# Find the image of the point (3, 8) with respect to the line $x +3y = 7$ assuming the line to be a plane mirror

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

Toolbox:
• 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)