If $2 x+3 y=5$ is the perpendicular bisector of the line segment joining the points $A\left(1, \frac{1}{3}\right)$ and $B$, then $B$ is equal to

• A

  $\left(\frac{21}{13}, \frac{49}{39}\right)$
• B

  $\left(\frac{17}{13}, \frac{31}{39}\right)$
• C

  $\left(\frac{7}{13}, \frac{49}{39}\right)$
• Correct answer
  D

  $\left(\frac{21}{13}, \frac{31}{39}\right)$

Solution

Let $\quad l_1 \equiv 2 x+3 y=5$ Since, line $A B \perp I_1$ $\therefore$ Slope of $l_1$ is $$ \begin{array}{r} m_1 \text { say }=\frac{-2}{3} \\ \therefore \text { Slope of } A B=\frac{-1}{(-2 / 3)}=\frac{3}{2} \end{array} $$
Equation of line $A B$ is $$ \begin{array}{cc} & \left(y-\frac{1}{3}\right)=\frac{3}{2}(x-1) \\ \Rightarrow \quad & 3 x-2 y=\frac{7}{3} \end{array} $$ Equation of line $I_1$ is $$ 2 x+3 y=5 $$ From Eqs. (i) and (ii), we get $$ \Rightarrow \quad \begin{array}{rl} 13 x & =17 \\ x & x=\frac{17}{13} \end{array} $$ From Eq. (i), we get $$ 3 y=5-\frac{34}{13} \Rightarrow y=\frac{65-34}{13 \times 3}=\frac{31}{39} $$ So, mid-point $P \rightarrow\left(\frac{17}{13}, \frac{31}{39}\right)$ Coordinate of point $B$ $$ \begin{aligned} & =\left(\frac{17}{13} \times 2-1, \frac{31}{39} \times 2-\frac{1}{3}\right) \\ & =\left(\frac{34-13}{13}, \frac{62-13}{39}\right)=\left(\frac{21}{13}, \frac{49}{39}\right) \end{aligned} $$

Hence, the correct answer is option: 'D'.