If the image of the point (2,1) in a line is (4,3) then the equation of line is
(i) `x+y+5=0`
(ii) `x-y+5=0`
(iii) `x+y-5=0`
(iv) `x-y-5=0`

Options

A. `x+y+5=0`
B. `x-y+5=0`
C. `x+y-5=0`
D. `x-y-5=0`

Answer

To find the equation of the line given that the image of the point (2,1) is (4,3), we can follow these steps:

### Step 1: Find the midpoint of the segment connecting the point and its image.
The midpoint \( R \) of points \( P(2, 1) \) and \( Q(4, 3) \) can be calculated using the midpoint formula:

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

Substituting the coordinates:

\[ R = \left( \frac{2 + 4}{2}, \frac{1 + 3}{2} \right) = \left( \frac{6}{2}, \frac{4}{2} \right) = (3, 2) \]

### Step 2: Calculate the slope of the line segment \( PQ \).
The slope \( m \) of the line segment connecting points \( P(2, 1) \) and \( Q(4, 3) \) is given by:

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

Substituting the coordinates:

\[ m = \frac{3 - 1}{4 - 2} = \frac{2}{2} = 1 \]

### Step 3: Determine the slope of the line \( LM \).
Since the line \( LM \) is perpendicular to the line segment \( PQ \), the product of their slopes is -1. Therefore, if the slope of \( PQ \) is 1, the slope of \( LM \) (denoted as \( m_{LM} \)) is:

\[ m_{LM} = -\frac{1}{m_{PQ}} = -\frac{1}{1} = -1 \]

### Step 4: Use the point-slope form to find the equation of line \( LM \).
The point-slope form of a line is given by:

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

Using point \( R(3, 2) \) and slope \( m_{LM} = -1 \):

\[ y - 2 = -1(x - 3) \]

### Step 5: Simplify the equation.
Expanding and rearranging gives:

\[ y - 2 = -x + 3 \]

Adding \( x \) and 2 to both sides:

\[ x + y - 5 = 0 \]

### Final Equation:
Thus, the equation of the line is:

\[ x + y - 5 = 0 \]

This corresponds to option (iii).

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