Draw the graphs of \(2x + y = 4\) and \(2x - y = 4\). Find the vertices of the triangle formed by these two lines and the y-axis, and its area.
Vertices: \((0,4)\), \((0,-4)\), \((2,0)\). Area \(= 8\) square units.
Step 1: Find where each line cuts the y-axis.
On the y-axis, \(x = 0\).
For the first line: \(2x + y = 4\).
Put \(x = 0\):
\(2(0) + y = 4 \Rightarrow y = 4\).
So the point is \((0, 4)\).
For the second line: \(2x - y = 4\).
Put \(x = 0\):
\(2(0) - y = 4 \Rightarrow -y = 4 \Rightarrow y = -4\).
So the point is \((0, -4)\).
Step 2: Find where the two lines meet (their intersection).
We have:
\(2x + y = 4\) … (i)
\(2x - y = 4\) … (ii)
Add (i) and (ii):
\((2x + y) + (2x - y) = 4 + 4\)
\(4x = 8\)
So, \(x = 2\).
Put \(x = 2\) in (i):
\(2(2) + y = 4\)
\(4 + y = 4\)
\(y = 0\).
So the point is \((2, 0)\).
Step 3: Identify the triangle.
The three vertices are \((0, 4)\), \((0, -4)\), and \((2, 0)\).
Step 4: Find the area.
The base is on the y-axis between \((0, 4)\) and \((0, -4)\).
Length of base = \(4 - (-4) = 8\).
The height is the distance from the y-axis to the point \((2, 0)\).
This distance is just the x-coordinate, which is \(2\).
Area of triangle = \(\tfrac{1}{2} \times \text{base} \times \text{height}\)
= \(\tfrac{1}{2} \times 8 \times 2\)
= \(8\) square units.