Let $A=(4,0), B=(0,12)$ be two points in the plane. The locus of a point $C$ such that the area of triangle $A B C$ is 18 sq. units is -

• A

  $(y+3 x+12)^{2}=81$
• B

  $(y+3 x+81)^{2}=12$
• Correct answer
  C

  $(y+3 x-12)^{2}=81$
• D

  $(y+3 x-81)^{2}=12$

Solution

$\frac{1}{2}\left|\begin{array}{ccc}1 & x & y \\ 1 & 0 & 12 \\ 1 & 4 & 0\end{array}\right|=\pm 18$ $1(-48)-x(-12)+y(4)=\pm 36$ $12 x+4 y-48=\pm 36$ $3 x+y-12=\pm 9$ $(3 x+y-12)^{2}=81$

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