Question

The image of the point (3, 5) in the line x - y + 1 = 0, lies on :

• A. (x-4)² + (y+2)² = 16
• B. (x-2)² + (y-4)² = 4
• C. (x-4)² + (y-4)² = 8
• D. (x-2)² + (y-2)² = 12

Hint:

The image of a point is found by dropping a perpendicular to the line and moving the same distance on the other side.

Answer 1

The Correct Option is B

Step 1: Image formula: $\frac{x-x_1}{a} = \frac{y-y_1}{b} = -2 \frac{ax_1 + by_1 + c}{a^2 + b^2}$.
Step 2: $\frac{x-3}{1} = \frac{y-5}{-1} = -2 \frac{3-5+1}{1^2 + (-1)^2} = -2 \frac{-1}{2} = 1$.
Step 3: $x-3 = 1 \Rightarrow x=4$ and $y-5 = -1 \Rightarrow y=4$. Image is $(4, 4)$.
Step 4: Check $(4, 4)$ in options: (B) $(4-2)^2 + (4-4)^2 = 2^2 + 0 = 4$. This is correct.