Let the coordinates of two points P and Q be (1, 2) and (7, 5) respectively. The line PQ is rotated thorugh `315^(@)` is clockwise direction about the point of trisection of PQ which is nearer to Q. The equation of the line in the new position is

Options

A. `2x-y-6=0`
B. `x-y-1=0`
C. `3x-y-11=0`
D. `3x-y-9=0`

Answer

To solve the problem step by step, we will follow these instructions:

### Step 1: Find the coordinates of points P and Q
Given:
- Point P = (1, 2)
- Point Q = (7, 5)

### Step 2: Find the point of trisection nearer to Q
The coordinates of the point of trisection can be found using the formula for the trisection point. The point of trisection that is nearer to Q divides the segment PQ in the ratio 2:1.

Using the section formula:
\[ \text{Trisection point} = \left( \frac{2x_2 + x_1}{3}, \frac{2y_2 + y_1}{3} \right) \]
Substituting \( P(1, 2) \) and \( Q(7, 5) \):
\[ \text{Trisection point} = \left( \frac{2 \cdot 7 + 1}{3}, \frac{2 \cdot 5 + 2}{3} \right) = \left( \frac{14 + 1}{3}, \frac{10 + 2}{3} \right) = \left( \frac{15}{3}, \frac{12}{3} \right) = (5, 4) \]

### Step 3: Find the slope of line PQ
The slope \( m \) of line PQ can be calculated using the formula:
\[ m = \frac{y_2 - y_1}{x_2 - x_1} \]
Substituting the coordinates of P and Q:
\[ m = \frac{5 - 2}{7 - 1} = \frac{3}{6} = \frac{1}{2} \]

### Step 4: Determine the angle of rotation
The line PQ is rotated through \( 315^\circ \) clockwise. To find the equivalent counterclockwise angle:
\[ \text{Counterclockwise angle} = 360^\circ - 315^\circ = 45^\circ \]

### Step 5: Calculate the new slope after rotation
Using the formula for the slope after rotation:
\[ m' = \frac{m + \tan(\theta)}{1 - m \tan(\theta)} \]
where \( m = \frac{1}{2} \) and \( \theta = 45^\circ \) (where \( \tan(45^\circ) = 1 \)):
\[ m' = \frac{\frac{1}{2} + 1}{1 - \frac{1}{2} \cdot 1} = \frac{\frac{3}{2}}{\frac{1}{2}} = 3 \]

### Step 6: Write the equation of the new line
Using the point-slope form of the equation of a line:
\[ y - y_1 = m'(x - x_1) \]
Substituting the point (5, 4) and the new slope (3):
\[ y - 4 = 3(x - 5) \]
Expanding this:
\[ y - 4 = 3x - 15 \]
\[ y = 3x - 11 \]

### Step 7: Write the final equation in standard form
Rearranging the equation:
\[ 3x - y - 11 = 0 \]

### Final Answer
The equation of the line in the new position is:
\[ 3x - y - 11 = 0 \]