A light beam emanating from the point A(3, 10 | vedclass

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Question

(A) Let the image of point $A(3, 10)$ with respect to the line $2x + y - 6 = 0$ be $A'(x', y')$.Using the formula $\frac{x' - 3}{2} = \frac{y' - 10}{1

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$4x - 3y + 18 = 0$ and $y = 6$

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(A) Let the image of point $A(3, 10)$ with respect to the line $2x + y - 6 = 0$ be $A'(x', y')$.Using the formula $\frac{x' - 3}{2} = \frac{y' - 10}{1} = -2 \frac{2(3) + 10 - 6}{2^2 + 1^2} = -2 \frac{10}{5} = -4$.So,$x' - 3 = -8 \Rightarrow x' = -5$ and $y' - 10 = -4 \Rightarrow y' = 6$.Thus,$A' = (-5, 6)$.The reflected beam passes through $B(5, 6)$ and $A'(-5, 6)$. Since both points have $y = 6$,the equation of the reflected beam is $y = 6$.The incident beam passes through $A(3, 10)$ and the point of incidence $P$. The point $P$ is the intersection of $AB'$ (where $B'$ is the image of $B$) or simply the intersection of the line $A'B$ with the mirror line.The line $A'B$ is $y = 6$. The intersection of $y = 6$ and $2x + y - 6 = 0$ is $2x + 6 - 6 = 0 \Rightarrow x = 0$. So $P = (0, 6)$.The incident beam passes through $A(3, 10)$ and $P(0, 6)$.The slope is $m = \frac{6 - 10}{0 - 3} = \frac{-4}{-3} = \frac{4}{3}$.The equation is $y - 6 = \frac{4}{3}(x - 0)$ $\Rightarrow 3y - 18 = 4x$ $\Rightarrow 4x - 3y + 18 = 0$.

$A$ light beam emanating from the point $A(3, 10)$ reflects from the line $2x + y - 6 = 0$ and then passes through the point $B(5, 6)$. The equations of the incident and reflected beams are respectively:

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