The lines `y=ax+b an y=bx+a` are graphed in the xy-plane. If a and b are non-zero constants and a+b=0, which statements must be true?

Options

A. The lines are parallel
B. The lines intersect at right angled
C. The lines have the same x-intercept.
D. The lines have the same y-intercept.

Answer

To solve the problem, we need to analyze the given equations of the lines and the condition that \( a + b = 0 \).

### Step 1: Write down the equations of the lines
The equations given are:
1. \( y = ax + b \)
2. \( y = bx + a \)

### Step 2: Substitute the condition \( a + b = 0 \)
From the condition \( a + b = 0 \), we can express \( b \) in terms of \( a \):
\[ b = -a \]

### Step 3: Substitute \( b \) into the second equation
Now, substitute \( b = -a \) into the second equation:
\[ y = (-a)x + a \]
This simplifies to:
\[ y = -ax + a \]

### Step 4: Rewrite both equations
Now we have:
1. \( y = ax + b \) (where \( b \) is still \( b \))
2. \( y = -ax + a \)

### Step 5: Find the x-intercepts of both lines
To find the x-intercepts, we set \( y = 0 \) in both equations.

For the first equation:
\[ 0 = ax + b \implies ax = -b \implies x = -\frac{b}{a} \]

For the second equation:
\[ 0 = -ax + a \implies -ax = -a \implies x = 1 \]

### Step 6: Analyze the x-intercepts
Since we have \( a + b = 0 \) or \( b = -a \), we can substitute \( b \) into the x-intercept of the first equation:
\[ x = -\frac{-a}{a} = 1 \]

Thus, both lines have the same x-intercept, which is \( x = 1 \).

### Step 7: Conclusion
The lines \( y = ax + b \) and \( y = bx + a \) intersect at the same x-intercept, which is \( x = 1 \).

### Final Answer
The statement that must be true is: **The lines have the same x-intercept.**

---