Q86 / 160 · AP EAMCET 2006 · Mathematics

The equation of the line passing through the point of intersection of the lines $x-3 y+2=0$ and $2 x+5 y-7=0$ and perpendicular to the line $3 x+2 y+5=0$ is :

  • Correct answer
    A
    $2 x-3 y+1=0$
  • B
    $6 x-9 y+11=0$
  • C
    $2 x-3 y+5=0$
  • D
    $3 x-2 y+1=0$

Solution

The equation of lines are $x-3 y+2=0$ ...(i) and $2 x+5 y-7=0$ ...(ii) On solving Eqs. (i) and (ii), we get $x=1, y=1$ $\therefore$ The co-ordinates of a point of intersection of given lines are $(1,1)$. The equation of line perpendicular to $3 x+2 y+5=0$ is $2 x-3 y+\lambda=0$. Which passes through $(1,1)$. $\therefore \quad 2-3+\lambda=0$ $\Rightarrow \quad \lambda=1$ $\therefore$ Required equation of line is $2 x-3 y+1=0$

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

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