State the equation of the line which has the y-intercept equal to `4/(3)` and is perpendicular to `3x-4y+11=0`

Answer

To find the equation of the line that has a y-intercept of \( \frac{4}{3} \) and is perpendicular to the line given by the equation \( 3x - 4y + 11 = 0 \), we can follow these steps:

### Step 1: Convert the given line equation to slope-intercept form
The given line is \( 3x - 4y + 11 = 0 \). We need to rearrange this into the form \( y = mx + c \), where \( m \) is the slope.

\[ 3x - 4y + 11 = 0 \implies -4y = -3x - 11 \implies y = \frac{3}{4}x + \frac{11}{4} \]

### Step 2: Identify the slope of the given line
From the equation \( y = \frac{3}{4}x + \frac{11}{4} \), we can see that the slope \( m_1 \) of the given line is \( \frac{3}{4} \).

### Step 3: Find the slope of the perpendicular line
For two lines to be perpendicular, the product of their slopes must equal \(-1\). If the slope of the given line is \( m_1 = \frac{3}{4} \), then the slope \( m_2 \) of the line we are looking for can be calculated as follows:

\[ m_1 \cdot m_2 = -1 \implies \frac{3}{4} \cdot m_2 = -1 \implies m_2 = -\frac{4}{3} \]

### Step 4: Write the equation of the required line
We know that the required line has a y-intercept \( c = \frac{4}{3} \) and a slope \( m_2 = -\frac{4}{3} \). We can use the slope-intercept form \( y = mx + c \):

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

### Step 5: Convert to standard form
To convert this equation into standard form \( Ax + By + C = 0 \), we can rearrange it:

\[ y + \frac{4}{3}x - \frac{4}{3} = 0 \]

To eliminate the fractions, we can multiply the entire equation by 3:

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

Rearranging gives:

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

### Final Answer
The equation of the required line is:

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

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