A ray of light incident along a line meets an | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let the slope of the incident ray be $m$. The mirror line is $7x - y + 1 = 0$,which has a slope $m_1 = 7$. The reflected ray is $y + 2x = 1$,which

Vedclass · Accepted Answer

$41x - 38y + 38 = 0$

Vedclass · Answer

(C) Let the slope of the incident ray be $m$. The mirror line is $7x - y + 1 = 0$,which has a slope $m_1 = 7$. The reflected ray is $y + 2x = 1$,which has a slope $m_2 = -2$. The point of incidence is $(0, 1)$.Since the angle of incidence equals the angle of reflection,the angle between the incident ray and the mirror is equal to the angle between the reflected ray and the mirror.Using the formula $	an 	heta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} ight|$,we have:$\left| \frac{m - 7}{1 + 7m} ight| = \left| \frac{7 - (-2)}{1 + 7(-2)} ight| = \left| \frac{9}{1 - 14} ight| = \left| \frac{9}{-13} ight| = \frac{9}{13}$.$\frac{m - 7}{1 + 7m} = \frac{9}{13}$ or $\frac{m - 7}{1 + 7m} = -\frac{9}{13}$.Case $1$: $13(m - 7) = 9(1 + 7m)$ $\Rightarrow 13m - 91 = 9 + 63m$ $\Rightarrow -50m = 100$ $\Rightarrow m = -2$. This is the slope of the reflected ray.Case $2$: $13(m - 7) = -9(1 + 7m)$ $\Rightarrow 13m - 91 = -9 - 63m$ $\Rightarrow 76m = 82$ $\Rightarrow m = \frac{41}{38}$.The equation of the incident line passing through $(0, 1)$ with slope $m = \frac{41}{38}$ is:$y - 1 = \frac{41}{38}(x - 0)$ $\Rightarrow 38y - 38 = 41x$ $\Rightarrow 41x - 38y + 38 = 0$.

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

Similar Questions

Let $A(1, 1)$ be a point. $B$ is the image of $A$ with respect to the line $x + 2y + 2 = 0$. If $C$ is the foot of the perpendicular from $B$ on the line $3x + 4y - 10 = 0$,then $AC$ is equal to

If $(h, k)$ is the image of the point $(2, -3)$ with respect to the line $5x - 3y = 2$,then $h + k =$

The coordinates of the foot of the perpendicular drawn from the point $(-2, 3)$ to the line $3x - y - 1 = 0$ are:

If the image of point $P(\lambda, 2\lambda)$ in the line mirror $x - y + 2 = 0$ is $Q(\mu, 3\lambda)$,then choose the correct option.

If the foot of the perpendicular drawn from the origin to a straight line is $(3, -4)$,then the equation of the line is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module