If k is a positive constant such that `0 lt k lt 1`, which of the following could be the graph of `y-x=k(x+y)` ?

Options

A.
B.
C.
D.

Answer

To solve the equation \( y - x = k(x + y) \) where \( k \) is a positive constant such that \( 0 < k < 1 \), we will follow these steps:

### Step 1: Rearranging the Equation
Start with the given equation:
\[ y - x = k(x + y) \]
We can rearrange this to isolate \( y \):
\[ y - k y = k x + x \]
This simplifies to:
\[ y(1 - k) = (k + 1)x \]

### Step 2: Solving for \( y \)
Now, divide both sides by \( (1 - k) \) (since \( k < 1 \), \( 1 - k > 0 \)):
\[ y = \frac{(k + 1)}{(1 - k)} x \]

### Step 3: Analyzing the Slope
The slope of the line is given by:
\[ \text{slope} = \frac{(k + 1)}{(1 - k)} \]
Since \( k \) is a positive constant less than 1, both \( k + 1 \) and \( 1 - k \) are positive. Therefore, the slope is positive.

### Step 4: Finding the y-intercept
Since the equation is in the form \( y = mx + b \) (where \( b \) is the y-intercept), and there is no constant term added to \( y \), we can conclude that the line passes through the origin (0,0).

### Step 5: Conclusion
Since the slope is positive and the line passes through the origin, the graph of the equation \( y - x = k(x + y) \) will be a straight line through the origin with a positive slope.

Now, we can eliminate options that do not fit this description:
- Any option with a y-intercept (not passing through the origin) can be eliminated.
- Any option with a negative slope can also be eliminated.

Thus, the correct graph would be the one that shows a straight line through the origin with a positive slope.