If the image of point P(2, 3) in a line L is Q(4, 5), then the image of point R(0, 0) in the same line is:

To solve this problem, we are given that the image of point P(2, 3) in a line L is Q(4, 5). We need to find the image of point R(0, 0) in the same line L. The image of a point in a line means that the line acts as the perpendicular bisector of the segment joining the point and its image. Therefore, for any point and its image, the line L must satisfy two conditions: it passes through the midpoint of the segment, and it is perpendicular to the segment.

First, we find the equation of line L using points P(2, 3) and Q(4, 5). The midpoint of segment PQ is calculated as follows:

Midpoint M = $$\left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) = \left( \frac{2 + 4}{2}, \frac{3 + 5}{2} \right) = \left( \frac{6}{2}, \frac{8}{2} \right) = (3, 4)$$.

Next, we find the slope of PQ. The slope is given by the change in y-coordinates divided by the change in x-coordinates:

Slope of PQ = $$\frac{y_2 - y_1}{x_2 - x_1} = \frac{5 - 3}{4 - 2} = \frac{2}{2} = 1$$.

Since line L is perpendicular to PQ, its slope is the negative reciprocal of the slope of PQ. Therefore, slope of L = $$-\frac{1}{1} = -1$$.

Now, line L passes through the midpoint M(3, 4) and has a slope of -1. Using the point-slope form of a line equation:

$$y - y_1 = m(x - x_1)$$

Substituting the values:

$$y - 4 = -1(x - 3)$$

Simplifying:

$$y - 4 = -x + 3$$

$$y = -x + 3 + 4$$

$$y = -x + 7$$

So, the equation of line L is $$y = -x + 7$$, which can be rewritten as $$x + y - 7 = 0$$.

Now, we need to find the image of point R(0, 0) in this line. Let the image be S(x, y). Since L is the perpendicular bisector of segment RS, the midpoint of RS must lie on L, and RS must be perpendicular to L.

First, the midpoint of RS (with R(0, 0) and S(x, y)) is:

Midpoint = $$\left( \frac{0 + x}{2}, \frac{0 + y}{2} \right) = \left( \frac{x}{2}, \frac{y}{2} \right)$$.

This midpoint must satisfy the equation of L: $$x + y - 7 = 0$$. Substituting the midpoint coordinates:

$$\frac{x}{2} + \frac{y}{2} - 7 = 0$$

Multiplying both sides by 2 to eliminate denominators:

$$x + y - 14 = 0 \quad \text{(Equation 1)}$$.

Next, RS must be perpendicular to L. The slope of L is -1 (from the equation $$y = -x + 7$$). The slope of RS is given by $$\frac{y - 0}{x - 0} = \frac{y}{x}$$.

Since the lines are perpendicular, the product of their slopes is -1:

$$(\text{slope of L}) \times (\text{slope of RS}) = -1$$

$$(-1) \times \left( \frac{y}{x} \right) = -1$$

$$-\frac{y}{x} = -1$$

Multiplying both sides by -1:

$$\frac{y}{x} = 1$$

$$y = x \quad \text{(Equation 2)}$$.

Now, we substitute Equation 2 into Equation 1:

$$x + y - 14 = 0$$

$$x + x - 14 = 0$$ (since y = x)

$$2x - 14 = 0$$

$$2x = 14$$

$$x = 7$$

Then, from Equation 2, y = x = 7.

Therefore, the image of R(0, 0) is (7, 7).

Comparing with the options:

A. (2, 2)

B. (4, 5)

C. (3, 4)

D. (7, 7)

The point (7, 7) corresponds to Option D.

Hence, the correct answer is Option D.