A ray of light passing through the point P(2, | vedclass

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Question

(A) Let $P'(2, -3)$ be the reflection of $P(2, 3)$ across the $x$-axis $(y=0)$.Since the ray reflects at $A$ on the $x$-axis,$P', A,$ and $Q$ are coll

Vedclass · Accepted Answer

$31$

Vedclass · Answer

(A) Let $P'(2, -3)$ be the reflection of $P(2, 3)$ across the $x$-axis $(y=0)$.Since the ray reflects at $A$ on the $x$-axis,$P', A,$ and $Q$ are collinear.The equation of line $P'Q$ passing through $(2, -3)$ and $(5, 4)$ is:$y - (-3) = \frac{4 - (-3)}{5 - 2}(x - 2)$$y + 3 = \frac{7}{3}(x - 2) \implies 3y + 9 = 7x - 14 \implies 7x - 3y = 23$.At point $A$,$y = 0$,so $7x = 23 \implies x = \frac{23}{7}$. Thus,$A = (\frac{23}{7}, 0)$.$R$ divides $AQ$ in ratio $2:1$. $A = (\frac{23}{7}, 0)$ and $Q = (5, 4)$.$R = (\frac{2(5) + 1(23/7)}{2+1}, \frac{2(4) + 1(0)}{2+1}) = (\frac{10 + 23/7}{3}, \frac{8}{3}) = (\frac{93/7}{3}, \frac{8}{3}) = (\frac{31}{7}, \frac{8}{3})$.The bisector of $\angle PAQ$ is the line perpendicular to the $x$-axis passing through $A$ (since the angle of incidence equals the angle of reflection). This is the line $x = \frac{23}{7}$.The foot of the perpendicular $M$ from $R(\frac{31}{7}, \frac{8}{3})$ to the line $x = \frac{23}{7}$ is $M(\frac{23}{7}, \frac{8}{3})$.Thus,$\alpha = \frac{23}{7}$ and $\beta = \frac{8}{3}$.$7\alpha + 3\beta = 7(\frac{23}{7}) + 3(\frac{8}{3}) = 23 + 8 = 31$.

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