Question

Drag and drop the points to graph and solve the system of equations.
y = 1/2x - 1 (blue line)

y = -1/2x + 3 (red line)

What are their points on the line?

Answer 1

To graph the equations \( y = \frac{1}{2}x - 1 \) (blue line) and \( y = -\frac{1}{2}x + 3 \) (red line), we first need to identify some points for each line.

### For the blue line \( y = \frac{1}{2}x - 1 \):
1. **When \( x = 0 \)**:
\[
y = \frac{1}{2}(0) - 1 = -1 \quad \Rightarrow \quad (0, -1)
\]

2. **When \( x = 2 \)**:
\[
y = \frac{1}{2}(2) - 1 = 0 \quad \Rightarrow \quad (2, 0)
\]

3. **When \( x = 4 \)**:
\[
y = \frac{1}{2}(4) - 1 = 1 \quad \Rightarrow \quad (4, 1)
\]

The points on the blue line are \( (0, -1) \), \( (2, 0) \), and \( (4, 1) \).

### For the red line \( y = -\frac{1}{2}x + 3 \):
1. **When \( x = 0 \)**:
\[
y = -\frac{1}{2}(0) + 3 = 3 \quad \Rightarrow \quad (0, 3)
\]

2. **When \( x = 2 \)**:
\[
y = -\frac{1}{2}(2) + 3 = 2 \quad \Rightarrow \quad (2, 2)
\]

3. **When \( x = 4 \)**:
\[
y = -\frac{1}{2}(4) + 3 = 1 \quad \Rightarrow \quad (4, 1)
\]

The points on the red line are \( (0, 3) \), \( (2, 2) \), and \( (4, 1) \).

### Finding the Intersection Point:
To solve the system of equations, we can set the two equations equal to each other:

\[
\frac{1}{2}x - 1 = -\frac{1}{2}x + 3
\]

To eliminate the fractions, multiply through by 2:

\[
x - 2 = -x + 6
\]

Now combine like terms:

\[
x + x = 6 + 2
\]
\[
2x = 8 \implies x = 4
\]

Now substitute \( x = 4 \) back into either equation to find \( y \):

Using the blue line:
\[
y = \frac{1}{2}(4) - 1 = 2 - 1 = 1
\]

Thus, the intersection point is \( (4, 1) \).

### Summary:
- **Points on Blue Line**: \( (0, -1) \), \( (2, 0) \), \( (4, 1) \)
- **Points on Red Line**: \( (0, 3) \), \( (2, 2) \), \( (4, 1) \)
- **Intersection Point**: \( (4, 1) \)