Question

The line $L_1\equiv 4x+3y-12=0$ intersects the $x-$ and $y-$axes at $A$ and $B$, respectively. A variable line perpendicular to $L_1$ intersects the $x-$ and the $y-$axes at $P$ and $Q$, respectively. Then the locus of the circumcentre of triangle $ABQ$ is

• A. $3x-4y+2=0$
• B. $4x+3y+7=0$
• C. $6x-8y+7=0$
• D. $Noneofthese$

Answer 1

Answer: (C)

$6x-8y+7=0$

The circumcentre of the triangle ABQ will lie on the perpendicular bisector of the side AB.
Thus the locus will be a straight line with slope $m=\frac{3}{4}$ and passing through the point (3/2, 2).
Hence the locus is given by
$\frac{y-2}{x-\frac{3}{2}}=\frac{3}{4}$
$4y-8=3x-\frac{9}{2}$
$3x-4y+8-\frac{9}{2}=0$
$3x-4y+\frac{7}{2}=0$
$6x-8y+7=0$