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:

Notice that the mirror line \(x + y + 5 = 0\) is perfectly symmetric with a slope of \(-1\). Because our starting point \((1,1)\) lies directly on the diagonal line \(y = x\), its reflected image must naturally stay locked along that exact same path where \(y = x\). As soon as you discover \(x_2 = -6\), you can instantly infer \(y_2 = -6\). For any point \((k,k)\), its diagonal distance to the origin is always simply \(|k|\sqrt{2}\), which evaluates directly to \(6\sqrt{2}\) in under 5 seconds!

Answer 1

The Correct Option is C

Step 1 : Understanding the Question:
This 2D coordinate geometry problem asks us to find the distance from the origin to the reflection (image) of the point $(1, 1)$ across the mirror line $x + y + 5 = 0$. To solve this, we must first determine the coordinates of the image point. Once the coordinates of this reflected point are established, we can calculate its straight-line distance to the origin $(0, 0)$.
Step 2 : Key Formulas and Approach:
The coordinates of the image point $(x_2, y_2)$ of a given point $(x_1, y_1)$ with respect to a line $ax + by + c = 0$ can be determined using the standard transformation formula:
\[ \frac{x_2 - x_1}{a} = \frac{y_2 - y_1}{b} = -2 \left( \frac{a x_1 + b y_1 + c}{a^2 + b^2} \right) \]
We will substitute the coordinates of the point and the coefficients of the line into this formula to solve for $x_2$ and $y_2$.
After finding the image coordinates, we will compute the distance from the origin $(0, 0)$ to $(x_2, y_2)$ using the standard Euclidean distance formula:
\[ D = \sqrt{x_2^2 + y_2^2} \]
Step 3 : Detailed Explanation:

We start by identifying the parameters from the problem: the given point is $(x_1, y_1) = (1, 1)$, and the equation of the line is $1x + 1y + 5 = 0$. This gives the coefficients $a = 1$, $b = 1$, and $c = 5$.

Next, we calculate the term on the right-hand side of the image formula: $-2 \left( \frac{1(1) + 1(1) + 5}{1^2 + 1^2} \right) = -2 \left( \frac{7}{2} \right) = -7$.

We now set up two linear equations to find the coordinates of the image point $(x_2, y_2)$.

For the x-coordinate, we have: $\frac{x_2 - 1}{1} = -7 \implies x_2 - 1 = -7 \implies x_2 = -6$.

For the y-coordinate, we have: $\frac{y_2 - 1}{1} = -7 \implies y_2 - 1 = -7 \implies y_2 = -6$.

This gives the coordinates of the reflected image point as $(-6, -6)$.

Finally, we calculate the distance from the origin $(0, 0)$ to the image point $(-6, -6)$: $D = \sqrt{(-6 - 0)^2 + (-6 - 0)^2} = \sqrt{(-6)^2 + (-6)^2} = \sqrt{36 + 36} = \sqrt{72}$.

Simplifying the square root of 72 gives: $D = \sqrt{36 \times 2} = 6\sqrt{2}$ units.

Step 4 : Final Answer:
The distance from the origin to the image point is $6\sqrt{2}$, which corresponds to Option (C).