Question

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

• A. $\frac{21}{13},\frac{49}{39}$
• B. $\frac{17}{13},\frac{31}{39}$
• C. $\frac{17}{13},\frac{49}{39}$
• D. $\frac{7}{13},\frac{49}{39}$

Answer 1

The correct option is A $\frac{21}{13},\frac{49}{39}$

First we can write the equation in the form of $y=mx+c$

therefore, $y=-\frac{2}{3}x+\frac{5}{3}$

therefore the perpendicular line with slope $m_1$ is

$\Rightarrow$ $m{m}_1=-1$

$\Rightarrow$$\frac{-2}{3}\times m_1=-1$

$\Rightarrow$$m_1=\frac{3}{2}$

therefore the equation for given line is

$y-\frac{1}{3}=\frac{3}{2}(x-1)$

$6y-2=9x-9$

$9x-6y=7$ (!)

$2x+3y=5$

The point of intersection is $(\frac{17}{13},\frac{31}{39})$

This is the midpoint of A and B$(h,k).$

So, $\frac{h+1}{2}=\frac{17}{13}$

$\frac{k+\frac{1}{3}}{2}=\frac{17}{13}$

B is $(\frac{21}{13},\frac{49}{39})$

Option $A$ is the correct answer.