Coordinates of the points A, B, C,D are (13,7), `(-5,2), (7,3) and (3,7)` respectively. Let AB and CD meet at P, then PA:PB is equal to

Options

A. `6:7`
B. `6:7`
C. `7:6`
D. None of these

Answer

To solve the problem, we need to find the point of intersection \( P \) of the lines \( AB \) and \( CD \) and then determine the ratio \( PA:PB \).

### Step 1: Determine the equations of lines AB and CD

**Coordinates:**
- \( A(13, 7) \)
- \( B(-5, 2) \)
- \( C(7, 3) \)
- \( D(3, 7) \)

**Equation of line AB:**
Using the two-point form of the equation of a line, we have:
\[ y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1) \]
Substituting \( A(13, 7) \) and \( B(-5, 2) \):
\[ y - 7 = \frac{2 - 7}{-5 - 13}(x - 13) \]
\[ y - 7 = \frac{-5}{-18}(x - 13) \]
\[ y - 7 = \frac{5}{18}(x - 13) \]
Multiplying through by 18 to eliminate the fraction:
\[ 18(y - 7) = 5(x - 13) \]
\[ 18y - 126 = 5x - 65 \]
Rearranging gives:
\[ 5x - 18y + 61 = 0 \quad \text{(Equation 1)} \]

**Equation of line CD:**
Using the same method for points \( C(7, 3) \) and \( D(3, 7) \):
\[ y - 3 = \frac{7 - 3}{3 - 7}(x - 7) \]
\[ y - 3 = \frac{4}{-4}(x - 7) \]
\[ y - 3 = -1(x - 7) \]
\[ y - 3 = -x + 7 \]
Rearranging gives:
\[ x + y - 10 = 0 \quad \text{(Equation 2)} \]

### Step 2: Find the intersection point P of lines AB and CD

To find \( P \), we solve the system of equations formed by Equation 1 and Equation 2:
1. \( 5x - 18y + 61 = 0 \)
2. \( x + y - 10 = 0 \)

From Equation 2, we can express \( x \) in terms of \( y \):
\[ x = 10 - y \]
Substituting this into Equation 1:
\[ 5(10 - y) - 18y + 61 = 0 \]
\[ 50 - 5y - 18y + 61 = 0 \]
\[ 111 - 23y = 0 \]
\[ 23y = 111 \implies y = \frac{111}{23} \]
Substituting \( y \) back to find \( x \):
\[ x = 10 - \frac{111}{23} = \frac{230 - 111}{23} = \frac{119}{23} \]
Thus, the coordinates of point \( P \) are \( P\left(\frac{119}{23}, \frac{111}{23}\right) \).

### Step 3: Find the ratio \( PA:PB \)

Using the section formula, if \( P \) divides \( AB \) in the ratio \( m:n \), then:
\[ P = \left(\frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n}\right) \]
Let \( PA:PB = m:n \). We can set \( m = k \) and \( n = 1 \) for simplicity.

Using the coordinates of \( A \) and \( B \):
\[ \frac{(k \cdot (-5) + 1 \cdot 13)}{k + 1} = \frac{119}{23} \]
\[ \frac{(-5k + 13)}{k + 1} = \frac{119}{23} \]
Cross-multiplying:
\[ 23(-5k + 13) = 119(k + 1) \]
\[ -115k + 299 = 119k + 119 \]
\[ 299 - 119 = 119k + 115k \]
\[ 180 = 234k \implies k = \frac{180}{234} = \frac{30}{39} = \frac{10}{13} \]

Thus, the ratio \( PA:PB \) is \( 10:13 \).

### Final Answer:
The ratio \( PA:PB \) is \( 10:13 \).

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