Key Concepts and Formulas

• Reflection of a point across a line: The image $(x', y')$ of a point $(x_1, y_1)$ across the line $Ax + By + C = 0$ is given by: $\frac{x' - x_1}{A} = \frac{y' - y_1}{B} = \frac{-2(Ax_1 + By_1 + C)}{A^2 + B^2}$
• Two-point form of a line: The equation of a line passing through points $(x_1, y_1)$ and $(x_2, y_2)$ is given by: $y - y_1 = \frac{y_2 - y_1}{x_2 - x_1} (x - x_1)$
• Law of Reflection: The angle of incidence is equal to the angle of reflection. This implies that the incident ray, the point of reflection, and the image of the reflected point with respect to the reflecting line are collinear.

Step-by-Step Solution

Step 1: Find the image of point B(7, 2) with respect to the line 2x + y = 6

We need to find the coordinates of the image $B'=(x', y')$ of point $B=(7,2)$ with respect to the line $L: 2x+y-6=0$. We will use the formula for the image of a point across a line.

Here, $(x_1, y_1) = (7, 2)$, and $A=2, B=1, C=-6$.

First, calculate $Ax_1 + By_1 + C$: $Ax_1 + By_1 + C = 2(7) + 1(2) - 6 = 14 + 2 - 6 = 10$

Next, calculate $A^2 + B^2$: $A^2 + B^2 = 2^2 + 1^2 = 4 + 1 = 5$

Now, substitute these values into the image formula: $\frac{x' - 7}{2} = \frac{y' - 2}{1} = \frac{-2(10)}{5}$ $\frac{x' - 7}{2} = \frac{y' - 2}{1} = -4$

Equating each part to $-4$ to find $x'$ and $y'$:

1. For $x'$: $\frac{x' - 7}{2} = -4$ $x' - 7 = -8$ $x' = -8 + 7 = -1$
2. For $y'$: $\frac{y' - 2}{1} = -4$ $y' - 2 = -4$ $y' = -4 + 2 = -2$

So, the image of point $B=(7,2)$ with respect to the line $2x+y-6=0$ is $B' = (-1,-2)$.

Why this step? The reflection principle tells us that the incident ray will pass through the image of the reflected point. Finding this image simplifies the problem to finding the equation of a line through two points.

Step 2: Determine the equation of the incident ray passing through A(3, 10) and B'(-1, -2)

The incident ray passes through point $A=(3,10)$ and the image point $B'=(-1,-2)$. We can find the equation of this line using the two-point form.

Using the formula for the equation of a line passing through $(x_1, y_1)$ and $(x_2, y_2)$: Let $(x_1, y_1) = (3,10)$ and $(x_2, y_2) = (-1,-2)$.

Substitute the coordinates into the formula: $y - 10 = \frac{-2 - 10}{-1 - 3} (x - 3)$ $y - 10 = \frac{-12}{-4} (x - 3)$ $y - 10 = 3 (x - 3)$

Now, simplify the equation to the standard form $Ax+By+C=0$: $y - 10 = 3x - 9$ $3x - y - 9 + 10 = 0$ $3x - y + 1 = 0$

Thus, the equation of the incident ray is $3x - y + 1 = 0$.

Why this step? Knowing two points on the incident ray allows us to directly calculate its equation using the standard two-point form.

Step 3: Identify the values of a and b

We are given that the equation of the incident ray is $ax+by+1=0$. From our calculation, the equation of the incident ray is $3x-y+1=0$.

By comparing these two equations, we can directly equate the coefficients of $x$ and $y$:

• $a = 3$
• $b = -1$

Why this step? We need to extract the values of a and b to compute the final expression. This step is a straightforward coefficient comparison.

Step 4: Calculate the value of a^2 + b^2 + 3ab

The problem asks for the value of $a^2+b^2+3ab$. Substitute the values $a=3$ and $b=-1$ into the expression: $a^2+b^2+3ab = (3)^2 + (-1)^2 + 3(3)(-1)$ $= 9 + 1 - 9$ $= 1$

Why this step? This is the final computation to obtain the answer.

Common Mistakes & Tips

• Sign errors: Be careful with signs when applying the image formula and the two-point form.
• Incorrect Image Formula: Make sure to remember the correct formula for finding the image of a point.
• Not simplifying the equation: Ensure that the equation of the line is simplified before comparing coefficients.

Summary

The problem involves finding the equation of an incident ray after reflection. We used the reflection principle to find the image of a point on the reflected ray with respect to the reflecting line. Then, we found the equation of the line passing through the initial point and the image point, which represents the incident ray. Finally, we compared coefficients to find the values of $a$ and $b$ and calculated the required expression.

The final answer is \boxed{1}. The correct option is 1.