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

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

• 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)