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
ABC is 18 sq. units is

• The locus of point C will be 2 parallel lines
  L
  1
  and
  L
  2
  .
  This is because the distance between
  L
  1
  and
  L
  or
  L
  2
  and
  L
  is equal to h.
  Thus area formed by corresponding triangle will have area
  =
  18
  sq. units.
  Equation of line
  L
  :
  3
  x
  +
  y
  −
  12
  =
  0
  Let equations of line
  L
  1
  and
  L
  2
  be defined by parameter
  c
  :
  3
  x
  +
  y
  +
  c
  =
  0
  .... eqn(i)
  d
  =
  √
  (
  4
  −
  0
  )
  2
  +
  (
  12
  −
  0
  )
  2
  =
  4
  √
  10
  units
  Area
  =
  1
  2
  ×
  d
  ×
  h
  =
  18
  ⇒
  h
  =
  18
  ×
  2
  4
  √
  10
  =
  9
  √
  10
  Using distance formula for lines:
  |
  c
  −
  (
  −
  12
  )
  |
  √
  (
  3
  2
  +
  1
  2
  )
  =
  h
  =
  9
  √
  1
  0
  ⇒
  |
  c
  +
  12
  |
  =
  9
  ⇒
  c
  =
  9
  −
  12
  =
  −
  3
  OR
  c
  =
  −
  9
  −
  12
  =
  −
  21
  Substituting back in eqn(i) gives,
  ⇒
  y
  +
  3
  x
  −
  12
  =
  −
  9
  OR
  y
  +
  3
  x
  −
  12
  =
  9
  ⇒
  
  (
  y
  +
  3
  x
  −
  12
  )
  2
  =
  81
• 5 years agoHelpfull: Yes(0) No(0)