The correct answer is **(B)**. To find the coordinates of the reflected point $B(x_2, y_2)$ for a given point $A(x_1, y_1)$ reflected across a line $L: ax + by + c = 0$, we utilize two key properties of reflection: 1. The midpoint of the segment $AB$ lies on the line $L$. 2. The line segment $AB$ is perpendicular to the line $L$. Given point $A = (4, 3)$ and line $L: x - y + 1 = 0$. Let the reflected point be $B = (x_2, y_2)$. **Step 1: Verify if point A lies on the line L.** Substitute the coordinates of $A(4, 3)$ into the equation of line $L$: $4 - 3 + 1 = 2$. Since $2 \neq 0$, point $A$ does not lie on line $L$. Thus, $B$ will be different from $A$. **Step 2: Use the midpoint property.** The midpoint $M$ of $AB$ has coordinates $M = \left(\frac{4+x_2}{2}, \frac{3+y_2}{2}\right)$. Since $M$ lies on line $L$, its coordinates must satisfy the equation of $L$: $\frac{4+x_2}{2} - \frac{3+y_2}{2} + 1 = 0$ Multiply by 2 to clear denominators: $(4+x_2) - (3+y_2) + 2 = 0$ $4 + x_2 - 3 - y_2 + 2 = 0$ $x_2 - y_2 + 3 = 0 \quad \ldots (1)$ **Step 3: Use the perpendicularity property.** The slope of line $L: x - y + 1 = 0$ can be found by rearranging it to $y = x + 1$. The slope $m_L = 1$. The slope of line segment $AB$ is $m_{AB} = \frac{y_2 - 3}{x_2 - 4}$. Since $AB \perp L$, the product of their slopes must be $-1$: $m_{AB} \cdot m_L = -1$ $\left(\frac{y_2 - 3}{x_2 - 4}\right) \cdot (1) = -1$ $y_2 - 3 = -(x_2 - 4)$ $y_2 - 3 = -x_2 + 4$ $x_2 + y_2 = 7 \quad \ldots (2)$ **Step 4: Solve the system of linear equations.** We have two equations with two variables: 1. $x_2 - y_2 = -3$ 2. $x_2 + y_2 = 7$ Adding equation (1) and equation (2): $(x_2 - y_2) + (x_2 + y_2) = -3 + 7$ $2x_2 = 4$ $x_2 = 2$ Substitute $x_2 = 2$ into equation (2): $2 + y_2 = 7$ $y_2 = 5$ Thus, the coordinates of point $B$ are $(2, 5)$. Alternatively, one can use the formula for reflection of a point $(x_1, y_1)$ across a line $ax + by + c = 0$ to a point $(x_2, y_2)$: $\frac{x_2 - x_1}{a} = \frac{y_2 - y_1}{b} = -2 \frac{ax_1 + by_1 + c}{a^2 + b^2}$ Given $A = (4, 3)$, so $x_1 = 4, y_1 = 3$. Given $L: x - y + 1 = 0$, so $a = 1, b = -1, c = 1$. Substitute these values into the formula: $\frac{x_2 - 4}{1} = \frac{y_2 - 3}{-1} = -2 \frac{1(4) + (-1)(3) + 1}{1^2 + (-1)^2}$ $\frac{x_2 - 4}{1} = \frac{y_2 - 3}{-1} = -2 \frac{4 - 3 + 1}{1 + 1}$ $\frac{x_2 - 4}{1} = \frac{y_2 - 3}{-1} = -2 \frac{2}{2}$ $\frac{x_2 - 4}{1} = \frac{y_2 - 3}{-1} = -2$ From the first part: $x_2 - 4 = -2 \implies x_2 = 2$ From the second part: $\frac{y_2 - 3}{-1} = -2 \implies y_2 - 3 = 2 \implies y_2 = 5$ Both methods yield the same result: $B = (2, 5)$. Comparing this with the given options: (A) $(5, 2)$ (B) $(2, 5)$ (C) $(3, 4)$ (D) $(5, 4)$ The correct option is (B).