Question

A ray of light incident along a line meets another line $7x - y + 1 = 0$ at the point $(0,1)$ and is then reflected along the line $y + 2x = 1$. Then the equation of the line of inciden is

• A. $41x - 25y + 25 = 0$
• B. $41x + 38y - 38 = 0$
• C. $41x - 38y + 38 = 0$
• D. $41x + 25y - 25 = 0$

Hint:

Light Reflection: Use angle formulas for slope differences and apply perpendicularity condition. If slope of mirror is $m$, reflected and incident rays make equal angles with the mirror.

Answer 1

The Correct Option is C

Mirror line: $7x - y + 1 = 0 \Rightarrow y = 7x + 1 \Rightarrow \text{slope } m = 7$
Reflected line: $y + 2x = 1 \Rightarrow y = -2x + 1 \Rightarrow \text{slope } m_R = -2$ Angle between reflected ray and mirror: \[ \tan\theta = \left|\frac{-2 - 7}{1 + (-2)(7)}\right| = \left|\frac{-9}{-13}\right| = \frac{9}{13} \] Let slope of incidence line be $m_I$. Then: \[ \left|\frac{m_I - 7}{1 + 7m_I}\right| = \frac{9}{13} \] Solving: \[ \frac{m_I - 7}{1 + 7m_I} = -\frac{9}{13} \Rightarrow m_I = \frac{41}{38} \] It passes through $(0,1)$: \[ y - 1 = \frac{41}{38}x \Rightarrow 38y - 38 = 41x \Rightarrow 41x - 38y + 38 = 0 \]

Answer 2

Step 1: Identify given data
The ray of light hits the line \(7x - y + 1 = 0\) at point \( (0,1) \). After reflection, it travels along the line \( y + 2x = 1 \). We need to find the equation of the incident ray's line.

Step 2: Find the slope of the reflecting line
Rewrite \( y + 2x = 1 \) as \( y = -2x + 1 \), so slope \( m_r = -2 \).

Step 3: Find slope of the reflecting surface
Line \(7x - y + 1 = 0 \Rightarrow y = 7x + 1\), slope of surface \( m_s = 7 \).

Step 4: Use reflection law for slopes
Let \( m_i \) be slope of incident ray.
The formula relating slopes for reflection is:
\[ m_r = \frac{2 m_s + m_i - m_s^2 m_i}{1 + 2 m_s m_i - m_s^2} \] Or equivalently:
\[ m_r = \frac{2 m_s - m_i (1 - m_s^2)}{1 + 2 m_s m_i - m_s^2} \] But the easier formula for reflection is:
\[ \tan \theta_r = \tan(2\alpha - \theta_i) \] where \(\alpha = \arctan m_s\), \(\theta_i = \arctan m_i\), \(\theta_r = \arctan m_r\).

Step 5: Calculate angles
\[ \alpha = \arctan(7) \] \[ \theta_r = \arctan(-2) \] Reflection formula:
\[ \theta_r = 2\alpha - \theta_i \implies \theta_i = 2\alpha - \theta_r \]

Step 6: Calculate \(\theta_i\)
\[ \theta_i = 2 \arctan(7) - \arctan(-2) \] Calculate approximate values:
\(\arctan(7) \approx 81.87^\circ\)
\(\arctan(-2) \approx -63.43^\circ\)
\[ \theta_i \approx 2 \times 81.87^\circ - (-63.43^\circ) = 163.74^\circ + 63.43^\circ = 227.17^\circ \] Slope \( m_i = \tan 227.17^\circ \approx \tan (227.17^\circ - 180^\circ) = \tan 47.17^\circ \approx 1.09 \) but since it's in third quadrant, slope is positive.

Step 7: Find equation of incident ray passing through \((0,1)\)
Equation:
\[ y - 1 = m_i (x - 0) \Rightarrow y = 1.09 x + 1 \] Multiply by 100 for integers:
\[ 100 y = 109 x + 100 \] Rewrite:
\[ 109 x - 100 y + 100 = 0 \] Close to given answer \(41x - 38y + 38 = 0\), so more precise calculation or using formula for reflection of slopes:

Step 8: Use formula for reflected slope
\[ m_i = \frac{2 m_s - m_r (1 - m_s^2)}{1 + 2 m_s m_r - m_s^2} \] Substitute:
\[ m_i = \frac{2 \times 7 - (-2)(1 - 7^2)}{1 + 2 \times 7 \times (-2) - 7^2} = \frac{14 + 2(1 - 49)}{1 - 28 - 49} = \frac{14 + 2(-48)}{-76} = \frac{14 - 96}{-76} = \frac{-82}{-76} = \frac{82}{76} = \frac{41}{38} \]

Step 9: Write incident ray equation
Passing through \((0,1)\) with slope \(\frac{41}{38}\):
\[ y - 1 = \frac{41}{38} x \implies 38y - 38 = 41 x \implies 41x - 38 y + 38 = 0 \]

Final answer:
\[ \boxed{41x - 38 y + 38 = 0} \]