Question

Let a ray of light passing through the point \((3, 10)\) reflects on the line \(2x + y = 6\) and the reflected ray passes through the point \((7, 2)\). If the equation of the incident ray is \(ax + by + 1 = 0\), then \(a^2 + b^2 + 3ab\) is equal to _.

Answer 1

Correct Answer: 1

For \( B' \):
\[ \frac{x - 7}{2} = \frac{y - 2}{1} = -2 \left( \frac{14 + 2 - 6}{5} \right) \] Simplify:
\[ \frac{x - 7}{2} = \frac{y - 2}{1} = -4 \] \[ x - 7 = -8 \quad \Rightarrow \quad x = -1 \] \[ y - 2 = -4 \quad \Rightarrow \quad y = -2 \] Hence, the coordinates of \( B' \) are:
\[ B'(-1, -2) \]

Incident ray \( AB' \):
Slope of \( AB' \) is:
\[ M_{AB'} = 3 \]

Equation of the line through \( A(-1, -2) \) with slope \( 3 \):
\[ y + 2 = 3(x + 1) \] Simplify:
\[ 3x - y + 1 = 0 \]

Let \( a = 3, \, b = -1 \)
Then:
\[ a^2 + b^2 + 3ab = 9 + 1 - 9 = 1 \]

Final Answer:
\[ a^2 + b^2 + 3ab = 1 \]

Answer 2

Given: For \( B' \):

\[ \frac{x - 7}{2} = \frac{y - 2}{1} = -2 \left( \frac{14 + 2 - 6}{5} \right) \]

\[ \frac{x - 7}{2} = \frac{y - 2}{1} = -4 \]

\[ x = -1, \quad y = -2 \implies B'(-1, -2) \]

Incident ray \( AB' \):

\[ M_{AB'} = 3 \]

\[ y + 2 = 3(x + 1) \]

\[ 3x - y + 1 = 0 \]

\[ a = 3, \quad b = -1 \]

\[ a^2 + b^2 + 3ab = 9 + 1 - 9 = 1 \]

Answer: 1