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AS-Level Maths: Core 1 for Edexcel

AS-Level Maths: Core 1 for Edexcel. C1.5 Coordinate geometry. This icon indicates the slide contains activities created in Flash. These activities are not editable. For more detailed instructions, see the Getting Started presentation. 1 of 33. © Boardworks Ltd 2005.

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AS-Level Maths: Core 1 for Edexcel

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  1. AS-Level Maths:Core 1for Edexcel C1.5 Coordinate geometry This icon indicates the slide contains activities created in Flash. These activities are not editable. For more detailed instructions, see the Getting Started presentation. 1 of 33 © Boardworks Ltd 2005

  2. The distance between two points • 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 Contents 2 of 33 © Boardworks Ltd 2005

  3. 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 coordinates of a point P are written in the form P(x, y). 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.

  4. 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. Generalization for the distance between two points The horizontal distance between the points is . x2 – x1 The vertical distance between the points is . y2 – y1 The distance between the points A(x1, y1) and B(x2, y2) is What is the distance between two general points with coordinates A(x1, y1) and B(x2, y2)? Using Pythagoras’ Theorem, the square of the distance between the points A(x1, y1) and B(x2, y2) is

  6. 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)? x1 y1 x2 y2 A(5, –1) B(–4, 5)

  7. The mid-point of a line segment • 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 Contents 7 of 33 © Boardworks Ltd 2005

  8. Finding the mid-point of a line segment

  9. Generalization for the mid-point of a line y (x2, y2) is the mean of the x-coordinates. (x1, y1) is the mean of the y-coordinates. 0 x In general, the coordinates of the mid-point of the line segment joining (x1, y1) and (x2, y2) are given by:

  10. 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: Equating the y-coordinates: –3 + a = 2 4 + b = –4 a = 5 b = –8 The coordinates of the point P are (5, –8)

  11. Calculating the gradient of a straight line • 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 Contents 11 of 33 © Boardworks Ltd 2005

  12. Calculating gradients

  13. Finding the gradient from two given points the gradient = y (x2, y2) (x1, y1) change in y change in x 0 x 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: y2 – y1 Draw a right-angled triangle between the two points on the line as follows: x2 – x1

  14. The equation of a straight line • 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 Contents 14 of 33 © Boardworks Ltd 2005

  15. The equation of a straight line y m 1 c 0 x 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-interceptand it has the coordinates (0, c). For example, the line y = 3x + 4 has a gradient of 3 and crosses they-axis at the point (0, 4).

  16. The equation of a straight line A line passes through the point (0, –4) and has a gradient of . What is the equation of the line? Using y = mx + c with and c = –4 we can write the equation of the line as 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:

  17. Finding the equation of a line y P(x, y) A(2, 5) 0 x • 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. We can then write the gradient as y– 5 x– 2 But the gradient is 2 so

  18. 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.

  19. Finding the equation of a line y P(x, y) A(x1, y1) So 0 x Suppose a line passes through A(x1, y1) with gradient m. Let P(x, y) be any other point on the line. y– y1 x– x1 This can be rearranged to give y– y1 = m(x– x1). In general: The equation of a line through A(x1, y1) with gradient m is y– y1 = m(x– x1)

  20. Finding the equation of a line y P(x, y) B(5, 4) 0 x A(3, –2) • Finding the equation of a line given two points on the line 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 = The gradient of AB, mAB = But AP and AB are parts of the same line so their gradients must be equal.

  21. Finding the equation of a line = = 3 Putting mAP equal to mAB gives the equation 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.

  22. Finding the equation of a line The equation of a line through A(x1, y1) and B(x2, y2) is 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 gradient of AP = the gradient of AB. y P(x, y) B(x2, y2) So A(x1, y1) Or 0 x

  23. 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 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. Parallel and perpendicular lines • 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 Contents 24 of 33 © Boardworks Ltd 2005

  25. Parallel lines –6x + 2 y = 3 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 formy = mx + c. 3y + 6x = 2 3y = –6x + 2 y = –2x + 2/3 The gradient, m, is –2 for both lines and so they are parallel.

  26. Exploring perpendicular lines

  27. Perpendicular lines In general, if the gradient of a line is m, then the gradient of the line perpendicular to it is . 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. Let’s call the mid-point of AB point M, so M is the point

  28. Perpendicular lines mAB = Using this and the fact that it passes through the point we can use y – y1 = m(x – x1) to write 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. 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. Sketching straight line graphs y 6 4 0 x 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 To find the y-intercept put x = 0 in the equation of the line: 2y– 12 = 0 y = 6 • the x-intercept To find the x-intercept put y = 0 in the equation of the line: 3x– 12 = 0 x = 4

  30. Examination-style questions • 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 Contents 30 of 33 © Boardworks Ltd 2005

  31. 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. y l1 l2 • Find the equation of the line l2. • Find the length of AB. x 0 A B

  32. Examination-style question y = x + So the gradient of l1 is . Since l2 is perpendicular to l1 its gradient is – . y – 4 = – (x – 2) 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

  33. Examination-style question When y = 0 we have When y = 0 we have 3x + 6 = 0 4x – 20 = 0 b) The point A lies on the line with equation 3x – 4y + 6 = 0. x = –2 So A is the point (–2, 0). The point B lies on the line with equation 4x + 3y – 20 = 0. x = 5 So B is the point (5, 0).  The length of AB is 5 – (–2) = 7

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