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

Concept: To find the coordinates of an image point \((x_2, y_2)\) reflected across a mirror line equation \(ax + by + c = 0\) from an initial point \((x_1, y_1)\), we use the standard vector coordinate projection identity: \[ \frac{x_2 - x_1}{a} = \frac{y_2 - y_1}{b} = -2 \left( \frac{ax_1 + by_1 + c}{a^2 + b^2} \right) \] Once the explicit coordinate components of the reflection image point are isolated, we determine its geometric distance to the origin \((0,0)\) using the basic distance formula: \[ D = \sqrt{x_2^2 + y_2^2} \]

Step 1: Calculating the image coordinates $(x_2, y_2)$.
From the problem parameters, the initial point is \((x_1, y_1) = (1, 1)\) and the line equation is \(1x + 1y + 5 = 0\), giving coefficients \(a = 1, b = 1, c = 5\). Let's evaluate the scaling constant on the right side of our vector formula: \[ \text{Scaling Factor} = -2 \left( \frac{1(1) + 1(1) + 5}{1^2 + 1^2} \right) = -2 \left( \frac{1 + 1 + 5}{1 + 1} \right) = -2 \left( \frac{7}{2} \right) = -7 \] Now, equate this scale value separately to find both spatial components:
• For the x-coordinate: \[ \frac{x_2 - 1}{1} = -7 \quad \implies \quad x_2 - 1 = -7 \quad \implies \quad x_2 = -6 \]
• For the y-coordinate: \[ \frac{y_2 - 1}{1} = -7 \quad \implies \quad y_2 - 1 = -7 \quad \implies \quad y_2 = -6 \] Thus, the coordinates of the image point are precisely \((-6, -6)\).

Step 2: Calculating the distance from the origin $(0,0)$ to the image point $(-6, -6)$.
Using the standard Euclidean 2D distance formula relative to the origin: \[ D = \sqrt{(-6 - 0)^2 + (-6 - 0)^2} \] \[ D = \sqrt{(-6)^2 + (-6)^2} = \sqrt{36 + 36} = \sqrt{72} \] Simplifying the radical into its prime factors: \[ D = \sqrt{36 \times 2} = 6\sqrt{2} \text{ units} \]