Let PS be the median of the triangle with vertices P(2,2), Q(6,-1) an R(7,3). The equation of the line passing through (1,-1) and parallel to PS is

Options

A. `2x-9y-7=0`
B. `2x-9y-11=0`
C. `2x+9y-11=0`
D. `2x+9y+7=0`

Answer

To solve the problem, we need to find the equation of the line passing through the point (1, -1) and parallel to the median PS of triangle PQR with vertices P(2, 2), Q(6, -1), and R(7, 3).

### Step 1: Find the coordinates of the midpoint S of side QR
The midpoint S of the line segment QR can be calculated using the midpoint formula:
\[ S\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) \]
Here, \(Q(6, -1)\) and \(R(7, 3)\):
\[ S\left(\frac{6 + 7}{2}, \frac{-1 + 3}{2}\right) = S\left(\frac{13}{2}, \frac{2}{2}\right) = S\left(\frac{13}{2}, 1\right) \]

### Step 2: Find the slope of the median PS
The slope of line PS can be calculated using the formula:
\[ \text{slope} = \frac{y_2 - y_1}{x_2 - x_1} \]
Using points P(2, 2) and S\(\left(\frac{13}{2}, 1\right)\):
\[ \text{slope of PS} = \frac{1 - 2}{\frac{13}{2} - 2} = \frac{-1}{\frac{13}{2} - \frac{4}{2}} = \frac{-1}{\frac{9}{2}} = -\frac{2}{9} \]

### Step 3: Write the equation of the line parallel to PS through point (1, -1)
Since the line we are looking for is parallel to PS, it will have the same slope, which is \(-\frac{2}{9}\). We can use the point-slope form of the equation of a line:
\[ y - y_1 = m(x - x_1) \]
Substituting \(m = -\frac{2}{9}\), \(x_1 = 1\), and \(y_1 = -1\):
\[ y - (-1) = -\frac{2}{9}(x - 1) \]
This simplifies to:
\[ y + 1 = -\frac{2}{9}(x - 1) \]

### Step 4: Rearranging the equation
Now, we can rearrange this equation:
\[ y + 1 = -\frac{2}{9}x + \frac{2}{9} \]
Subtracting 1 from both sides:
\[ y = -\frac{2}{9}x + \frac{2}{9} - 1 \]
Converting -1 to a fraction:
\[ y = -\frac{2}{9}x + \frac{2}{9} - \frac{9}{9} \]
This simplifies to:
\[ y = -\frac{2}{9}x - \frac{7}{9} \]

### Step 5: Writing in standard form
To write this in standard form \(Ax + By + C = 0\):
\[ \frac{2}{9}x + y + \frac{7}{9} = 0 \]
Multiplying through by 9 to eliminate the fraction:
\[ 2x + 9y + 7 = 0 \]

### Final Answer
The equation of the line passing through (1, -1) and parallel to PS is:
\[ 2x + 9y + 7 = 0 \]