1. Straight Line Graph revision

2. x
Plotting graphs
Choose 3 values for x (0, 2, 4)
Substitute into equation to find each y value
Plot 3 coordinates
Draw line x x
y = 2x + 3
x = 0, y = 3
x = 2, y = 7
x = 3, y = 9

3. 6 +3 3 Finding equation of the line Find y intercept
Find gradient by drawing triangle through nice points
change in y ÷ change in x
(6 ÷ 3 = 2)
y = 2x + 3 6 +3 3

4. x 10 x 5 Equation from 2 points A (-2, -1) and B (3, 9) x coordinate
(-2 - 3) = -5
y coordinate
(-1 - 9) ÷ 2 = -10
change in y ÷ change in x (-10 ÷ -5) = 2
y = 2x + c
Substitute in one coordinate
9 = 2(3) + c
c = 3
y = 2x + 3 10 x 5

5. x x Midpoint of 2 points A (-2, -1) and B (3, 9)
Average 2 x parts and 2 y parts
x coordinate
(-2 + 3) ÷ 2 = ½
y coordinate
(-1 + 9) ÷ 2 = 4
Midpoint = (½, 4) x

6. x 10 x 5 125 Length between 2 points A (-2, -1) and B (3, 9)
Same as gradient  change in x & change in y or
x coordinate
(-2 - 3) = -5
y coordinate
(-1 - 9) ÷ 2 = -10
now
Do Pythagoras theory
(-10)2 + (-5)2 = 125 10 x 5
125

7. x Parallel Lines y = 2x + 3 Parallel line has same gradient y = 2x + c
Parallel line passing through (2, 0)
Substitute in coordinate
0 = 2(2) + c
0 = 4 + c
c = -4
y = 2x - 4 x

8. x Perpendicular Lines y = 2x + 3
Perpendicular line has negative reciprocal gradient
y = -½ x + c
Perpendicular line passing through (2, 0)
Substitute in coordinate
0 = -½(2) + c
0 = -4 + c
c = 4
y = 2x + 4 x

9. From each pair of points:
a) Midpoint b) length of line c) equation of line
d) parallel line passing through (2, 1) e) perpendicular line passing through (2, 1)
(3, 5) and (7,9)
(0, 2) and (11,5)
(-4, 3) and (7,-6)
(-5, 1) and (4,9)
(-5, -9) and (7,-6)

10. y = x2 + 2x + 3
y = x3 - x + 5 x -3 -2 -1 x2 2x 3
y = x2 + 2x + 3