The image of the point (3, 8) in the line x + | vedclass

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(C) The equation of the line perpendicular to $x + 3y = 7$ is of the form $3x - y + \lambda = 0$.Since this line passes through $(3, 8)$,we substitute

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$(-1, -4)$

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(C) The equation of the line perpendicular to $x + 3y = 7$ is of the form $3x - y + \lambda = 0$.Since this line passes through $(3, 8)$,we substitute the coordinates:$3(3) - 8 + \lambda = 0$ $\Rightarrow 9 - 8 + \lambda = 0$ $\Rightarrow \lambda = -1$.Thus,the equation of the perpendicular line is $3x - y - 1 = 0$.The point of intersection of $x + 3y = 7$ and $3x - y - 1 = 0$ is the foot of the perpendicular. Solving these equations:From the second equation,$y = 3x - 1$. Substituting into the first: $x + 3(3x - 1) = 7$ $\Rightarrow x + 9x - 3 = 7$ $\Rightarrow 10x = 10$ $\Rightarrow x = 1$.Then $y = 3(1) - 1 = 2$. So,the foot of the perpendicular is $(1, 2)$.Let the image of the point $(3, 8)$ be $(x_1, y_1)$. Since $(1, 2)$ is the midpoint of the segment joining $(3, 8)$ and $(x_1, y_1)$:$\frac{3 + x_1}{2} = 1$ $\Rightarrow 3 + x_1 = 2$ $\Rightarrow x_1 = -1$.$\frac{8 + y_1}{2} = 2$ $\Rightarrow 8 + y_1 = 4$ $\Rightarrow y_1 = -4$.Therefore,the image of the point is $(-1, -4)$.

The image of the point $(3, 8)$ in the line $x + 3y = 7$ is

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