Find the equation of a straight line passing through the point of intersection of the lines :
`3x+y-9=0` and `4x+3y-7=0` and perpendicular to the line `5x-4y+1=0`

Answer

To find the equation of a straight line that passes through the point of intersection of the lines \(3x + y - 9 = 0\) and \(4x + 3y - 7 = 0\), and is perpendicular to the line \(5x - 4y + 1 = 0\), we will follow these steps:

### Step 1: Find the point of intersection of the two lines

We have the equations:
1. \(3x + y - 9 = 0\) (Equation 1)
2. \(4x + 3y - 7 = 0\) (Equation 2)

We can solve these equations simultaneously. From Equation 1, we can express \(y\) in terms of \(x\):
\[ y = 9 - 3x \]
Now, substitute this expression for \(y\) into Equation 2:
\[ 4x + 3(9 - 3x) - 7 = 0 \]
Expanding this gives:
\[ 4x + 27 - 9x - 7 = 0 \]
Combining like terms:
\[ -5x + 20 = 0 \]
Solving for \(x\):
\[ 5x = 20 \implies x = 4 \]
Now substitute \(x = 4\) back into the expression for \(y\):
\[ y = 9 - 3(4) = 9 - 12 = -3 \]
Thus, the point of intersection is \((4, -3)\).

### Step 2: Determine the slope of the given line

The line \(5x - 4y + 1 = 0\) can be rearranged to slope-intercept form \(y = mx + b\):
\[ -4y = -5x - 1 \implies y = \frac{5}{4}x + \frac{1}{4} \]
The slope \(m_1\) of this line is \(\frac{5}{4}\).

### Step 3: Find the slope of the perpendicular line

The slope \(m_2\) of a line perpendicular to another line is given by:
\[ m_2 = -\frac{1}{m_1} \]
Thus,
\[ m_2 = -\frac{1}{\frac{5}{4}} = -\frac{4}{5} \]

### Step 4: Use the point-slope form to find the equation of the line

Using the point-slope form of the equation of a line, which is:
\[ y - y_1 = m(x - x_1) \]
where \((x_1, y_1)\) is the point of intersection \((4, -3)\) and \(m = -\frac{4}{5}\):
\[ y - (-3) = -\frac{4}{5}(x - 4) \]
This simplifies to:
\[ y + 3 = -\frac{4}{5}x + \frac{16}{5} \]
Rearranging gives:
\[ y = -\frac{4}{5}x + \frac{16}{5} - 3 \]
Converting \(-3\) to a fraction with a denominator of 5:
\[ y = -\frac{4}{5}x + \frac{16}{5} - \frac{15}{5} = -\frac{4}{5}x + \frac{1}{5} \]

### Step 5: Write the final equation in standard form

To express this in standard form \(Ax + By + C = 0\):
\[ \frac{4}{5}x + y - \frac{1}{5} = 0 \]
Multiplying through by 5 to eliminate the fraction:
\[ 4x + 5y - 1 = 0 \]

Thus, the final equation of the line is:
\[ \boxed{4x + 5y - 1 = 0} \]