1. Question 16

2. Question 16:
The graphs of line a and line b are shown on this coordinate grid. Match each line with its equation.
The students do not need a calculator for this problem. They are just matching the possible equations to the lines drawn. On the next slide, the graph will be shown with the possible equations.

3. Question 16 Cont
y = -2x + 3 y = 2x + 3 y = 3x – 2 y = - ½x + 3 y = -1/3x – 2
Line b 1 1 -2 3
Line a
The students need to match line a and line b with the correct equation.
The things you need to look at for the line are the y-intercept (where the line crosses the y-axis) and the slope (how steep the line is). Once you have these two things, you have your line.
To find the corresponding equation for line a, you need to first find the y-intercept. The y-intercept is where the line crosses the y-axis (the vertical axis). This line crosses the y-axis at -2 and that is shown with a point. We now have it narrowed down to two equations and we can cross out the rest of them. We then want to find the slope, we will look how many spaces we go up, and then how many spaces we go right until we find the next time where the lines cross through the corner of the grid behind it. In this case, we went up 3 spaces and over one. That makes our slope 3 / 1, or just 3. We have now found the equation of this line: 3x – 2 .
Now to find the equation of the second line, we do the same thing. We first look at where the line crosses the y-axis. In this case, it is 3. So we can cross out anything that shows the y-intercept as negative two. Then we find the slope the same way we did for the previous problem. We end up getting the slope to be -2/1. This means that the slope is negative 2. Our final answer for our equation is: y = -2x + 3