# Co-ordinate geometry Objectives: Students should be able to

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

Co-ordinate geometry A 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)

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)

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). x coordinate: y coordinate:  2b = 10  5a  2b = 10  7.5 = 2.5  b = 1.25 Therefore a = 1.5 and b = 1.25

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.

Example Q P

Negative gradient On the line MN, as we move from A to A1, 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.

The gradient of a line joining ( x1 , x2 ) to ( y1 , y2 )

Therefore PQ is parallel to RS
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

The distance between two points
Find the distance AB AN = 4 NB = 3 A(1, 3) N(5, 3) Pythagoras: AB2 = 25 AB = 5 The distance AB between two points (x1, y1) and B(x2, y2) is given by the formula AB =

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

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.  Find the vertices of the triangle by solving simultaneously the equations for each pair of lines (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).

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

 Draw a sketch, labelling the points
y x  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.

a) Show that ABC is isosceles.  Draw a sketch
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.  Draw a sketch y x A(–4, 1) C(3, 2) B(–2, –3) O Tip: Use the sketch to identify which two sides are likely to be the same length.  Work out AC2 and BC2

© Pearson Education Ltd. 2005
b) Find the coordinates of the midpoint of the base. y x A(–4, 1) C(3, 2) B(–2, –3) O  Identify the base and substitute into the mid-point formula The base of the triangle is AB. Let the mid-point of AB be M. © Pearson Education Ltd. 2005

 Use the formula for the area of a triangle
c) Find the area of ABC. Show M on the diagram y x A(–4, 1) C(3, 2) B(–2, –3) O M(–3, –1) 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

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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