Q68 / 180 · KCET 2017 · Mathematics

Equation of line passing through the point \( (1,2) \) and perpendicular to the line
\( y=3 x-1 \)

  • Correct answer
    A
    \( x+3 y-7=0 \)
  • B
    \( x+3 y+7=0 \)
  • C
    \( x+3 y=0 \)
  • D
    \( x-3 y=0 \)

Solution

Given line, $y=3 x-1 \rightarrow(1)$ General equation of line is given by $y=m x+c$ Here $m=3$ and $c=-1$. Condition for two perpendicular lines is $m_{1} \cdot m_{2}=-1$ So, $3 m_{2}=-1 \Rightarrow m_{2}=-\frac{1}{3}$ Standard equation of line passing through point $\left(x_{1}, y_{1}\right)$ is given by $\left(y-y_{1}\right)=m\left(x-x_{1}\right)$ Here $\left(x_{1}, y_{1}\right)=(1,2)$ So, $(y-2)=-\frac{1}{3}(x-1)$ $\Rightarrow-3 y+6=x-1$ $\Rightarrow x+3 y-7=0$

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

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