Statement I The differential equation of all non-vertical lines in a plane is `(d^(2)x)/(dy^(2))=0.`
Satement II The general equation of all non-vertical lines in a plane is ax+by=1, where `b!=0.`

Options

A. Statement I is true ,and Statement II is the correct explanation for Statement I.
B. Statement I is true, Statement II is true and Statement II is the correct explanation for Statment I
C. Statement I is true, Statement II is false.
D. Statement I is false, Statement II is true.

Answer

To solve the problem, we will analyze both statements step by step.

### Step 1: Understanding Statement II
The general equation of all non-vertical lines in a plane is given as:
\[ ax + by = 1, \quad b \neq 0 \]

This means that the line has a slope and is not vertical (i.e., it does not have an undefined slope).

### Step 2: Differentiating the Equation
We will differentiate the equation \( ax + by = 1 \) with respect to \( y \):
1. Differentiate both sides:
\[ \frac{d}{dy}(ax + by) = \frac{d}{dy}(1) \]
2. Applying the product rule:
\[ a \frac{dx}{dy} + b = 0 \]
(since the derivative of a constant is zero).

### Step 3: Rearranging the First Derivative
From the differentiation, we can express \( \frac{dx}{dy} \):
\[ a \frac{dx}{dy} = -b \]
\[ \frac{dx}{dy} = -\frac{b}{a} \]

### Step 4: Second Differentiation
Now, we will differentiate \( \frac{dx}{dy} \) again with respect to \( y \):
1. Differentiate \( \frac{dx}{dy} = -\frac{b}{a} \):
\[ \frac{d^2x}{dy^2} = 0 \]
(since \( -\frac{b}{a} \) is a constant).

### Step 5: Conclusion on Statement I
The result \( \frac{d^2x}{dy^2} = 0 \) indicates that the second derivative of \( x \) with respect to \( y \) is zero, which confirms that the differential equation of all non-vertical lines is indeed:
\[ \frac{d^2x}{dy^2} = 0 \]

### Final Conclusion
Both statements are true:
- Statement I is true because the second derivative of \( x \) with respect to \( y \) for non-vertical lines is zero.
- Statement II is true as it correctly represents the general equation of non-vertical lines.