A line intersects the straight lines `5x-y-4=0` and `3x-4y-4=0` at `A` and `B` , respectively. If a point `P(1,5)` on the line `A B` is such that `A P: P B=2:1` (internally), find point `Adot`

Explanation

To solve the problem step by step, we will follow the given instructions and use the section formula to find the coordinates of point A.

### Step 1: Identify the equations of the lines
The equations of the lines are given as:
1. \( 5x - y - 4 = 0 \) (Line 1)
2. \( 3x - 4y - 4 = 0 \) (Line 2)

### Step 2: Find the coordinates of points A and B
Assume point A lies on Line 1. We can express the coordinates of point A as:
- Let \( A(t, 5t - 4) \) where \( t \) is the x-coordinate of point A.

For point B on Line 2, we can express the coordinates as:
- Let \( B(r, \frac{3r - 4}{4}) \) where \( r \) is the x-coordinate of point B.

### Step 3: Use the section formula
Point P divides line segment AB in the ratio \( 2:1 \) internally. The coordinates of point P are given as \( P(1, 5) \).

Using the section formula, the coordinates of point P can be expressed as:
\[ P_x = \frac{m \cdot x_2 + n \cdot x_1}{m + n} \]
\[ P_y = \frac{m \cdot y_2 + n \cdot y_1}{m + n} \]
where \( m = 2 \) and \( n = 1 \).

### Step 4: Set up equations for x and y coordinates
For the x-coordinate:
\[ 1 = \frac{2r + t}{2 + 1} \implies 1 = \frac{2r + t}{3} \implies 2r + t = 3 \quad \text{(Equation 1)} \]

For the y-coordinate:
\[ 5 = \frac{2 \cdot \frac{3r - 4}{4} + 1 \cdot (5t - 4)}{3} \]
Multiplying through by 3:
\[ 15 = 2 \cdot \frac{3r - 4}{4} + (5t - 4) \]
\[ 15 = \frac{3r - 4}{2} + 5t - 4 \]
Multiplying through by 2 to eliminate the fraction:
\[ 30 = 3r - 4 + 10t - 8 \]
\[ 30 = 3r + 10t - 12 \implies 3r + 10t = 42 \quad \text{(Equation 2)} \]

### Step 5: Solve the system of equations
We have the following system of equations:
1. \( 2r + t = 3 \)
2. \( 3r + 10t = 42 \)

From Equation 1, we can express \( t \) in terms of \( r \):
\[ t = 3 - 2r \]

Substituting \( t \) into Equation 2:
\[ 3r + 10(3 - 2r) = 42 \]
\[ 3r + 30 - 20r = 42 \]
\[ -17r + 30 = 42 \]
\[ -17r = 12 \implies r = -\frac{12}{17} \]

Now substituting \( r \) back to find \( t \):
\[ t = 3 - 2\left(-\frac{12}{17}\right) = 3 + \frac{24}{17} = \frac{51}{17} + \frac{24}{17} = \frac{75}{17} \]

### Step 6: Find the coordinates of point A
The coordinates of point A are:
\[ A\left(t, 5t - 4\right) = \left(\frac{75}{17}, 5\left(\frac{75}{17}\right) - 4\right) \]
Calculating the y-coordinate:
\[ 5t - 4 = 5 \cdot \frac{75}{17} - 4 = \frac{375}{17} - \frac{68}{17} = \frac{307}{17} \]
Thus, the coordinates of point A are:
\[ A\left(\frac{75}{17}, \frac{307}{17}\right) \]

### Final Answer
The coordinates of point A are \( \left(\frac{75}{17}, \frac{307}{17}\right) \).

---