If the image of the point (3, 8) in the line | vedclass

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(C) Let the line $L$ be $x + 3y = 7$ and point $P$ be $(3, 8)$. Let $Q(h, k)$ be the image of point $P$ in the line $L$. Since $L$ acts as a mirror,th

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-$5$

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(C) Let the line $L$ be $x + 3y = 7$ and point $P$ be $(3, 8)$. Let $Q(h, k)$ be the image of point $P$ in the line $L$. Since $L$ acts as a mirror,the line $PQ$ is perpendicular to $L$ and the midpoint $R$ of $PQ$ lies on $L$. The slope of line $L$ is $m_1 = -1/3$. Since $PQ \perp L$,the slope of $PQ$ is $m_2 = 3$. The equation of line $PQ$ passing through $(3, 8)$ with slope $3$ is $y - 8 = 3(x - 3) \Rightarrow y = 3x - 1$. The midpoint $R$ of $PQ$ is $(\frac{h+3}{2}, \frac{k+8}{2})$. Since $R$ lies on $x + 3y = 7$,we have $\frac{h+3}{2} + 3(\frac{k+8}{2}) = 7 \Rightarrow h + 3k = -13$. Substituting $k = 3h - 1$ into the equation: $h + 3(3h - 1) = -13$ $\Rightarrow 10h = -10$ $\Rightarrow h = -1$. Then $k = 3(-1) - 1 = -4$. Thus,$(\alpha, \beta) = (-1, -4)$. Therefore,$\alpha + \beta = -1 + (-4) = -5$.

If the image of the point $(3, 8)$ in the line $x + 3y = 7$ is $(\alpha, \beta)$,then $\alpha + \beta =$

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