The point P(2,1) is shifted through a distance `3sqrt(2)` units measured parallel to the line x+y=1 in the direction of decreasing ordinates, to reach at Q. The image of Q with respect to given line is

Options

A. (3,-4)
B. (-3,2)
C. (0,-1)
D. none of these

Explanation

To solve the given problem step by step, we will follow the instructions provided in the video transcript while ensuring clarity in each step.

### Step 1: Identify the Given Information
We have the following information:
- Point P = (2, 1)
- Distance = \(3\sqrt{2}\)
- Line equation = \(x + y = 1\)

### Step 2: Find the Direction of the Line
The line \(x + y = 1\) can be rewritten in slope-intercept form as:
\[ y = -x + 1 \]
From this, we can identify the slope \(m = -1\).

### Step 3: Determine the Angle of the Line
The angle \(\theta\) corresponding to the slope can be calculated:
\[ \tan(\theta) = -1 \]
This means:
\[ \theta = 135^\circ \]

### Step 4: Calculate the Cosine and Sine Values
Using the angle \(\theta = 135^\circ\):
- \(\cos(135^\circ) = -\frac{1}{\sqrt{2}}\)
- \(\sin(135^\circ) = \frac{1}{\sqrt{2}}\)

### Step 5: Shift the Point P to Point Q
We need to move from point P(2, 1) in the direction of decreasing ordinates (downward) by a distance of \(3\sqrt{2}\).

Using the parametric equations:
\[ x = x_1 + d \cdot \cos(\theta) \]
\[ y = y_1 + d \cdot \sin(\theta) \]
where \(d = -3\sqrt{2}\) (negative because we are moving downward).

Substituting the values:
\[ x_Q = 2 + (-3\sqrt{2}) \cdot \left(-\frac{1}{\sqrt{2}}\right) = 2 + 3 = 5 \]
\[ y_Q = 1 + (-3\sqrt{2}) \cdot \left(\frac{1}{\sqrt{2}}\right) = 1 - 3 = -2 \]

Thus, the coordinates of point Q are:
\[ Q = (5, -2) \]

### Step 6: Find the Image of Point Q with Respect to the Line
To find the image of point Q with respect to the line \(x + y = 1\), we will use the formula:
\[ \frac{H - x_1}{a} = \frac{K - y_1}{b} = \frac{d}{\sqrt{a^2 + b^2}} \]
where \(a = 1\), \(b = 1\) (coefficients of the line equation), and \(d\) is the perpendicular distance from point Q to the line.

### Step 7: Calculate the Perpendicular Distance
The distance \(d\) from point Q(5, -2) to the line \(x + y - 1 = 0\) is given by:
\[ d = \frac{|1(5) + 1(-2) - 1|}{\sqrt{1^2 + 1^2}} = \frac{|5 - 2 - 1|}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2} \]

### Step 8: Find the Coordinates of the Image
Using the formula for the image:
\[ H = x_Q - 2 \cdot \frac{a \cdot d}{\sqrt{a^2 + b^2}} = 5 - 2 \cdot \frac{1 \cdot \sqrt{2}}{\sqrt{2}} = 5 - 2 = 3 \]
\[ K = y_Q - 2 \cdot \frac{b \cdot d}{\sqrt{a^2 + b^2}} = -2 - 2 \cdot \frac{1 \cdot \sqrt{2}}{\sqrt{2}} = -2 - 2 = -4 \]

Thus, the coordinates of the image \(Q'\) are:
\[ Q' = (3, -4) \]

### Final Answer
The image of point Q with respect to the line \(x + y = 1\) is:
\[ \boxed{(3, -4)} \]