A line `5x + 3y + 15 = 0` meets y-axis at point P. Find the co-ordinates of point P. Find the equation of a line through P and perpendicular to `x - 3y + 4 = 0`. 

Answer

To solve the problem step by step, we will first find the coordinates of point P where the line \(5x + 3y + 15 = 0\) meets the y-axis, and then we will find the equation of a line through point P that is perpendicular to the line \(x - 3y + 4 = 0\).

### Step 1: Find the coordinates of point P

1. A line meets the y-axis where \(x = 0\).
2. Substitute \(x = 0\) into the equation of the line \(5x + 3y + 15 = 0\):
\[ 5(0) + 3y + 15 = 0 \]
This simplifies to:
\[ 3y + 15 = 0 \]
3. Solve for \(y\):
\[ 3y = -15 \implies y = \frac{-15}{3} = -5 \]
4. Therefore, the coordinates of point P are:
\[ P(0, -5) \]

### Step 2: Find the slope of the line \(x - 3y + 4 = 0\)

1. Rearrange the equation \(x - 3y + 4 = 0\) into slope-intercept form \(y = mx + c\):
\[ -3y = -x - 4 \implies y = \frac{1}{3}x + \frac{4}{3} \]
2. The slope \(m_2\) of this line is \(\frac{1}{3}\).

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

1. The slope of a line perpendicular to another is the negative reciprocal of the original slope. Therefore, if \(m_2 = \frac{1}{3}\), then:
\[ m_3 = -\frac{1}{m_2} = -3 \]

### Step 4: Write the equation of the line through point P with slope \(m_3\)

1. Use the point-slope form of the equation of a line, which is given by:
\[ y - y_1 = m(x - x_1) \]
Here, \(P(0, -5)\) gives us \(x_1 = 0\) and \(y_1 = -5\), and \(m = -3\).
2. Substitute these values into the point-slope form:
\[ y - (-5) = -3(x - 0) \]
This simplifies to:
\[ y + 5 = -3x \]
3. Rearranging gives:
\[ 3x + y + 5 = 0 \]

### Final Answer

The coordinates of point P are \( (0, -5) \) and the equation of the line through P that is perpendicular to \(x - 3y + 4 = 0\) is:
\[ 3x + y + 5 = 0 \]