A light ray emerging from the point source placed at P( 1,3) is reflected at a point Q in the axis of x. If the reflected ray passes through the point R (6, 7), then the abscissa of Q is:

Question

A light ray emerging from the point source placed at P( 1,3) is reflected at a point Q in the axis of x. If the reflected ray passes through the point R (6, 7), then the abscissa of Q is: (A) 1 (B) 3 (C) (7/2) (D) (5/2)

Tardigrade Admin · Accepted Answer

Let abcissa of Q = x ∴ Q = (x, 0) tan θ = (0-7/x-6), tan (180°-θ) = (0-3/x-1) Now, tan (180°- θ)=- tan θ ∴ (-3/x-1) = (-7/x-6) ⇒ x = (5/2)

Q. A light ray emerging from the point source placed at $P (1, 3)$ is reflected at a point $Q$ in the axis of $x$ . If the reflected ray passes through the point $R (6, 7)$ , then the abscissa of $Q$ is:

$1$

$3$

$\frac{7}{2}$

$\frac{5}{2}$

Let abcissa of $Q = x$
$∴ Q = (x, 0)$
$t an θ = \frac{0 - 7}{x - 6}, t an (18 0^{\circ} - θ) = \frac{0 - 3}{x - 1}$
Now, tan $(18 0^{\circ} - θ) = - t an θ$
$∴ \frac{- 3}{x - 1} = \frac{- 7}{x - 6} \Rightarrow x = \frac{5}{2}$

Q. A light ray emerging from the point source placed at P(1,3) is reflected at a point Q in the axis of x. If the reflected ray passes through the point R(6,7), then the abscissa of Q is:

1

3

27​

25​

Let abcissa of Q=x ∴Q=(x,0) tanθ=x−60−7​,tan(180∘−θ)=x−10−3​ Now, tan (180∘−θ)=−tanθ ∴x−1−3​=x−6−7​⇒x=25​

Q. A light ray emerging from the point source placed at $P (1, 3)$ is reflected at a point $Q$ in the axis of $x$ . If the reflected ray passes through the point $R (6, 7)$ , then the abscissa of $Q$ is:

$1$

$3$

$\frac{7}{2}$

$\frac{5}{2}$

Let abcissa of $Q = x$
$∴ Q = (x, 0)$
$t an θ = \frac{0 - 7}{x - 6}, t an (18 0^{\circ} - θ) = \frac{0 - 3}{x - 1}$
Now, tan $(18 0^{\circ} - θ) = - t an θ$
$∴ \frac{- 3}{x - 1} = \frac{- 7}{x - 6} \Rightarrow x = \frac{5}{2}$