What will be the equation of the straight line that passes through the intersection of the straight lines 2x - 3y + 4 = 0 and 3x + 4y - 5 = 0 and is perpendicular to the straight line 3x - 4y = 5?

Options

A. `8x + 6y =(32)/(7)`
B. `4x + 3y = (84)/(17)`
C. `4x + 3y = (62)/(17)`
D. `8x + 6y = (58)/(17)`

Answer

To solve the problem, we need to find the equation of a straight line that passes through the intersection of the two given lines and is perpendicular to a third line. Here’s how we can do this step by step:

### Step 1: Find the intersection point of the lines
We have two lines:
1. \( 2x - 3y + 4 = 0 \) (Equation 1)
2. \( 3x + 4y - 5 = 0 \) (Equation 2)

To find the intersection point, we can solve these two equations simultaneously.

**Multiply Equation 1 by 3:**
\[ 3(2x - 3y + 4) = 0 \implies 6x - 9y + 12 = 0 \quad \text{(Equation 3)} \]

**Multiply Equation 2 by 2:**
\[ 2(3x + 4y - 5) = 0 \implies 6x + 8y - 10 = 0 \quad \text{(Equation 4)} \]

Now we have:
- Equation 3: \( 6x - 9y + 12 = 0 \)
- Equation 4: \( 6x + 8y - 10 = 0 \)

Next, we can eliminate \( x \) by subtracting Equation 4 from Equation 3:
\[ (6x - 9y + 12) - (6x + 8y - 10) = 0 \]
This simplifies to:
\[ -17y + 22 = 0 \implies 17y = 22 \implies y = \frac{22}{17} \]

Now substitute \( y = \frac{22}{17} \) back into one of the original equations to find \( x \). Using Equation 1:
\[ 2x - 3\left(\frac{22}{17}\right) + 4 = 0 \]
\[ 2x - \frac{66}{17} + \frac{68}{17} = 0 \implies 2x + \frac{2}{17} = 0 \implies 2x = -\frac{2}{17} \implies x = -\frac{1}{17} \]

So, the intersection point is:
\[ \left(-\frac{1}{17}, \frac{22}{17}\right) \]

### Step 2: Find the slope of the line that is perpendicular
The line we need to find is perpendicular to the line given by:
\[ 3x - 4y = 5 \]
To find the slope of this line, we can rearrange it into slope-intercept form \( y = mx + b \):
\[ -4y = -3x + 5 \implies y = \frac{3}{4}x - \frac{5}{4} \]
Thus, the slope \( m_1 \) of this line is \( \frac{3}{4} \).

The slope \( m_2 \) of the line we are looking for, which is perpendicular to this line, is given by:
\[ m_2 = -\frac{1}{m_1} = -\frac{4}{3} \]

### Step 3: Use the point-slope form to find the equation of the line
We can use the point-slope form of the line equation:
\[ y - y_1 = m(x - x_1) \]
Where \( (x_1, y_1) \) is the intersection point \( \left(-\frac{1}{17}, \frac{22}{17}\right) \) and \( m = -\frac{4}{3} \).

Substituting these values:
\[ y - \frac{22}{17} = -\frac{4}{3}\left(x + \frac{1}{17}\right) \]

### Step 4: Simplify the equation
Distributing the slope:
\[ y - \frac{22}{17} = -\frac{4}{3}x - \frac{4}{51} \]

Now, adding \( \frac{22}{17} \) to both sides:
\[ y = -\frac{4}{3}x + \frac{22}{17} - \frac{4}{51} \]

To combine the constants, we need a common denominator (which is 51):
\[ \frac{22}{17} = \frac{66}{51} \quad \text{(since } 22 \times 3 = 66\text{)} \]
So,
\[ y = -\frac{4}{3}x + \left(\frac{66}{51} - \frac{4}{51}\right) = -\frac{4}{3}x + \frac{62}{51} \]

### Step 5: Rearranging to standard form
Multiplying through by 51 to eliminate the fraction:
\[ 51y = -68x + 62 \]
Rearranging gives:
\[ 68x + 51y = 62 \]

### Final Equation
Thus, the equation of the straight line is:
\[ 68x + 51y = 62 \]