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Co-ordinate geometry Objectives: Students should be able to * find the distance between two points * find the gradient of a line * find the mid-point of a line

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Co-ordinate geometry Co-ordinates Co-ordinates are a means of describing a position relative to some fixed points, or origin. In two dimensions you need two pieces of information; in three dimensions, you need three pieces of information. x y A B The coordinate of A ? (4, 3) The coordinate of B ?( -4, -1)

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The mid – point of a line segment Example 1: Find the midpoint of PQ where P is the point ( 2, 4) and Q is the point (4, 8). Midpoint = = (1, 2) The coordinates of the midpoint are (1, 2)

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The mid – point of a line segment Example 2. Find a and b if the point (3, 5) is the midpoint of the line joining (3a, 2a 4b) and (a, 3a + 2b). 2b = 10 5a 2b = 10 7.5 = 2.5 b = 1.25 x coordinate: y coordinate: Therefore a = 1.5 and b = 1.25

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The gradient of a line The direction of a straight line is given by its gradient. The gradient of a line is the amount by which the y coordinate increases if we move along the line far enough to increase the x coordinate by one unit. i.e. the gradient of a line is a measure of its steepness. The steeper the line, the larger the gradient.

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Example P Q

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Negative gradient On the line MN, as we move from A to A 1, the x coordinate increases by 1 unit but the y coordinate decreases by g units. But a decrease of g units may be regarded as an increase of (– g) units. Thus the gradient of MN will be –g.

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The gradient of a line joining ( x 1, x 2 ) to ( y 1, y 2 ) y 2 – y 1 x 2 – x 1

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Example If A and B are the points (1, 3) and (2, 1) find the gradient of AB. Example If P,Q, R and S are points (2, 3), (4, 8), (-3, -2) and (1, 8), respectively, show that PQ is parallel to RS. Therefore PQ is parallel to RS

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The distance between two points A(1, 3) B(5, 6) AB = N(5, 3) Find the distance AB AN = 4 NB = 3 Pythagoras: AB 2 = 25 AB = 5 The distance AB between two points (x 1, y 1 ) and B(x 2, y 2 ) is given by the formula

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Example: For the pair of points A(–2, 6), B(4, –2), calculate (a)The midpoint of AB (b) The gradient of AB (c) The distance AB A(-2, 6) B(4, -2) x y (a) Mid-point = =(1, 2) (b) Gradient = AB = (c) =10

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Find the vertices of the triangle by solving simultaneously the equations for each pair of lines Example A triangle is formed by three straight lines, y =, 2x + y + 5 = 0 and x + 3y – 5 = 0. Prove that the triangle is isosceles. (1) (2) Let the point of intersection of line (1) and line (2) be P. Substitute y from (1) into (2). Substitute in (1) P is the point (–2, –1).

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Let the point of intersection of line (1) and line (3) be Q. Substitute y from (1) into (3). At Q, Substitute in (1) Q is the point (2, 1). Let the point of intersection of line (2) and line (3) be R. (3) Substitute in (2) R is the point (–4, 3). (1) (2) (3) (2) × 3 (4) (4) – (3)

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Draw a sketch, labelling the points Work out the squares of the lengths and compare Tip: Since we can see that two lengths are the same, there is no need to work out the third length. y x

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Draw a sketch Work out AC 2 and BC 2 Example In the triangle ABC, A, B and C are the points (–4, 1), (–2, –3) and (3, 2), respectively. a) Show that ABC is isosceles. Tip: Use the sketch to identify which two sides are likely to be the same length. y x A(–4, 1) C(3, 2) B(–2, –3) O

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Identify the base and substitute into the mid-point formula © Pearson Education Ltd b) Find the coordinates of the midpoint of the base. The base of the triangle is AB. Let the mid-point of AB be M. y x A(–4, 1) C(3, 2) B(–2, –3) O

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Show M on the diagram c) Find the area of ABC. Tip: Since triangle ABC is isosceles, MC is perpendicular to AB, by symmetry. Find AB and MC Use the formula for the area of a triangle Area of triangle ABC y x A(–4, 1) C(3, 2) B(–2, –3) O M(–3, –1)

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The properties of lines Parallel lines have equal gradients. Lines parallel to the x-axis have gradient zero. Lines parallel to the y-axis have infinite gradient. Two lines are perpendicular if the product of their gradient is –1.

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