Determine, graphically, the vertices of the triangle formed by the lines \(y = x\), \(3y = x\), and \(x + y = 8\).
Vertices: \((0,0)\), \((4,4)\), and \((6,2)\).
Step 1: Find the intersection of \(y = x\) and \(3y = x\).
From the first line: \(y = x\).
From the second line: \(x = 3y\).
Put \(y = x\) into \(x = 3y\):
\(x = 3x\).
Subtract \(x\) from both sides:
\(x - 3x = 0\).
\(-2x = 0\).
So, \(x = 0\).
Then, \(y = 0\).
First vertex is \((0,0)\).
Step 2: Find the intersection of \(y = x\) and \(x + y = 8\).
From the first line: \(y = x\).
Substitute \(y = x\) into \(x + y = 8\):
\(x + x = 8\).
\(2x = 8\).
So, \(x = 4\).
Since \(y = x\), we get \(y = 4\).
Second vertex is \((4,4)\).
Step 3: Find the intersection of \(3y = x\) and \(x + y = 8\).
From the first line: \(x = 3y\).
Substitute into \(x + y = 8\):
\(3y + y = 8\).
\(4y = 8\).
So, \(y = 2\).
Now, \(x = 3y = 3 \times 2 = 6\).
Third vertex is \((6,2)\).
Final Answer:
The vertices of the triangle are \((0,0)\), \((4,4)\), and \((6,2)\).