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Coordinate Geometry Regions on the Number Plane. By Mr Porter 0 246-2 X-axis Y-axis -2 2 4

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Assumed Knowledge. (1)Students are able to graph linear relationships of the form: a)y = mx + b b)ax + by + c =0 (2) Students can substitute points into the linear relationships and evaluate. Examples: (a) Graph 2x – y + 4 = 0 Method 1: Intercepts Where does the line cross the x-axis and y-axis? Y-axis intercept: When x = 0 2(0) – y + 4 = 0 0 1 Y-axis X-axis 1 2 3 4 2345 -3 -2 5 – y + 4 = 0 y = 4 X-axis intercept: When y = 0 2x – (0) + 4 = 0 2x + 4 = 0 x = -2 2x – y + 4 = 0 Draw line!

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Assumed Knowledge. Examples: (b) Graph 2x – 3y + 3 = 0 Method 2: Intercept - Gradient Rearrange the equation into standard form: y = mx + b 0 1 Y-axis X-axis 1 2 3 4 2345 -3 -2 5 2x – 3y + 3 = 0, move the 3y, as its negative. 2x + 3 = 3y, divide by 3 Y-intercept is at y = 1, point (0,1) Gradient 2 / 3, means 3 units right, 2 units up, repeat. Or 3 units LEFT, 2 units DOWN, repeat. 2x – 3y + 3 = 0 Draw line! Method 3: Plot points. x-303 y13 Construct a table of points using Use multiple of the denominator. Graphing Straight Lines is an essential requirement for this 2 unit Mathematics Course.

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Assumed Knowledge. Line Types For the following inequality equations, a broken line is drawn. ax + by + c < 0 ax + by + c > 0 For the following inequality equations, a solid line is drawn. ax + by + c ≥ 0 ax + by + c ≤ 0 Points of Intersection Open circle at the Points of Intersection Closed circle at the Points of Intersection Equal is part of the inequality!

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Graph the Region Indicated. Example 1: Graph shade the region indicated by: x x - 3 x < 2 is a vertical broken line at x = 2 0 1 Y-axis X-axis 1 2 3 4 2345 -3 -2 5 -2 -3 y > x - 3, using intercepts (x = 0, y = -3) and (y = 0, x = 3) A broken line. x = 2 y = x - 3 Region 3 Region 4 Region 2 Region 1 The number plane is divide into 4 areas – regions. Point of intersection of the 2 lines is an open circle. Now, we need to test a point in each region ( but not on the lines). Construct a truth table. Always select (0,0) if it is in a region, or points on the two axes (as one value will be zero) Point (x,y)x < 2y > x - 3 (0,0) (4,0) (3,2) (1,-3) 0 < 2 True 4 < 2 False 3 < 2 False 1 < 2 True 0 > 0 - 3 True 0 > 4 - 3 False 2 > 3 - 3 True -3 > 1 - 3 False The truth table tells us the point (0,0) is in both regions, hence region 2 needs to be shaded. You do not need to show the truth table, it can be done using a calculator. You do not need grid/graph paper for these types of questions.

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Graph the Region Indicated. Example 2: Graph shade the region indicated by: y ≥ 2 - x and y > x - 3 y > x - 3, using intercepts (x = 0, y = -3) and (y = 0, x = 3) A broken line. y ≥ 2 - x, using intercepts (x = 0, y = 2) and (y = 0, x = 2) A solid line. Y-axis X-axis y = 2 - x y = x – 3 Point of intersection of the 2 lines is an open circle. -3 2 2 3 Select test points, test in y ≥ 2 – x and y ≥ x - 3 (0,0) (0,3) (4,0) (2,-3) Calculator = F + T Calculator = T + F Calculator = T + TCalculator = F + F The truth table tells us the point (0,3) is in both regions, hence region containing (0,3) to be shaded. Example 3: Graph shade the region indicated by: 4x – y ≤ 0 and x + 2y ≥ 0 4x – y ≤ 0, using intercepts (x = 0, y = 0). A solid line, sloping up. x + 2y ≥ 0, using intercepts (x = 0, y = 0). A solid line, sloping down. Y-axis X-axis Point of intersection of the 2 lines is an solid circle. 4x – y = 0 x + 2y = 0 Select test points, test in 4x – y ≤ 0 and x + 2y ≥ 0. (You can NOT test the point (0,0). It lies on both lines. (0,3) (0,-3) (3,0) (-3,0) Calculator = T + T Calculator = F + T Calculator = F + FCalculator = T + F The truth table tells us the point (0,3) is in both regions, hence region containing (0,3) to be shaded.

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0 (1,1) 2 (-1,3) Example 4: Graph shade the region indicated by: x > -1, y ≤ 2 - x and y > x Y-axis X-axis x > -1 is a vertical broken line at x = -1 y ≤ 2 - x, using intercepts (x = 0, y = 2) and (y = 0, x = 2), a solid line. y > x, using intercepts (x = 0, y = 0), a broken line. 2 (-1,-1) Select test points, test in x > - 1, y ≤ 2 – x and y ≥ x - 3 (-2,0) (0,1) (4,0) (0,-1) Calculator = F + T + T Calculator = T + T + F Calculator = T + T + TCalculator = T + F + T The truth table tells us the point (0,1) is in both regions, hence region containing (0,1) to be shaded. 0 2 -2 Y-axis X-axis Example 5: Graph shade the region indicated by: y +1 ≥0, x + y ≤ 0 and y < x + 2 y + 1 ≥ 0 is a horizontal solid line at y = -1 x + y ≤ 0, using intercepts (x = 0, y = 0), a solid line, sloping down. y < x + 2, using intercepts (x = 0, y = 2) and when (y = 0, x = -2), a broken line. Select test points, test in y + 1 ≥ 0, x + y ≤ 0 and y ≤ x + 2 (0,-2) (1,0) (0,4) (-1,0) Calculator = F + T + T Calculator = T + F + F Calculator = T + F + TCalculator = T + T + T (0,4)Calculator = T + F + F Point of intersection of the 2 lines is an open circle. Once we have T +T +T, we can stop testing!. The truth table tells us the point (-1,0) is in both regions, hence region containing (-1,0) to be shaded. Point of intersection of the 2 lines is an solid circle. Point of intersection of the 2 lines is an open circle. x = -1 y = 2 – x y = x y + 1 =0 x + y =0 y < x + 2

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