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**CO-ORDINATE GEOMETRY OF THE LINE**

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**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

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**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. 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. The y-coordinate, or ordinate, tells us the vertical distance from the x-axis to the point.

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**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. 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.

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**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 This slide shows the generalization for finding the distance between two points. The distance between the points A(x1, y1) and B(x2, y2) is

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**What is the distance between the points A(5, –1) and B(–4, 5)?**

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) 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, (to two decimal places).

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**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

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**Finding the mid-point of a line segment**

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.

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**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: (x2, y2) (x1, y1) x y is the mean of the x-coordinates. 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). is the mean of the y-coordinates.

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**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: The answer can be checked by verifying that (1, –2) is the mid-point of the line segment joining (–3, 4) and (5, –8). –3 + a = 2 4 + b = –4 a = 5 b = –8 The coordinates of the point P are (5, –8)

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**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

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**Calculating gradients**

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.

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

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**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

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**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. x y 1 m c 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). 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. For example, the line y = 3x + 4 has a gradient of 3 and crosses the y-axis at the point (0, 4).

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**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: A line passes through the point (0, –4) and has a gradient of . What is the equation of the line? The equation of this line can be rearranged into the form 5y – 2x + 20 = 0. Using y = mx + c with and c = –4 we can write the equation of the line as

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**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) We can then write the gradient as P(x, y) y – 5 x – 2 But the gradient is 2 so

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**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.

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**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) P(x, y) y – y1 x – x1 So This can be rearranged to give y – y1 = m(x – x1). In general: This result should be memorized. The equation of a line through A(x1, y1) with gradient m is y – y1 = m(x – x1)

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**Finding the equation of a line**

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. x y B(5, 4) A(3, –2) The gradient of AP, mAP = P(x, y) The gradient of AB, mAB = 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. But AP and AB are parts of the same line so their gradients must be equal.

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**Finding the equation of a line**

Putting mAP equal to mAB gives the equation = = 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.

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

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**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. 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. 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.

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**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

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**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 3y = –6x + 2 y = –6x + 2 3 y = –2x + 2/3 The gradient, m, is –2 for both lines and so they are parallel.

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**Exploring perpendicular lines**

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.

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Perpendicular lines If the gradients of two lines have a product of –1 then they are perpendicular. In general, if the gradient of a line is m, then the gradient of the line perpendicular to it is . 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. 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. Let’s call the mid-point of AB point M, so M is the point

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Perpendicular lines The gradient of the line joining A(–2, 2) and B(4, –1) is mAB = The gradient of the perpendicular bisector of AB is therefore 2. Using this and the fact that it passes through the point we can use y – y1 = m(x – x1) to write 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. 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.

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**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 x y To find the y-intercept put x = 0 in the equation of the line: 6 2y – 12 = 0 y = 6 the x-intercept 4 To find the x-intercept put y = 0 in the equation of the line: 3x – 12 = 0 x = 4

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**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

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**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 l2 l1 Find the equation of the line l2. Find the length of AB. A B x

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**Examination-style question**

a) Rearranging the equation of l1 to the form y = mx + c gives 3x – 4y + 6 = 0 4y = 3x + 6 y = x + So the gradient of l1 is . Since l2 is perpendicular to l1 its gradient is – . Using y – y1 = m(x – x1) with this gradient and the point (2, 4) we can write the equation of l2 as: y – 4 = – (x – 2) 3y – 12 = –4x + 8 4x + 3y – 20 = 0

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**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 x = –2 So A is the point (–2, 0). The point B lies on the line with equation 4x + 3y – 20 = 0. When y = 0 we have 4x – 20 = 0 x = 5 So B is the point (5, 0). The length of AB is 5 – (–2) = 7

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C1: Parallel and Perpendicular Lines

C1: Parallel and Perpendicular Lines

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