A line passing through the points (a, 2a) and (-2, 3) is perpendicular to the line 4x + 3y + 5 = 0. Find the value of a?

Options

A. -14/3
B. 18/5
C. 14/3
D. -18/5

Answer

To solve the problem step by step, we need to find the value of \( a \) such that the line passing through the points \( (a, 2a) \) and \( (-2, 3) \) is perpendicular to the line given by the equation \( 4x + 3y + 5 = 0 \).

### Step 1: Find the slope of the line passing through the points \( (a, 2a) \) and \( (-2, 3) \).

The formula for the slope \( m \) between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by:

\[ m = \frac{y_2 - y_1}{x_2 - x_1} \]

For our points, we have:

- \( (x_1, y_1) = (a, 2a) \)
- \( (x_2, y_2) = (-2, 3) \)

Substituting these values into the slope formula:

\[ m_1 = \frac{3 - 2a}{-2 - a} \]

### Step 2: Find the slope of the line given by the equation \( 4x + 3y + 5 = 0 \).

To find the slope of this line, we can rearrange it into the slope-intercept form \( y = mx + b \):

\[ 3y = -4x - 5 \implies y = -\frac{4}{3}x - \frac{5}{3} \]

Thus, the slope \( m_2 \) of this line is:

\[ m_2 = -\frac{4}{3} \]

### Step 3: Use the property of perpendicular lines.

Two lines are perpendicular if the product of their slopes is \( -1 \):

\[ m_1 \cdot m_2 = -1 \]

Substituting the slopes we found:

\[ \left(\frac{3 - 2a}{-2 - a}\right) \cdot \left(-\frac{4}{3}\right) = -1 \]

### Step 4: Simplify the equation.

Removing the negative signs, we have:

\[ \frac{3 - 2a}{-2 - a} \cdot \frac{4}{3} = 1 \]

Cross-multiplying gives:

\[ 3(3 - 2a) = 4(-2 - a) \]

Expanding both sides:

\[ 9 - 6a = -8 - 4a \]

### Step 5: Solve for \( a \).

Rearranging the equation:

\[ 9 + 8 = -4a + 6a \]

This simplifies to:

\[ 17 = 2a \]

Dividing both sides by 2:

\[ a = \frac{17}{2} \]

### Conclusion

The value of \( a \) is \( \frac{17}{2} \).

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