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 the outlined process in the transcript:

### Step 1: Identify 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 point of trisection R divides the line segment PQ in the ratio 2:1 (since it is nearer to Q). We can use the section formula to find the coordinates of R.

Using the section formula:
\[ R = \left(\frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n}\right) \]
where \(m = 2\), \(n = 1\), \(P(x_1, y_1) = (1, 2)\), and \(Q(x_2, y_2) = (7, 5)\).

Calculating the x-coordinate of R:
\[ x_R = \frac{2 \cdot 7 + 1 \cdot 1}{2 + 1} = \frac{14 + 1}{3} = \frac{15}{3} = 5 \]

Calculating the y-coordinate of R:
\[ y_R = \frac{2 \cdot 5 + 1 \cdot 2}{2 + 1} = \frac{10 + 2}{3} = \frac{12}{3} = 4 \]

Thus, the coordinates of point R are \(R(5, 4)\).

### 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: Calculate the new slope after rotation
The line is rotated \(315^\circ\) clockwise. To find the new slope, we first convert this to an angle measured counterclockwise:
\[ 360^\circ - 315^\circ = 45^\circ \]

Using the formula for the tangent of the sum of angles:
\[ \tan(\theta_1) = \frac{\tan(45^\circ) + \tan(\theta_2)}{1 - \tan(45^\circ) \tan(\theta_2)} \]
where \(\tan(45^\circ) = 1\) and \(\tan(\theta_2) = \frac{1}{2}\).

Substituting the values:
\[ \tan(\theta_1) = \frac{1 + \frac{1}{2}}{1 - 1 \cdot \frac{1}{2}} = \frac{\frac{3}{2}}{\frac{1}{2}} = 3 \]

### Step 5: Write the equation of the new line
Now we have the slope of the new line \(m = 3\) and it passes through point R(5, 4). Using the point-slope form of the equation of a line:
\[ y - y_1 = m(x - x_1) \]
Substituting the values:
\[ y - 4 = 3(x - 5) \]

Expanding this:
\[ y - 4 = 3x - 15 \]
\[ y = 3x - 11 \]

Rearranging gives:
\[ 3x - y - 11 = 0 \]

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