Question

The distance from the origin to the image of $(1,1)$ with respect to the line $x + y + 5 = 0$ is:

• A. $7\sqrt{2}$
• B. $3\sqrt{2}$
• C. $6\sqrt{2}$
• D. $4\sqrt{2}$

Hint:

Using the reflection formula helps determine the image of a point easily.

Answer 1

The Correct Option is C

Using the formula for the image of a point $(x_1,y_1)$ with respect to the line $Ax + By + C = 0$: \[ x' = x_1 - \frac{2A(Ax_1 + By_1 + C)}{A^2 + B^2}, \quad y' = y_1 - \frac{2B(Ax_1 + By_1 + C)}{A^2 + B^2} \] Substituting values, we get the image as $(-6,-6)$. The required distance from the origin: \[ D = \sqrt{(-6 - 0)^2 + (-6 - 0)^2} = \sqrt{36 + 36} = 6\sqrt{2} \]

Answer 2

Step 1: Understand the problem
We are given a point P(1, 1) and a line: x + y + 5 = 0.
We are to find the distance from the origin (0, 0) to the image of point P with respect to this line.

Step 2: Use formula for reflection of a point over a line
If a point \( P(x_1, y_1) \) is reflected over the line \( ax + by + c = 0 \), the coordinates of the image \( P'(x', y') \) are:
\[ x' = x_1 - \frac{2a(ax_1 + by_1 + c)}{a^2 + b^2}, \quad y' = y_1 - \frac{2b(ax_1 + by_1 + c)}{a^2 + b^2} \]
Here, a = 1, b = 1, c = 5, and point is (1, 1).

Step 3: Plug values into the formula
First, compute: \( ax_1 + by_1 + c = 1(1) + 1(1) + 5 = 7 \)
Also, \( a^2 + b^2 = 1^2 + 1^2 = 2 \)

Now compute reflected coordinates:
x' = 1 - (2 × 1 × 7) / 2 = 1 - 7 = -6
y' = 1 - (2 × 1 × 7) / 2 = 1 - 7 = -6

So the image of (1, 1) is (–6, –6).

Step 4: Find distance from origin to this image point
Distance from origin = √[ (–6)² + (–6)² ] = √(36 + 36) = √72 = 6√2

Final Answer:
The required distance is 6√2.