1. Coordinate Geometry Regions on the Number Plane.2 4 6 -2
X-axis
Y-axis
By Mr Porter

2. Assumed Knowledge.
Students are able to graph linear relationships of the form:
y = mx + 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 1
Y-axis
X-axis 2 3 4 5 -3 -2 -1
Where does the line cross the x-axis and y-axis?
2x – y + 4 = 0
Draw line!
Y-axis intercept: When x = 0
– y + 4 = 0
y = 4
2(0) – y + 4 = 0
X-axis intercept: When y = 0
2x + 4 = 0
x = -2
2x – (0) + 4 = 0

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

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

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

6. Graph the Region Indicated.
Example 3: Graph shade the region indicated by: 4x – y ≤ 0 and x + 2y ≥ 0
Example 2: Graph shade the region indicated by:
y ≥ 2 - x and y > x - 3
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 ≥ 2 - x , using intercepts (x = 0, y = 2) and (y = 0, x = 2)
A solid line.
Y-axis
Point of intersection of the 2 lines is an solid circle.
y > x - 3 , using intercepts (x = 0, y = -3) and (y = 0, x = 3)
A broken line.
4x – y = 0
Point of intersection of the 2 lines is an open circle.
x + 2y = 0
Y-axis
X-axis
X-axis
y = 2 - x
y = x – 3 2 2 3
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. -3
(0,3)
Calculator = T + T
(3,0)
Calculator = F + T
Select test points, test in y ≥ 2 – x and y ≥ x - 3
(0,0)
Calculator = F + T
(4,0)
Calculator = T + F
(0,-3)
Calculator = F + F
(-3,0)
Calculator = T + F
(0,3)
Calculator = T + T
(2,-3)
Calculator = F + F
The truth table tells us the point (0,3) is in both regions, hence region containing (0,3) to be shaded.
The truth table tells us the point (0,3) is in both regions, hence region containing (0,3) to be shaded.

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