The line `(x+1)/1 =(y-1)/2=(z-1)/0`

Options

A. is perpendicular to z-axis
B. is parallel to z-axis
C. lies in xy-plane
D. None of these

Answer

To determine the nature of the line given by the equation \((x+1)/1 = (y-1)/2 = (z-1)/0\), we can analyze it step by step.

### Step 1: Rewrite the line equation in parametric form
The given line equation can be expressed in parametric form by introducing a parameter \( \lambda \):

\[ \frac{x + 1}{1} = \frac{y - 1}{2} = \frac{z - 1}{0} = \lambda \]

From this, we can derive the equations for \(x\), \(y\), and \(z\):

\[ x + 1 = \lambda \implies x = \lambda - 1 \]
\[ y - 1 = 2\lambda \implies y = 2\lambda + 1 \]
\[ z - 1 = 0 \implies z = 1 \]

### Step 2: Analyze the values of \(x\), \(y\), and \(z\)
From the parametric equations, we see that:

- \(x = \lambda - 1\) can take any value depending on \(\lambda\).
- \(y = 2\lambda + 1\) can also take any value depending on \(\lambda\).
- \(z = 1\) is constant.

### Step 3: Determine the orientation of the line
Since \(z\) is constant (equal to 1), this means that the line does not change in the \(z\)-direction. Therefore, the line lies in the plane where \(z = 1\), which is parallel to the \(xy\)-plane.

### Step 4: Check if the line is perpendicular to the \(z\)-axis
A line that is parallel to the \(xy\)-plane is indeed perpendicular to the \(z\)-axis. This is because any movement along the line does not affect the \(z\)-coordinate, thus indicating that the line does not rise or fall in the \(z\)-direction.

### Conclusion
The line represented by the equation \((x + 1)/1 = (y - 1)/2 = (z - 1)/0\) is perpendicular to the \(z\)-axis.