Draw the graphs of the equations `y =-1, y =3 and y = 4x - 5.` Also, find the area of the quadrilateral formed by the lines and the y-axis.

Answer

To solve the problem, we will follow these steps:

### Step 1: Draw the lines \( y = -1 \) and \( y = 3 \)

1. **Graph of \( y = -1 \)**:
- This is a horizontal line that crosses the y-axis at \( y = -1 \).
- It will extend infinitely in both directions along the x-axis.

2. **Graph of \( y = 3 \)**:
- This is another horizontal line that crosses the y-axis at \( y = 3 \).
- It also extends infinitely in both directions along the x-axis.

### Step 2: Draw the line \( y = 4x - 5 \)

1. **Find the intercepts**:
- **When \( x = 0 \)**:
\[ y = 4(0) - 5 = -5 \quad \text{(y-intercept)} \]
So, the point is \( (0, -5) \).
- **When \( y = 0 \)**:
\[ 0 = 4x - 5 \implies 4x = 5 \implies x = \frac{5}{4} = 1.25 \quad \text{(x-intercept)} \]
So, the point is \( \left(\frac{5}{4}, 0\right) \).

2. **Plot the points**:
- Plot the points \( (0, -5) \) and \( \left(\frac{5}{4}, 0\right) \) on the graph.
- Draw the line through these points.

### Step 3: Identify the vertices of the quadrilateral formed

1. **Find the intersection points**:
- **Intersection of \( y = -1 \) and \( y = 4x - 5 \)**:
\[ -1 = 4x - 5 \implies 4x = 4 \implies x = 1 \]
So, the point is \( (1, -1) \).

- **Intersection of \( y = 3 \) and \( y = 4x - 5 \)**:
\[ 3 = 4x - 5 \implies 4x = 8 \implies x = 2 \]
So, the point is \( (2, 3) \).

2. **Vertices of the quadrilateral**:
- The vertices are \( (0, -1) \), \( (0, 3) \), \( (1, -1) \), and \( (2, 3) \).

### Step 4: Calculate the area of the quadrilateral

1. **Identify the shape**:
- The quadrilateral formed is a trapezium with bases along the lines \( y = -1 \) and \( y = 3 \).

2. **Calculate the height and bases**:
- Height \( h = 3 - (-1) = 4 \).
- Base \( a = 1 \) (distance from \( (0, -1) \) to \( (1, -1) \)).
- Base \( b = 2 \) (distance from \( (0, 3) \) to \( (2, 3) \)).

3. **Area of the trapezium**:
\[ \text{Area} = \frac{1}{2} \times h \times (a + b) = \frac{1}{2} \times 4 \times (1 + 2) = \frac{1}{2} \times 4 \times 3 = 6 \text{ square units} \]

### Final Answer:
The area of the quadrilateral formed by the lines and the y-axis is **6 square units**.

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