Question

The reflection of the point $(4, -13)$ about the line $5x + y + 6 = 0$ is:

• A. $(-1, -14)$
• B. $(-5, -14)$
• C. $(-1, -5)$
• D. $(-2, -7)$

Hint:

In reflection problems, always compute $ax_1 + by_1 + c$ first. If this value is large or messy, simplify before substitution to avoid arithmetic errors.

Answer 1

The Correct Option is A

Step 1: Identify given values Given point: \[ (x_1, y_1) = (4, -13) \] Given line: \[ 5x + y + 6 = 0 \] Thus, \[ a = 5,\quad b = 1,\quad c = 6 \]

Step 2: Compute substitution expression We first compute: \[ S = ax_1 + by_1 + c \] Substituting values: \[ S = 5(D) + 1(-13) + 6 \] \[ S = 20 - 13 + 6 = 13 \] This scalar value represents the signed perpendicular influence of the point with respect to the line.

Step 3: Compute denominator \[ a^2 + b^2 = 5^2 + 1^2 = 25 + 1 = 26 \]

Step 4: Compute reflected x-coordinate \[ x_2 = 4 - \frac{2 \cdot 5 \cdot 13}{26} \] \[ x_2 = 4 - \frac{130}{26} \] \[ x_2 = 4 - 5 = -1 \] Thus, the x-coordinate of the image point is $-1$.

Step 5: Compute reflected y-coordinate \[ y_2 = -13 - \frac{2 \cdot 1 \cdot 13}{26} \] \[ y_2 = -13 - \frac{26}{26} \] \[ y_2 = -13 - 1 = -14 \] Thus, the y-coordinate of the image point is $-14$.

Step 6: Final verification reasoning The midpoint of $(4,-13)$ and $(-1,-14)$ is: \[ \left(\frac{4 + (-1)}{2}, \frac{-13 + (-14)}{2}\right) = \left(\frac{3}{2}, \frac{-27}{2}\right) \] Substituting into the line: \[ 5\left(\frac{3}{2}\right) + \left(\frac{-27}{2}\right) + 6 \] \[ = \frac{15 - 27 + 12}{2} = 0 \] Since the midpoint satisfies the equation of the line, the reflection is confirmed. Final Answer: \[ {(-1, -14)} \]