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____

Solution:

Given points P(1, 3) Q (x, 0) ,R ( 6, 7)

$\begin{array}{l} S l o p e o f Q R = \frac{7 - 0}{6 - x} = \frac{7}{6 - x} \\ ∴ \tan θ = \frac{4}{5} a n d S l o p e o f P Q = \frac{0 - 3}{x - 1} \\ \Rightarrow Tan (180 - θ) = \frac{- 3}{x - 1} \\ \Rightarrow - Tan θ (180 - θ) = \frac{- 3}{x - 1} \\ \Rightarrow - \tan θ = \frac{- 3}{x - 1} \\ \Rightarrow \frac{- 7}{6 - x} = \frac{- 3}{x - 1} \\ \Rightarrow 7 x - 7 = 18 - 34 \\ \Rightarrow 10 x = 25 \\ \Rightarrow x = \frac{25}{10} \Rightarrow 2.5 \end{array}$

Solution:

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