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COORDINATE GEOMETRY

COORDINATE GEOMETRY. Week commencing Monday 16 th November 2009 Learning Intention: To be able to determine if two lines are parallel or perpendicular. Contents: 1. Parallel and Perpendicular Lines 2. Examples 3. Assignment 4. COORDINATE GEOMETRY. PARALLEL AND PERPENDICULAR LINES

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COORDINATE GEOMETRY

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  1. COORDINATE GEOMETRY Week commencing Monday 16th November 2009 Learning Intention: To be able to determine if two lines are parallel or perpendicular. Contents: 1.Parallel and Perpendicular Lines 2.Examples 3.Assignment 4

  2. COORDINATE GEOMETRY PARALLEL AND PERPENDICULAR LINES Parallel Lines have the same gradient. Perpendicular Lines are at right angles to each other.

  3. COORDINATE GEOMETRY PARALLEL AND PERPENDICULAR LINES Parallel Two lines y = m1x + c1 and y = m2x + c2 are parallelif m1 = m2. Perpendicular If m1 x m2 = -1 then the two lines y = m1x + c1 and y = m2x + c2 are perpendicular. Therefore, if the gradient of line 1 = m, then the gradient of line 2 = -1/m

  4. COORDINATE GEOMETRY EQUATION OF A LINE GIVEN TWO POINTS Examples: Work out the gradient of the line that is perpendicular to the lines with these gradients: (i) 3 (ii) -4 (iii) -½ Solutions: (i) m = 3 therefore perpendicular m = -1/3 (ii) m = -4 therefore perpendicular m = -1/-4 = 1/4 (iii) m = -½ therefore perpendicular m = -1/-½ = 2

  5. COORDINATE GEOMETRY EQUATION OF A LINE GIVEN TWO POINTS Example: Determine whether the lines y – 3x + 3 = 0 and 3y + x = 6 are parallel, perpendicular or neither. Solution: To compare the two lines we need to know the gradients of the two lines. Line 1: y – 3x + 3 = 0 rewrite in form y = mx + c y = 3x – 3 so m = 3 Line 2: 3y + x = 6 rewrite in form y = mx + c 3y = -x + 6 y = -1/3x + 6 so m =-1/3 Comparing gradients: 3 x -1/3 = -1, so lines are perpendicular

  6. COORDINATE GEOMETRY EQUATION OF A LINE GIVEN TWO POINTS Example: Line L is perpendicular to the line 2y – x + 3 = 0 at the point (4, ½). Determine the equation of the line L. Solution: To find the equation of a line we need a point and a gradient. We have a point and can find the gradient by first finding the gradient of the given line. 2y – x + 3 = 0 rewrite in form y = mx + c 2y = x – 3 y = ½x – 3/2 m = ½ Therefore perpendicular gradient = -2 Substitute into formula for equation of a line to get: y – ½ = -2(x – 4) y – ½ = -2x + 8 y = -2x + 8 + ½ = -2x + 17/2

  7. COORDINATE GEOMETRY Assignment 4 Follow the link for Assignment 4 in Moodle Course Area underneath Coordinate Geometry. Assignments to be completed by 5:00pm on Monday 23rd November 2009.

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