The line 4x + 5y + 20 = 0 meets x-axis at point A and y-axis at point B Find :
the co-ordinates of point P in AB such that AB : BP = 5:3.

Answer

To find the coordinates of point P on line segment AB such that the ratio AB:BP = 5:3, we will follow these steps:

### Step 1: Find the coordinates of point A (where the line meets the x-axis)
The equation of the line is given as:
\[ 4x + 5y + 20 = 0 \]

To find point A, we set \( y = 0 \):
\[ 4x + 5(0) + 20 = 0 \]
\[ 4x + 20 = 0 \]
\[ 4x = -20 \]
\[ x = -5 \]

Thus, the coordinates of point A are:
\[ A(-5, 0) \]

### Step 2: Find the coordinates of point B (where the line meets the y-axis)
Now, we find point B by setting \( x = 0 \):
\[ 4(0) + 5y + 20 = 0 \]
\[ 5y + 20 = 0 \]
\[ 5y = -20 \]
\[ y = -4 \]

Thus, the coordinates of point B are:
\[ B(0, -4) \]

### Step 3: Determine the coordinates of point P using the section formula
We need to find point P that divides segment AB in the ratio 5:3. According to the section formula, if a point P divides the line segment joining points A(x1, y1) and B(x2, y2) in the ratio m:n, then the coordinates of point P are given by:
\[ P\left( \frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n} \right) \]

Here, \( A(-5, 0) \) and \( B(0, -4) \), with \( m = 5 \) and \( n = 3 \).

Substituting the values:
\[ P\left( \frac{5(0) + 3(-5)}{5+3}, \frac{5(-4) + 3(0)}{5+3} \right) \]
\[ P\left( \frac{0 - 15}{8}, \frac{-20 + 0}{8} \right) \]
\[ P\left( \frac{-15}{8}, \frac{-20}{8} \right) \]
\[ P\left( -\frac{15}{8}, -\frac{20}{8} \right) \]

Thus, the coordinates of point P are:
\[ P\left( -\frac{15}{8}, -\frac{5}{2} \right) \]

### Final Answer:
The coordinates of point P are:
\[ P\left( -\frac{15}{8}, -\frac{5}{2} \right) \]