COORDINATE GEOMETRY Week commencing Monday 16 th November 2009 Learning Intention: To be able to determine if two lines are parallel or perpendicular.

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Presentation transcript:

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 LinesParallel and Perpendicular Lines 2.ExamplesExamples 3.Assignment 4Assignment 4

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

COORDINATE GEOMETRY PARALLEL AND PERPENDICULAR LINES Parallel Two lines y = m 1 x + c 1 and y = m 2 x + c 2 are parallel if m 1 = m 2. Perpendicular If m 1 x m 2 = -1 then the two lines y = m 1 x + c 1 and y = m 2 x + c 2 are perpendicular. Therefore, if the gradient of line 1 = m, then the gradient of line 2 = -1/m

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 = 3therefore perpendicular m = -1/3 (ii)m = -4therefore perpendicular m = -1/-4 = 1/4 (iii)m = -½ therefore perpendicular m = -1/-½ = 2

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 = 0rewrite in form y = mx + c y = 3x – 3so m = 3 Line 2:3y + x = 6rewrite in form y = mx + c 3y = -x + 6 y = -1/3x + 6so m =-1/3 Comparing gradients: 3 x -1/3 = -1, so lines are perpendicular

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 = 0rewrite in form y = mx + c 2y = x – 3 y = ½x – 3/2m = 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 ½ = -2x + 17/2

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 23 rd November 2009.