If the straight lines `2x+ 3y -3=0 and x+ky +7 =0` are perpendicular, then the value of k is

Options

A. `3/2`
B. `(-3)/(2)`
C. `2/3`
D. `(-2)/(3)`

Answer

To find the value of \( k \) such that the lines \( 2x + 3y - 3 = 0 \) and \( x + ky + 7 = 0 \) are perpendicular, we can follow these steps:

### Step 1: Find the slope of the first line
The equation of the first line is given as:
\[ 2x + 3y - 3 = 0 \]

To find the slope, we can rewrite this equation in the slope-intercept form \( y = mx + c \), where \( m \) is the slope.

Rearranging the equation:
\[ 3y = -2x + 3 \]
\[ y = -\frac{2}{3}x + 1 \]

Thus, the slope \( m_1 \) of the first line is:
\[ m_1 = -\frac{2}{3} \]

### Step 2: Find the slope of the second line
The equation of the second line is:
\[ x + ky + 7 = 0 \]

Again, we rearrange this into the slope-intercept form:
\[ ky = -x - 7 \]
\[ y = -\frac{1}{k}x - \frac{7}{k} \]

Thus, the slope \( m_2 \) of the second line is:
\[ m_2 = -\frac{1}{k} \]

### Step 3: Use the condition for perpendicular lines
For two lines to be perpendicular, the product of their slopes must equal \(-1\):
\[ m_1 \cdot m_2 = -1 \]

Substituting the values of \( m_1 \) and \( m_2 \):
\[ \left(-\frac{2}{3}\right) \cdot \left(-\frac{1}{k}\right) = -1 \]

This simplifies to:
\[ \frac{2}{3k} = -1 \]

### Step 4: Solve for \( k \)
To isolate \( k \), we can cross-multiply:
\[ 2 = -3k \]

Now, solving for \( k \):
\[ k = -\frac{2}{3} \]

### Final Answer
The value of \( k \) is:
\[ \boxed{-\frac{2}{3}} \]