Draw the graphs of \(x = 3\), \(x = 5\), and \(2x - y - 4 = 0\). Find the area of the quadrilateral formed by these lines and the x–axis.
Area \(= 8\) square units.
Step 1: Rewrite the slant line equation.
The line is given as \(2x - y - 4 = 0\).
Move terms around: \(y = 2x - 4\).
So, this is a straight line which goes upward (slope = 2) and cuts the y-axis at \(-4\).
Step 2: Understand the other two lines.
The equations \(x = 3\) and \(x = 5\) are vertical lines. They are parallel to the y-axis.
Step 3: Think about the region we want.
The x-axis is \(y = 0\).
So, the closed shape is formed between:
This shape is a trapezium (a quadrilateral with one pair of parallel sides).
Step 4: Find the heights of the trapezium.
We plug in x-values into the slant line \(y=2x-4\):
At \(x=3\): \(y = 2(3) - 4 = 6 - 4 = 2\).
At \(x=5\): \(y = 2(5) - 4 = 10 - 4 = 6\).
So the slant line is at height 2 above the x-axis at \(x=3\), and at height 6 above the x-axis at \(x=5\).
Step 5: Work out the distance between the vertical lines.
The vertical distance along the x-axis is \(5 - 3 = 2\) units.
Step 6: Apply the trapezium area formula.
Area of trapezium = \(\tfrac{1}{2} \times (\text{sum of parallel sides}) \times \text{distance}\).
Here, the two parallel sides are the heights: 2 units and 6 units.
So, area = \(\tfrac{1}{2} \times (2 + 6) \times 2 = \tfrac{1}{2} \times 8 \times 2 = 8\).
Final Answer: The area of the quadrilateral is \(8\) square units.