What is the equation of the line which passes through `(4, -5)` and is perpendicular to `3x+4y+5=0` ?

Options

A. `4x-3y-31=0`
B. `3x-4y-41=0`
C. `4x+3y-1=0`
D. `3x+4y+8=0`

Explanation

To find the equation of the line that passes through the point (4, -5) and is perpendicular to the line given by the equation \(3x + 4y + 5 = 0\), we can follow these steps:

### Step 1: Find the slope of the given line
The equation of the line is in the form \(Ax + By + C = 0\). Here, \(A = 3\), \(B = 4\), and \(C = 5\).

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

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

Thus, the slope of the given line is:

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

### Step 2: Find the slope of the perpendicular line
The slope of a line that is perpendicular to another line is the negative reciprocal of the slope of the original line. Therefore, the slope \(m'\) of the line we are looking for is:

\[ m' = -\frac{1}{m} = -\frac{1}{-\frac{3}{4}} = \frac{4}{3} \]

### Step 3: Use the point-slope form of the equation of a line
Now that we have the slope of the required line (\(\frac{4}{3}\)) and a point it passes through \((4, -5)\), we can use the point-slope form of the equation of a line:

\[ y - y_0 = m'(x - x_0) \]

Substituting \(y_0 = -5\), \(m' = \frac{4}{3}\), and \(x_0 = 4\):

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

This simplifies to:

\[ y + 5 = \frac{4}{3}(x - 4) \]

### Step 4: Simplify the equation
Now, we will simplify the equation:

\[ y + 5 = \frac{4}{3}x - \frac{16}{3} \]

Subtracting 5 from both sides:

\[ y = \frac{4}{3}x - \frac{16}{3} - 5 \]

Converting 5 to a fraction with a denominator of 3:

\[ 5 = \frac{15}{3} \]

So,

\[ y = \frac{4}{3}x - \frac{16}{3} - \frac{15}{3} \]

Combining the constants:

\[ y = \frac{4}{3}x - \frac{31}{3} \]

### Step 5: Rearranging to standard form
To express this in the standard form \(Ax + By + C = 0\), we can multiply through by 3 to eliminate the fraction:

\[ 3y = 4x - 31 \]

Rearranging gives:

\[ 4x - 3y - 31 = 0 \]

Thus, the equation of the line is:

\[ \boxed{4x - 3y - 31 = 0} \]