If the system of linear equations
x-2y + kz = 1, 2x + y+ z = 2, 3x-y-kz = 3 has a solution (x, y, z), `z ne 0`, then (x, y) lies on the straight line whose equation is

Options

A. 3x-4y-4=0
B. 3x-4y-1=0
C. 4x-3y-4=0
D. 4x-3y-1=0

Answer

To solve the problem, we need to analyze the given system of linear equations:

1. \( x - 2y + kz = 1 \) (Equation 1)
2. \( 2x + y + z = 2 \) (Equation 2)
3. \( 3x - y - kz = 3 \) (Equation 3)

We need to find the relationship between \( x \) and \( y \) given that \( z \neq 0 \).

### Step 1: Add Equation 1 and Equation 3

We start by adding Equation 1 and Equation 3:

\[ (x - 2y + kz) + (3x - y - kz) = 1 + 3 \]

This simplifies to:

\[ x + 3x - 2y - y + kz - kz = 4 \]

### Step 2: Combine Like Terms

Now, we combine the like terms:

\[ 4x - 3y = 4 \]

### Step 3: Rearrange the Equation

Next, we can rearrange this equation to express it in a standard linear form:

\[ 4x - 3y - 4 = 0 \]

### Step 4: Identify the Equation of the Line

The equation \( 4x - 3y - 4 = 0 \) represents a straight line in the \( xy \)-plane. We can express this in slope-intercept form \( y = mx + c \):

\[ 3y = 4x - 4 \]
\[ y = \frac{4}{3}x - \frac{4}{3} \]

Thus, the equation of the line is:

\[ y = \frac{4}{3}x - \frac{4}{3} \]

### Conclusion

Therefore, the solution indicates that the pair \( (x, y) \) lies on the straight line given by the equation \( 4x - 3y - 4 = 0 \) or in slope-intercept form \( y = \frac{4}{3}x - \frac{4}{3} \).

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