Find the equation of the straight line passing through the intersection of `x+2y-3=0` and `3x+4y-7=0` and perpendicular to `x+3y+4=0`

Answer

To find the equation of the straight line passing through the intersection of the lines \(x + 2y - 3 = 0\) and \(3x + 4y - 7 = 0\) and perpendicular to the line \(x + 3y + 4 = 0\), we can follow these steps:

### Step 1: Find the intersection point of the two lines
We need to solve the equations:

1. \(x + 2y - 3 = 0\) (Equation 1)
2. \(3x + 4y - 7 = 0\) (Equation 2)

We can express \(x\) from Equation 1:
\[ x = 3 - 2y \]

Now substitute \(x\) in Equation 2:
\[ 3(3 - 2y) + 4y - 7 = 0 \]
\[ 9 - 6y + 4y - 7 = 0 \]
\[ 2 - 2y = 0 \]
\[ 2y = 2 \implies y = 1 \]

Now substitute \(y = 1\) back into Equation 1 to find \(x\):
\[ x + 2(1) - 3 = 0 \implies x + 2 - 3 = 0 \implies x = 1 \]

Thus, the intersection point is \((1, 1)\).

### Step 2: Find the slope of the line \(x + 3y + 4 = 0\)
Rearranging this equation to slope-intercept form \(y = mx + c\):
\[ 3y = -x - 4 \implies y = -\frac{1}{3}x - \frac{4}{3} \]
The slope \(m_1\) of this line is \(-\frac{1}{3}\).

### Step 3: Find the slope of the required line
Since the required line is perpendicular to the line \(x + 3y + 4 = 0\), the slope \(m\) of the required line is the negative reciprocal of \(m_1\):
\[ m = -\frac{1}{m_1} = -\frac{1}{-\frac{1}{3}} = 3 \]

### Step 4: Use the point-slope form to find the equation of the line
Using the point \((1, 1)\) and the slope \(m = 3\):
\[ y - y_1 = m(x - x_1) \]
Substituting the values:
\[ y - 1 = 3(x - 1) \]
\[ y - 1 = 3x - 3 \]
\[ y = 3x - 2 \]

### Step 5: Write the equation in standard form
To convert \(y = 3x - 2\) into standard form:
\[ 3x - y - 2 = 0 \]

### Final Answer
The equation of the straight line is:
\[ 3x - y - 2 = 0 \]