What value of k will make the line containing points (k,3) and (-2,1) perpendicular to the line containing (5,k) and (1,0) ?

Options

A. `-4`
B. `-4/3`
C. `-1`
D. `-3/4`

Answer

To find the value of \( k \) that makes the line containing the points \( (k, 3) \) and \( (-2, 1) \) perpendicular to the line containing the points \( (5, k) \) and \( (1, 0) \), we will follow these steps:

### Step 1: Find the slope of the first line (L1)
The slope \( m_1 \) of the line through the points \( (k, 3) \) and \( (-2, 1) \) is given by the formula:
\[ m_1 = \frac{y_2 - y_1}{x_2 - x_1} \]
Here, \( (x_1, y_1) = (k, 3) \) and \( (x_2, y_2) = (-2, 1) \).
Substituting these values, we get:
\[ m_1 = \frac{1 - 3}{-2 - k} = \frac{-2}{-2 - k} = \frac{2}{2 + k} \]

### Step 2: Find the slope of the second line (L2)
The slope \( m_2 \) of the line through the points \( (5, k) \) and \( (1, 0) \) is given by:
\[ m_2 = \frac{y_2 - y_1}{x_2 - x_1} \]
Here, \( (x_1, y_1) = (5, k) \) and \( (x_2, y_2) = (1, 0) \).
Substituting these values, we have:
\[ m_2 = \frac{0 - k}{1 - 5} = \frac{-k}{-4} = \frac{k}{4} \]

### Step 3: Set the product of the slopes to -1
For the lines to be perpendicular, the product of their slopes must equal -1:
\[ m_1 \cdot m_2 = -1 \]
Substituting the expressions for \( m_1 \) and \( m_2 \):
\[ \frac{2}{2 + k} \cdot \frac{k}{4} = -1 \]

### Step 4: Solve for \( k \)
Now, we solve the equation:
\[ \frac{2k}{4(2 + k)} = -1 \]
Cross-multiplying gives:
\[ 2k = -4(2 + k) \]
Expanding the right side:
\[ 2k = -8 - 4k \]
Bringing all terms involving \( k \) to one side:
\[ 2k + 4k = -8 \]
\[ 6k = -8 \]
Dividing both sides by 6:
\[ k = -\frac{8}{6} = -\frac{4}{3} \]

### Final Answer
Thus, the value of \( k \) that makes the lines perpendicular is:
\[ \boxed{-\frac{4}{3}} \]