$\begin {array} {1 1} (A)\;(1,4) & \quad (B)\;(-1,-4) \\ (C)\;( \pm 1, \pm 4) & \quad (D)\;\text{ none of the above} \end {array}$

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- The coordinate of the midpoint of a line $\overline{AB} $ coordinate of A is $( x_1, y_1)$ and B is $(x_2, y_2)$ is $ \bigg( \large\frac{x_1+x_2}{2}$$, \large\frac{y_1+y_2}{2} \bigg)$
- If two lines are perpendicular then the product of the slopes is -1.

The equation of the given line is

$x+3y=7$-----(1)

Let a point B(a, b) be the image of the point A(3, 8).

It is given that equation (1) in the perpendicular bisector of AB.

Hence the slope of AB = $ \large\frac{y_2-y_1}{x_2-x_1}$

i.e., $m_1 = \large\frac{b-8}{a-3}$

Slope of the line (1) is $ \large\frac{-( coefficient \: of\: x)}{(coefficient \: of \: y)}$

(i.e.,) $m_2 = -\large\frac{1}{3}$

Since line (1) is perpendicular to AB, the product of their slopes is -1.

(i.e) $m_1m_2=-1$

$ \Rightarrow \large\frac{b-8}{a-3}$$ \times -\large\frac{1}{3}$$=-1$

On simplifying we get,

$ \large\frac{b-8}{3a-9}$$=1$

$ \Rightarrow b-8=3a-9$

$ \Rightarrow 3a-b=1$---------(2)

Mid point of AB = $ \bigg( \large\frac{a+3}{2}$$, \large\frac{b+8}{2} \bigg)$

The midpoint of the line segment AB will also satisfy line (1).

Hence $ \bigg( \large\frac{a+3}{2} \bigg)$$+3 \bigg( \large\frac{b+8}{2} \bigg)$$=7$

On simplifying we get,

$a+3+3b+24=14$

$ \Rightarrow a+3b=-13$-------(3)

Let us solve equation (2) and (3) for $a$ and $b$.

$ \qquad 3a-b=1$

$ ( \times 3) a+3b=-13$

$ \qquad 3a-b=1$

$ \qquad 3a+9b=-39$

$ \qquad \qquad -10b=40 \Rightarrow b = -4$

$ \therefore a = -1$

Hence $a=-1\: and \: b = -4$

Thus the image of the given point with respect to the given line is (-1, -4)

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