Q42 / 160 · TS EAMCET 2014 · Mathematics

If the image of $\left(\frac{-7}{5}, \frac{-6}{5}\right)$ in a line is $(1,2)$, then the equation of the line is

  • A
    $4 x+3 y=1$
  • B
    $3 x-y=0$
  • C
    $4 x-y=0$
  • Correct answer
    D
    $3 x+4 y=1$

Solution

The coordinate of mid-point of $P Q$ is $$ M\left(\frac{-\frac{7}{5}+1}{2}, \frac{-\frac{6}{5}+2}{2}\right) \text { i.e., } \quad\left(-\frac{1}{5}, \frac{2}{5}\right) $$ Now, slope of $P Q$ is $$ m=\frac{2+\frac{6}{5}}{1+\frac{7}{5}}=\frac{16}{12}=\frac{4}{3} $$ Since, the required line is perpendicular to $P Q$ and passing through $M\left(-\frac{1}{5}, \frac{2}{5}\right)$ $\therefore$ Equation of line is $$ \begin{array}{rlrl} & y-\frac{2}{5}=- & \frac{1}{4 / 3}\left(x+\frac{1}{5}\right) \\ & \frac{5 y-2}{5} & =-\frac{3}{4} \frac{(5 x+1)}{5} \\ & 20 y-8 & =-15 x-3 \\ \Rightarrow & & 20 y+15 x & =5 \\ \text { or } & & 4 y+3 x & =1 \end{array} $$

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

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