Let PS be the median of the triangle with vertices `P(2,""2),""Q(6,-1)"" and ""R(7,""3)` . The equation of the line passing through `(1,-1)` and parallel to PS is (1) `4x-7y-11""=""0` (2) `2x+""9y+""7""=""0` (3) `4x+""7y+""3""=""0` (4) `2x-9y-11""=""0`

Options

A. 4x-7y-1=0
B. 2x+9y+7=0
C. 4x+7y+3=0
D. 2x-9y-11=0

Explanation

To solve the problem, we need to find the equation of the line that passes through the point (1, -1) and is parallel to the median PS of triangle PQR, where the vertices are given as P(2, 2), Q(6, -1), and R(7, 3).

### Step 1: Find the midpoint S of segment QR
The midpoint S of a line segment with endpoints Q(x1, y1) and R(x2, y2) can be calculated using the midpoint formula:

\[ S\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) \]

Substituting the coordinates of Q and R:

\[ S\left(\frac{6 + 7}{2}, \frac{-1 + 3}{2}\right) = S\left(\frac{13}{2}, 1\right) \]

### Step 2: Calculate the slope of the median PS
The slope \( m \) of the line passing through two points \( P(x_1, y_1) \) and \( S(x_2, y_2) \) is given by:

\[ m = \frac{y_2 - y_1}{x_2 - x_1} \]

Using the coordinates of P(2, 2) and S(13/2, 1):

\[ m_{PS} = \frac{1 - 2}{\frac{13}{2} - 2} = \frac{-1}{\frac{13}{2} - \frac{4}{2}} = \frac{-1}{\frac{9}{2}} = -\frac{2}{9} \]

### Step 3: Use the point-slope form to find the equation of line AB
Since line AB is parallel to PS, it will have the same slope. The point-slope form of the equation of a line is given by:

\[ y - y_1 = m(x - x_1) \]

Substituting the point (1, -1) and the slope \( -\frac{2}{9} \):

\[ y - (-1) = -\frac{2}{9}(x - 1) \]

This simplifies to:

\[ y + 1 = -\frac{2}{9}(x - 1) \]

### Step 4: Rearranging the equation
Now, we can rearrange this equation to standard form:

\[ y + 1 = -\frac{2}{9}x + \frac{2}{9} \]

Multiplying through by 9 to eliminate the fraction:

\[ 9y + 9 = -2x + 2 \]

Rearranging gives:

\[ 2x + 9y + 7 = 0 \]

### Conclusion
The equation of the line passing through (1, -1) and parallel to PS is:

\[ 2x + 9y + 7 = 0 \]

This corresponds to option (2).