6 1 2 Using this and the fact that it passes through the point we can use y – y1 = m(x – x1) to write 1 2 (1, ) 2 1= 4( 1) y x 2 1= 4 4 y x 2 4 +3 = 0 y x So, the equation of the perpendicular bisector of the line joining the points A(–2, 2) and B(4, –1) is 2y – 4x + 3 = 0.
- 29. © Boardworks Ltd 2005 29 of 33 Sketching straight line graphs Suppose we want to sketch the straight line with the equation 2y + 3x – 12 = 0. It is sufficient to find two points on the line: the y-intercept the x-intercept To find the y-intercept put x = 0 in the equation of the line: 2y – 12 = 0 y = 6 To find the x-intercept put y = 0 in the equation of the line: 3x – 12 = 0 x = 4 0 x y 6 4
- 30. © Boardworks Ltd 2005 30 of 33 Contents © Boardworks Ltd 2005 30 of 33 The distance between two points The mid-point of a line segment Calculating the gradient of a straight line The equation of a straight line Parallel and perpendicular lines Examination-style questions Examination-style questions
- 31. © Boardworks Ltd 2005 31 of 33 Examination-style question The line l1 in the following diagram has equation 3x – 4y + 6 = 0 The line l2 is perpendicular to the line l1 and passes through the point (2, 4). The lines l1 and l2 cross the x-axis at the points A and B respectively. a) Find the equation of the line l2. b) Find the length of AB. l1 A 0 x y l2 B
- 32. © Boardworks Ltd 2005 32 of 33 Examination-style question a) Rearranging the equation of l1 to the form y = mx + c gives 3x – 4y + 6 = 0 4y = 3x + 6 Using y – y1 = m(x – x1) with this gradient and the point (2, 4) we can write the equation of l2 as: 3y – 12 = –4x + 8 4x + 3y – 20 = 0 So the gradient of l1 is . 3 4 y = x + 3 4 3 2 Since l2 is perpendicular to l1 its gradient is – . 4 3 4 3 y – 4 = – (x – 2)
- 33. © Boardworks Ltd 2005 33 of 33 Examination-style question b) The point A lies on the line with equation 3x – 4y + 6 = 0. When y = 0 we have 3x + 6 = 0 So A is the point (–2, 0). x = –2 The point B lies on the line with equation 4x + 3y – 20 = 0. When y = 0 we have 4x – 20 = 0 So B is the point (5, 0). x = 5 The length of AB is 5 – (–2) = 7
Editor's Notes
- #3: Introduce the system of Cartesian coordinates as providing us with a way to express the geometry of lines, curves and shapes algebraically. Explain that points are normally denoted by a capital letter followed by the coordinates of the point. Use the embedded Flash movie to demonstrate a variety of points in each of the four quadrants and on the axes. Observe that the x-coordinate of the point becomes negative as the point passes to the left of the y-axis and that the y-coordinate becomes negative as it passes below the x-axis.
- #4: This method for finding the shortest distance between two given points is the same as finding the length of a line segments joining the two points.
- #5: This slide shows the generalization for finding the distance between two points.
- #6: Point out that it doesn’t matter which point is called (x1, y1) and which point is called (x2, y2). It can help to write x1, y1, x2 and y2 above each coordinate as shown before substituting the values into the formula. The answer in this example is written in surd form. An alternative would be to write it to a given number of decimal places; for example, 10.82 (to two decimal places).
- #8: Use this activity to explore the mid-points of given line segments. Establish that the x-coordinate of the mid-point will be half way between the x-coordinates of the end points. This is the mean of the x-coordinates of the end-points. The y-coordinate of the mid-point will be half way between the y-coordinates of the end points. This is the mean of the y-coordinates of the end-points.
- #9: Talk through the generalization of the result for any two points (x1, y1) and (x2, y2). As for the generalization for the distance between two points, it doesn’t matter which point is called (x1, y1) and which point is called (x2, y2).
- #10: The answer can be checked by verifying that (1, –2) is the mid-point of the line segment joining (–3, 4) and (5, –8).
- #12: Establish that the gradient of a straight line is a measure of its slope. By looking at a variety of examples, establish that if we are given the coordinates of any two points on a line we can find its gradient by taking the y-coordinate of the first point subtracted from the y-coordinate of the second point and diving it by the x-coordinate of the first point subtracted from the x-coordinate of the second point. Demonstrate that if y increases as x increases the line will slope upwards and the gradient will be positive. If y decreases as x increases the line will slope downwards and the gradient will be negative. Also demonstrate the gradients of horizontal and vertical lines. Show that choosing different points on the same line will give the same gradient using equivalent fractions. For example, if the vertical distance between the end points is 6 and the horizontal distance between the end points is 4, the gradient of the line is 3/2. If we change the vertical distance to 9 and the horizontal distance to 6, the gradient of the line is still 3/2. Point out that it is often most useful to leave gradients as improper fractions. For example, the gradient 3/2 tells us that for every 2 squares we move along we move 3 up. Gradients can also be given as decimals. Hide the value of the gradient and ask pupils to tell you the gradients of given lines.
- #13: Explain how drawing a right-angled triangle on the line helps to calculate its gradient as shown in the previous activity. Also explain that since, for a straight line, the change in y is proportional to the corresponding change in x, the gradient will be the same no matter which two points we choose on a line. The gradient is usually denoted by the letter m.
- #15: This form is sometimes called the gradient intercept form. Only lines parallel to the y-axis cannot be written in the form y = mx + c.
- #16: The equation of this line can be rearranged into the form 5y – 2x + 20 = 0.
- #19: This result should be memorized.
- #20: The equation of this line could also be found by using the given points to find the gradient of the line and using the equation y – y1 = m(x – x1). The gradient could also be substituted into the equation y = mx + c. The value of c can then be found by substituting the values of x and y given by one of the points on the line.
- #22: Point out that in most examples x1, y1, x2 and y2 will be numbers given by the coordinates. x and y will be the only variables in the equation. Although it is useful to learn this equation, its use can be avoided, as mentioned previously. The two given points can be used to find the gradient of the line and the equation of the line y – y1 = m(x – x1) can be used. The gradient could also be substituted into the equation y = mx + c. The value of c can then be found by substituting the values of x and y given by one of the points on the line.
- #23: Point out that in the general form ax + by + c =0, the value c has a different meaning to the value of c in the form y = mx + c. If an equation is given in the form ax + by + c = 0, it can be rearranged to find the gradient and the y-intercept. This would give y = -a/bx – c/b where the gradient is -a/b and the y-intercept is -c/b.
- #26: The gradient of each line in this activity is given by the coefficient of x in the line’s equation. Modify the red line by dragging its defining points and observe the relationship between its gradient and the gradient of the line perpendicular to it.
- #27: The exception to this generalization that the product of the gradient of two perpendicular lines is -1 is when the two perpendicular lines in question are horizontal and vertical.
- #28: Point out that in the general form ax + by + c =0, the value c has a different meaning to the value of c in the form y = mx + c. If an equation is given in the form ax + by + c = 0, it can be rearranged to find the gradient and the y-intercept. This would give y = -a/bx – c/b where the gradient is -a/b and the y-intercept is -c/b.