Question

The image of the point $P(2,1)$ on the straight line $2x - 3y + 1 = 0$ is

• A. $\left(\frac{1}{13}, \frac{25}{13}\right)$
• B. $\left(\frac{15}{13}, \frac{25}{13}\right)$
• C. $\left(\frac{18}{13}, \frac{25}{13}\right)$
• D. $\left(\frac{21}{13}, \frac{25}{13}\right)$
• E. $\left(\frac{11}{13}, \frac{15}{13}\right)$

Hint:

For reflection problems, directly applying the formula is faster than geometric construction.

Answer 1

The Correct Option is B

Concept:
Reflection of a point $(x_1,y_1)$ in the line $ax + by + c = 0$ is given by: \[ x' = x_1 - \frac{2a(ax_1 + by_1 + c)}{a^2 + b^2} \] \[ y' = y_1 - \frac{2b(ax_1 + by_1 + c)}{a^2 + b^2} \]

Step 1: Identify values.
\[ a = 2, b = -3, c = 1 \] \[ (x_1, y_1) = (2,1) \]

Step 2: Compute $ax_1 + by_1 + c$.
\[ = 2(2) + (-3)(1) + 1 \] \[ = 4 - 3 + 1 = 2 \]

Step 3: Compute denominator.
\[ a^2 + b^2 = 4 + 9 = 13 \]

Step 4: Find $x'$ coordinate.
\[ x' = 2 - \frac{2 \cdot 2 \cdot 2}{13} \] \[ = 2 - \frac{8}{13} \] \[ = \frac{26 - 8}{13} = \frac{18}{13} \]

Step 5: Find $y'$ coordinate.
\[ y' = 1 - \frac{2(-3)(2)}{13} \] \[ = 1 + \frac{12}{13} \] \[ = \frac{25}{13} \]
Final Answer: \[ \boxed{\left(\frac{18}{13}, \frac{25}{13}\right)} \]