Let L be the line y = 2x , in the two dimensional plane .
Statement 1 : The image of the point (0,1) in L is the point (4/5 , 3/5)
Statement 2 : The points (0,1) and (4/5 , 3/5) lie on opposite sides of the line L and are at equal distance from it .

Options

A. Statement -1 is True , Statement - 2 is true , Statement- 2 is a correct explanation for statement - 15
B. Statement-1 is True , Statement-2 is True , Statement -2 is not a correct explanation for Statement - 1 .
C. Statement-1 is True , Statement - 2 is False .
D. Statement - 1 is False , Statement -2 is True .

Explanation

To solve the problem, we need to analyze the two statements regarding the line \( L: y = 2x \) and the points \( A(0, 1) \) and \( B\left(\frac{4}{5}, \frac{3}{5}\right) \).

### Step 1: Verify Statement 1
We need to check if the point \( B\left(\frac{4}{5}, \frac{3}{5}\right) \) is indeed the image of point \( A(0, 1) \) with respect to the line \( L \).

1. **Find the equation of the line perpendicular to \( L \) that passes through point \( A(0, 1) \)**:
- The slope of line \( L \) is \( 2 \). Therefore, the slope of the line perpendicular to \( L \) is \( -\frac{1}{2} \).
- The equation of the line passing through \( A(0, 1) \) with slope \( -\frac{1}{2} \) is:
\[ y - 1 = -\frac{1}{2}(x - 0) \implies y = -\frac{1}{2}x + 1 \]

2. **Find the intersection point of this perpendicular line with line \( L \)**:
- Set the equations equal to find the intersection:
\[ 2x = -\frac{1}{2}x + 1 \]
\[ 2.5x = 1 \implies x = \frac{2}{5} \]
- Substitute \( x \) back into the equation of line \( L \) to find \( y \):
\[ y = 2\left(\frac{2}{5}\right) = \frac{4}{5} \]
- Thus, the intersection point \( P \) is \( \left(\frac{2}{5}, \frac{4}{5}\right) \).

3. **Find the image point \( B \)**:
- The image point \( B \) can be found using the midpoint formula:
\[ P = \left(\frac{x_A + x_B}{2}, \frac{y_A + y_B}{2}\right) \]
- We know \( P = \left(\frac{2}{5}, \frac{4}{5}\right) \) and \( A(0, 1) \):
\[ \frac{0 + x_B}{2} = \frac{2}{5} \implies x_B = \frac{4}{5} \]
\[ \frac{1 + y_B}{2} = \frac{4}{5} \implies 1 + y_B = \frac{8}{5} \implies y_B = \frac{3}{5} \]
- Thus, the image point \( B \) is \( \left(\frac{4}{5}, \frac{3}{5}\right) \).

### Conclusion for Statement 1:
Since we have confirmed that the image of point \( A(0, 1) \) with respect to line \( L \) is indeed \( B\left(\frac{4}{5}, \frac{3}{5}\right) \), **Statement 1 is true**.

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### Step 2: Verify Statement 2
Now we need to check if points \( A(0, 1) \) and \( B\left(\frac{4}{5}, \frac{3}{5}\right) \) lie on opposite sides of line \( L \) and are at equal distances from it.

1. **Calculate the distance from point \( A(0, 1) \) to line \( L: y = 2x \)**:
- The formula for the distance \( d \) from a point \( (x_0, y_0) \) to the line \( Ax + By + C = 0 \) is:
\[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]
- Rewriting \( y = 2x \) in standard form gives \( 2x - y = 0 \) (where \( A = 2, B = -1, C = 0 \)):
\[ d_A = \frac{|2(0) - 1(1) + 0|}{\sqrt{2^2 + (-1)^2}} = \frac{|0 - 1|}{\sqrt{4 + 1}} = \frac{1}{\sqrt{5}} = \frac{1}{\sqrt{5}} \]

2. **Calculate the distance from point \( B\left(\frac{4}{5}, \frac{3}{5}\right) \) to line \( L \)**:
\[ d_B = \frac{|2\left(\frac{4}{5}\right) - 1\left(\frac{3}{5}\right) + 0|}{\sqrt{2^2 + (-1)^2}} = \frac{| \frac{8}{5} - \frac{3}{5}|}{\sqrt{5}} = \frac{\frac{5}{5}}{\sqrt{5}} = \frac{1}{\sqrt{5}} \]

### Conclusion for Statement 2:
Since both distances \( d_A \) and \( d_B \) are equal, and point \( A(0, 1) \) lies above the line while point \( B\left(\frac{4}{5}, \frac{3}{5}\right) \) lies below the line, **Statement 2 is also true**.

### Final Conclusion:
Both statements are true, and Statement 2 does not provide a correct explanation for Statement 1.

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