What is the minimum value of |c|, if the lines y=mx+4, x=m+c and y=3 are concurrent ?

Options

A. 0
B. 1
C. 2
D. 3

Answer

To find the minimum value of |c| such that the lines \( y = mx + 4 \), \( x = m + c \), and \( y = 3 \) are concurrent, we will follow these steps:

### Step 1: Set up the equations
We have the following three equations:
1. \( y = mx + 4 \) (Equation 1)
2. \( x = m + c \) (Equation 2)
3. \( y = 3 \) (Equation 3)

### Step 2: Substitute \( y = 3 \) into Equation 1
From Equation 3, we can substitute \( y = 3 \) into Equation 1:
\[ 3 = mx + 4 \]
Rearranging gives:
\[ mx = 3 - 4 \implies mx = -1 \implies x = -\frac{1}{m} \quad \text{(Equation 4)} \]

### Step 3: Substitute \( x = -\frac{1}{m} \) into Equation 2
Now we substitute Equation 4 into Equation 2:
\[ -\frac{1}{m} = m + c \]
Rearranging gives:
\[ c = -\frac{1}{m} - m \quad \text{(Equation 5)} \]

### Step 4: Express \( c \) in terms of \( m \)
From Equation 5, we can express \( c \) as:
\[ c = -m - \frac{1}{m} \]

### Step 5: Find the minimum value of |c|
To find the minimum value of \( |c| \), we need to analyze the expression:
\[ |c| = \left|-m - \frac{1}{m}\right| \]

Let \( f(m) = -m - \frac{1}{m} \). To find the critical points, we differentiate \( f(m) \):
\[ f'(m) = -1 + \frac{1}{m^2} \]
Setting the derivative to zero for critical points:
\[ -1 + \frac{1}{m^2} = 0 \implies \frac{1}{m^2} = 1 \implies m^2 = 1 \implies m = 1 \text{ or } m = -1 \]

### Step 6: Evaluate \( |c| \) at critical points
Now we evaluate \( |c| \) at \( m = 1 \) and \( m = -1 \):
1. For \( m = 1 \):
\[ c = -1 - 1 = -2 \implies |c| = 2 \]
2. For \( m = -1 \):
\[ c = 1 - 1 = 0 \implies |c| = 0 \]

### Step 7: Conclusion
The minimum value of \( |c| \) is \( 0 \).