The image of the point (3,8) in the line x+3y | 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 the point $(3,8)$,we have:$3(3)

Vedclass · Accepted Answer

$(-1, -4)$

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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 the point $(3,8)$,we have:$3(3) - 8 + \lambda = 0$ $\Rightarrow 9 - 8 + \lambda = 0$ $\Rightarrow \lambda = -1$.Thus,the equation of the perpendicular line is $3x-y-1=0$.The foot of the perpendicular is the intersection of $x+3y=7$ and $3x-y-1=0$.Solving these,we get $x=1$ and $y=2$.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 $(x_1, y_1)$,we have:$\frac{3+x_1}{2} = 1 \Rightarrow x_1 = -1$$\frac{8+y_1}{2} = 2 \Rightarrow y_1 = -4$Therefore,the image is $(-1, -4)$.

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

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